url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
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https://convertoctopus.com/37-2-kilometers-per-hour-to-knots
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crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00244.warc.gz
| 204,885,117 | 7,955 |
## Conversion formula
The conversion factor from kilometers per hour to knots is 0.53995680345662, which means that 1 kilometer per hour is equal to 0.53995680345662 knots:
1 km/h = 0.53995680345662 kt
To convert 37.2 kilometers per hour into knots we have to multiply 37.2 by the conversion factor in order to get the velocity amount from kilometers per hour to knots. We can also form a simple proportion to calculate the result:
1 km/h → 0.53995680345662 kt
37.2 km/h → V(kt)
Solve the above proportion to obtain the velocity V in knots:
V(kt) = 37.2 km/h × 0.53995680345662 kt
V(kt) = 20.086393088586 kt
The final result is:
37.2 km/h → 20.086393088586 kt
We conclude that 37.2 kilometers per hour is equivalent to 20.086393088586 knots:
37.2 kilometers per hour = 20.086393088586 knots
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.049784946236476 × 37.2 kilometers per hour.
Another way is saying that 37.2 kilometers per hour is equal to 1 ÷ 0.049784946236476 knots.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-seven point two kilometers per hour is approximately twenty point zero eight six knots:
37.2 km/h ≅ 20.086 kt
An alternative is also that one knot is approximately zero point zero five times thirty-seven point two kilometers per hour.
## Conversion table
### kilometers per hour to knots chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to knots
kilometers per hour (km/h) knots (kt)
38.2 kilometers per hour 20.626 knots
39.2 kilometers per hour 21.166 knots
40.2 kilometers per hour 21.706 knots
41.2 kilometers per hour 22.246 knots
42.2 kilometers per hour 22.786 knots
43.2 kilometers per hour 23.326 knots
44.2 kilometers per hour 23.866 knots
45.2 kilometers per hour 24.406 knots
46.2 kilometers per hour 24.946 knots
47.2 kilometers per hour 25.486 knots
| 553 | 2,049 |
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| 4.03125 | 4 |
CC-MAIN-2021-21
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latest
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https://estebantorreshighschool.com/faq-about-equations/discriminant-of-a-quadratic-equation.html
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## How do you find the discriminant of a quadratic equation?
The discriminant is the part under the square root in the quadratic formula, b²-4ac. If it is more than 0, the equation has two real solutions. If it’s less than 0, there are no solutions. If it’s equal to 0, there is one solution.
## How do you solve a discriminant?
a) Given. Find the discriminant Δ = b 2 – 4ac. For the equation to have one solution, the discriminant has to be equal to zero. The equation m 2 – 4 = 0 has two solutions. b) For the equation to have 2 real solution, the discriminant has to be greater than zero. The inequality m 2 – 4 > 0 has the following solution set.
## What are the 3 forms of a quadratic equation?
Here are the three forms a quadratic equation should be written in:1) Standard form: y = ax2 + bx + c where the a,b, and c are just numbers.2) Factored form: y = (ax + c)(bx + d) again the a,b,c, and d are just numbers.3) Vertex form: y = a(x + b)2 + c again the a, b, and c are just numbers.
## How do I find the discriminant of a parabola?
Quadratic EquationsThe number D = b2 – 4ac is called “discriminant”. If D < 0, then the quadratic equation has no real solutions(it has 2 complex solutions). The vertex of the parabola is x = − b 2 a displaystyle x = -frac{b}{2a} x=−2ab.Problem 3. Solve the equation: Problem 4. Solve the equation:
## Is the discriminant negative or positive?
A quadratic expression which always takes positive values is called positive definite, while one which always takes negative values is called negative definite. Quadratics of either type never take the value 0, and so their discriminant is negative.
## Can a quadratic equation have one real and one imaginary solution?
The statement should should read a quadratic equation with real coefficients can’t have only one imaginary root. The reason being in x2+ax+c=0 x 2 + a x + c = 0 because −a is sum of the roots and c is product of the roots. But a & c are both real numbers, that is impossible if only one of the roots were imaginary.
## What happens when the discriminant is 0?
The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation. A discriminant of zero indicates that the quadratic has a repeated real number solution. A negative discriminant indicates that neither of the solutions are real numbers.
## How do you know if a quadratic equation has no solution?
If the discriminant is less than 0, the equation has no real solution. Looking at the graph of a quadratic equation, if the parabola does not cross or intersect the x-axis, then the equation has no real solution. And no real solution does not mean that there is no solution, but that the solutions are not real numbers.
## Why does the discriminant work?
What are you using the discriminant for? The only use for it is to determinant the type of solutions and how many there are. And it works because it is exactly the crucial part of the quadratic formula that determine the outcome of the solutions.
## How many types of quadratic equations are there?
Two Different Forms of Quadratic Equations.
## How many forms of quadratic equations are there?
three different
You might be interested: Delta s equation
## What are the 4 methods in solving quadratic equation?
The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
## How does the discriminant relate to a graph?
A positive value indicates two solutions, zero indicates one solution, and a negative value indicates no real solution. The value of the discriminant can also be used to find the number of x-intercepts of the graph of y ax2 bx c.
## What is the formula of D in quadratic equation?
Mathwords: Discriminant of a Quadratic. The number D = b2 – 4ac determined from the coefficients of the equation ax2 + bx + c = 0. The discriminant reveals what type of roots the equation has. Note: b2 – 4ac comes from the quadratic formula.
### Releated
#### First order equation
What is first order difference equation? Definition A first-order difference equation is an equation. xt = f(t, xt−1), where f is a function of two variables. How do you solve first order equations? Here is a step-by-step method for solving them:Substitute y = uv, and. Factor the parts involving v.Put the v term equal to […]
#### Energy equation physics
What is the formula for energy? The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. How do you solve for energy in physics? In classical mechanics, kinetic energy (KE) […]
| 1,106 | 4,720 |
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http://slidegur.com/doc/197457/dp%E9%80%82%E7%94%A8%E9%97%AE%E9%A2%98%E7%AC%AC%E4%B8%80%E7%89%B9%E5%BE%81
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crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00546.warc.gz
| 135,367,041 | 28,316 |
### DP适用问题第一特征
```Dynamic Programming
DP适用问题第一特征：重叠子问题
DP适用问题第二特征：最优子结构
Fibonacci Number
Binomial Coefficients
Shortest paths in DAGs
Chain matrix multiplication
Longest Increasing Subsequences
DP问题分类
Knapsack 背包问题
Edit Distance
The Partition Problem
DP是在20世纪50年代由一位卓越的美国数学
DP适用问题的第一特征
Fibonacci Number
f(n) = f(n-1) + f(n-2)
f(0)=0, f(1)=1
n>1
Fibonacci Number
long long fib_r(int n)
{
if (n == 0) return 0;
if (n == 1) return 1;
return (fib_r(n-1) + fib_r(n-2));
}
Fibonacci Number
Fibonacci Number
long long fib_c(int n)
{
if (f[n] == -1)
f[n] = fib_c(n-1) + fib_c(n-2);
return (f[n]);
}
long long fib_c_driver(int n)
{
f[0] = 0;
f[1] = 1;
for (int i=2; i<=n; i++) f[i] = -1;
return (fib_c(n));
}
Fibonacci Number
long long fib_dp(int n)
{
long long f[MAXN + 1];
f[0] = 0;
f[1] = 1;
for (int i=2; i<=n; i++)
f[i] = f[i-1] + f[i-2];
return (f[n]);
}
Fibonacci Number
long long fib_ultimate(int n)
{
/* back2,back1 are last two values of f[n] */
long long back2 = 0, back1 = 1;
long long next;
if (n==0) return 0;
for (int i=2; i<n; i++) {
next = back1 + back2;
back2 = back1;
back1 = next;
}
return (back1+back2);
}
Binomial Coefficients
C(n,k)表示从n个不同物体中选出k个的组合数
C(n,k)=n!/((n−k)!k!)=C(n-1,k-1)+C(n-1,k)
Pascal’s triangle
Matrix representation
Binomial Coefficients
Const int MAXN = 100;
long long binomial_coefficient(int n, int k)
{
long long bc[MAXN+1][MAXN+1];
for (int i = 0; i<=n; i++) bc[i][0] = 1;
for (int j = 0; j<=n; j++) bc[j][j] = 1;
for (int i = 1; i<=n; i++)
for (int j=1; j<i; j++)
bc[i][j] = bc[i-1][j-1] + bc[i-1][j];
return ( bc[n][k] );
}
Shortest paths in DAGs
dist(D) = min{dist(B)+edge(BD), dist(C)+edge(CD)}
= min{dist(B)+1, dist(C)+3}
Shortest paths in DAGs
Topologically sort vertices in G;
Initialize all dist(.) values to INF
dist(s) = 0
for each v ∈ V , in linearized order do
dist(v) = min(u,v) ∈E { dist(u) + w(u,v) }
Example
6
r
5
s
0
2
t
1
u
7
–1
4
3
2
v
–2
w
Example
6
r
5
s
0
2
t
1
u
7
–1
4
3
2
v
–2
w
Example
6
r
5
s
0
2
t
2
1
u
7
6
–1
4
3
2
v
–2
w
Example
6
r
5
s
0
2
t
2
1
u
7
6
–1
v
6
–2
4
3
2
dist[u]=min{dist[s]+6, dist[t]+7}
w
4
Example
6
r
5
s
0
2
t
2
1
u
7
6
–1
4
3
2
v
5
–2
w
4
Example
6
r
5
s
0
2
t
2
1
u
7
6
–1
4
3
2
v
5
–2
w
3
Example
6
r
5
s
0
2
t
2
1
u
7
6
–1
4
3
2
v
5
–2
w
3
Dist[x]。
Dk(i) = opt { Dk-1(j) + cost(i,j) }
k阶段的i状态与k-1阶段的j状态有关
DP适用问题的第二特征
Ai×Ai+1×…×Ak和Ak+1×Ak+2×…×Ak的最优
Chain matrix multiplication
Chain matrix multiplication
Chain matrix multiplication
For 1≤ i≤k≤j ≤n, define
C(i,j) = minimum cost of multiplying Ai×Ai+1×…×Aj
for i = 1 to n: C(i, i) = 0
for s = 1 to n - 1:
for i = 1 to n - s:
j=i+s
C(i, j) = min{C(i, k) + C(k + 1, j) +mi-1·mk·mj : i≤k ≤j}
return C(1, n)
The subproblems constitute a two-dimensional table, each of
whose entries takes O(n) time to compute. The overall
running time is thus O(n3).
Chain matrix multiplication
Ai×Ai+1×…×Aj的一个最优加全部括号把乘积
Ai×Ai+1×…×Ak和Ak+1×Ak+2×…×Ak的最优
sicily1345 能量项链
Longest Increasing Subsequences
Longest increasing subsequence
Algorithm:
Longest Increasing Subsequences
L(0) = 1, P(.) = -1
For i = 1, 2, …, n
L(i) = 1 + max0<j<i{L(j)}, where Sj<Si
P(i) = j
L(i) is the length of the longest path ending at i (plus 1).
Time complexity is O(n2), the maximum being when the
input array is sorted in increasing order.
sicily1060 Bridging Signals （需要用特殊法
sicily1685 Missile（最长不单调子序列，数据
sicily1448 Antimonotonicity（最长不单调子序
01背包问题
01背包问题
f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}
for i=1..N
for v=V..0
f[v]=max{f[v], f[v-c[i]] + w[i]};
void ZeroOnePack(cost,weight){
for v=V..cost
f[v]=max{f[v],f[v-cost]+weight}
}
01背包问题的伪代码就可以这样写：
for i=1..N
ZeroOnePack(c[i],w[i]);
sicily1146 采药
sicily1342 开心的金明
sicily1782 Knapsack
f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i] | 0<=k*c[i]<=v}
O(VN)的算法
for i=1..N
for v=0..V
f[v]=max{f[v],f[v-cost]+weight}
void CompletePack(int cost,int weight)
{
for(int v=cost;v<=V;++v)
f[v] >?= f[v-cost]+weight;
}
01背包按照v=V..0的逆序来循环，要保证第i次
f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i] | 0≤k ≤ n[i]}
1,2,4,...,2k-1,n[i]-2k+1，且k是满足n[i]-2k+1>0的
for(int i=0;i<N;i++)
MultiplePack(c[i],w[i],n[i]);
void MultiplePack(cost,weight,amount)
{
if cost*amount>=V
CompletePack(cost,weight)
return
for(int k=1; k<amount; k++)
ZeroOnePack(k*cost,k*weight)
amount=amount-k
k=k*2
ZeroOnePack(amount*cost,amount*weight)
}
poj1014 Dividing
sicily1077 Cash Machine(可能不算多重背包，
f[k][v]=max{f[k-1][v],f[k-1][v-c[i]]+w[i] | 物品i属于
for 所有的组k
for v=V..0
for 所有的i属于组k
f[v]=max{f[v],f[v-c[i]]+w[i]}
“for v=V..0”这一层循环必须在“for 所有的i属
sicily1750 运动会
f'[0..V-c[i]]。那么这个主件及它的附件集合相当于Vc[i]+1个物品的物品组，其中费用为c[i]+k的物品的
sicily1346 金明的预算方案
j
a1,a2,…,an，求该序列形如的 a[k ]子段和的
k i
bi = bi-1 + ai
bi-1>0
ai
bi-1<=0
b1 = a1
ans = max{ bi }
ans=b=a[1];
for(i=2;i<=n;i++)
{
if (b>0) b+=a[i]; else b=a[i];
if (b>ans) ans=b;
}
poj1050 To the Max（最大子矩阵和问题）
sicily1091 Maximum Sum（最大m段子段和问
sicily1888 Circular Sequence （求一次最大
MAX，判断是否是整个串，如果是就求一次最
d[i][j] = d[i-1][j] + d[i-1][ j-c[i] ]
d[0][0]=1,d[0][1…s]=0
d[0]=1; d[1…s]=0;
for (i=1; i<=n; i++)
{
for (j=s; j>=c[i]; j--)
{
d[j] += d[j-v[i]];
}
}
m[i]枚，问用这n种硬币找零s元的方法数。
d[0]=1;d[1…s]=0;
for (i=1; i<=n; i++)
{
for (j=s; j>=c[i]; j--)
{
for(k=1;k<=m[i];k++)
{
if (j-k*c[i]>=0) d[j] += d[j-k*c[i]];
else break;
}
}
}
d[0]=1;d[1…s]=0;
for (i=1; i<=n; i++)
{
for (j=s; j>=c[i]; j--)
{
for(k=1;;k++)
{
if (j-k*c[i]>=0) d[j] += d[j-k*c[i]];
else break;
}
}
}
d[0]=1; d[1…s]=0;
for (i=1; i<=n; i++)
{
for (j=c[i];j<=s; j++)
{
d[j] += d[j-c[i]];
}
}
memset(vis,0,sizeof(vis));
memset(d,0,sizeof(d));
int dp(int s)
{
if(vis[s]) return d[s];
vis[s] = 1;
int& ans = d[s];
ans = -INF; // INF = 1<<30
for(int i=1;i<=n;i++)
if(s>=v[i]) ans >?= dp(s-v[i]) + 1;
return ans;
}
void print_ans(int *d, int s)
{
for(int i=1; i<=n; i++)
if (s>=v[i] && d[s]==d[s-v[i]]+1)
{
printf(“%d “, i);
print_ans(d, s-v[i]);
break;
}
}
min[0] = max[0] = 0;
for(int i=1; i<=s; i++) {
min[i]=INF; max[i]=-INF;
}
for(int j=1; j<=n; j++)
for(int i=1; i<=s; i++)
if( i>=v[j]) {
min[i] <?= min[i-v[j]] + 1;
max[i] >?= max[i-v[j]] + 1;
}
cout << min[s] << “ “ << max[s] << endl;
sicily 2014 Dairy Queen
sicily1005 Roll Playing Games（此题为搜索
sicily1564 HOUSING（可看成硬币问题）
sicily1902 Counting Problem
2
6
1
1
2
8
4
5
6
8
y个位置到达底层的最小路径得分。
Path(x,y)一定包含子问题D(x+1,y)或D(x+1,y+1)
D(x,y)＝min{D(x+1,y),D(x+1,y+1)}+a(x,y)
D(n,k)＝a(n,k)，k＝1，…，n
D(x,y)表示从第X层第y个位置到达底层的最小
sicily1563 GECKO
HOJ 1714 Minimax Triangulation
Party at Hali-Bula
n个人形成一个关系树，每个节点代表一个人，
Party at Hali-Bula
Party at Hali-Bula
DP, 用dp[i][0]表示不选择i点时，i点及其子树
Party at Hali-Bula
dp[i][0] = ∑max(dp[j][0], dp[j][1]) (j是i的儿子)
dp[i][1] = 1 + ∑dp[j][0] (j是i的儿子)
Party at Hali-Bula
== 0) 或 (dp[j][0] < dp[j][1] 且 dup[j][1] == 0) 或
(dp[j][0] == dp[j][1])，则dup[i][0] = 0
Strategic game
Strategic game
dproot[ i ]表示以i为根的子树，在i上放置一个
all[ i ]表示看守住整个以i为根的子树需要多少
Strategic game
dproot[i] = 1 + ∑all[j](j是i的儿子)；
all[i] = min( dproot[i], ∑dproot[j](j是i的儿子) )；
x=101100，即十进制整数44.
TSP问题。
n <= 16 (重要条件,状态压缩的标志)
TSP
5 ＝ 0000000000000101；（2进制表示）
31 ＝ 0000000000011111； （2进制表示）
5；
TSP
TSP
k表示i集合中去掉了j点的集合，s遍历集合k中
TSP
min( dp[( 1<<n ) – 1][j] ) ( 0 <= j < n )；
• 判断点j 是否属于集合i ：i & (1<<j)
• 在集合i 中去除点j ：i –(1<<j) 或者i & ( ~( 1 << j ) )
• 在集合i 中加入点j ：i | (1<<j)
• 问题描述：给出一个带权有向图G=(V, E)，
• 如果暴力穷举所有路径，复杂度不低于O(N!)
• 状态表示：用ans[j][i]表示以j点为起点，并且经
• 状态转移：
s 表示i 集合中去掉了j点的集合，k 遍历集合s 中
• 最后结果便是所有ans[j][i]的最大值.
2000）
Moving one of his feet from the central point to any side
points will consume 2 units of his strength. Moving from
one side point to another adjacent side point will
consume 3 units, such as from the top point to the left
point. Moving from one side point to the opposite side
point will consume 4 units, such as from the top point to
the bottom point. Yet, if he stays on the same point and
tread again, he will use 1 unit.
d[i][j][k]为跳到第k个舞步时，左脚在位置i，右脚在位
d[i,j,k]=min{ d[s[k],j,k+1]+cost(i,s[k]),
d[i,s[k],k+1]+cost(j,s[k]) }
d[i,j,n]=min{ cost(i,s[n]), cost(j,s[n]) }
Longest Common Substring
Longest Common Substring
c[i][j]记录序列Xi和Yj的最长公共子序列的长度。
c(i, j) = 0
c(i, j) = c(i-1, j-1) + 1
c(i, j) = max{c(i,j-1), c(i-1,j)}
if i=0 or j=0
if i, j>0 and xi = yj
if i, j>0 and xi ≠ yj
Edit Distance
When a spell checker encounters a possible misspelling, it looks in
its dictionary for other words that are close by. What is the
appropriate notion of closeness in this case?
A natural measure of the distance between two strings is the extent
to which they can be aligned, or matched up. Technically, an
alignment is simply a way of writing the strings one above the other.
For instance, here are two possible alignments of SNOWY and
SUNNY:
S–NOWY
–SNOW–Y
SUNN– Y
SUN– –NY
Cost: 3
Cost: 5
The ' – ' indicates a 'gap'; any number of these can be placed in
either string. The cost of an alignment is the number of columns in
which the letters differ. And the edit distance between two strings is
the cost of their best possible alignment.
Edit Distance
Our goal is to find the edit distance between two strings
x[1…m] and y[1…n]. What is a good subproblem?
How about looking at the edit distance between some
prefix x of the first string, x[1…i], and some prefix of the
second, y[1…j]? Call this subproblem E(i, j). Our goal
objective, then, is to compute E(m, n).
Edit Distance
For this to work, we need to somehow express E(i, j) in
terms of smaller subproblems.
Let's see—
what do we know about the best alignment
between x[1…i] and y[1…j]? Well, its rightmost column
can only be one of three things:
E(i, j) = min{1 + E(i-1, j), 1 + E(i, j-1), diff(i, j) + E(i-1, j-1)}
where diff(i, j) is defined to be 0 if x[i] = y[j] and 1
otherwise.
Edit Distance
For instance, in computing the edit distance between
EXPONENTIAL and POLYNOMIAL, subproblem E(4, 3)
corresponds to the prefixes EXPO and POL. The
rightmost column of their best alignment must be one of
the following:
Thus, E(4, 3) = min{1 + E(3, 3), 1 + E(4, 2), 1 + E(3, 2)}.
Edit Distance
The answers to all the subproblems E(i, j) form a twodimensional table.
Edit Distance
The algorithm for edit distance:
The overall running time is just the size of the table,
O(mn).
sicily 1822 决斗
A[i,j]=1，则i与j决斗时，i总是赢，如果A[i,j]=0，
meet[i,j]记录i和j是否相遇，能为1，否则为0，问题转化为能
meet[i,j]=1，否则meet[i][j]=0。
sicily上的一些dp题：
sicily1010 Zipper
sicily1011 Lenny's Lucky Lotto
sicily1019 Apple Tree （树型dp）
sicily1057 Rhyme Schemes
sicily1073 Pearls
sicily1123 The Longest Walk（状态压缩dp）
sicily1148 过河（路径压缩，贪心动态规划）
sicily1158 Pick numbers
sicily1222 单词选择（这题的hash很恶心）
sicily1264 Atomic Car Race （基本dp题）
sicily1404 Hie with the Pie（状态压缩dp）
sicily1687 Permutation
sicily1822 Fight Club
sicily1828 Minimal
```
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# Differentiating Scalar along a geodesic
I have been studying GR for sometime and doing exercises from Schutz and I have a question about differentiating along a geodesic. Here is what I know. The equation of geodesic in terms of four momentum is given as,$$p^\alpha p^{\beta}_{;\alpha}$$. Now if I want to differentiate a scalar along the geodesic I figured I have to do this, $$\frac{d\phi}{d\tau}$$ Here, $$\tau$$ is the proper time which is the parameter along the curve. The change of the scalar $$\phi$$ along the curve is equal to, $$\frac{d\phi}{d\tau}=\phi,_{\beta}U^\beta$$ Here, $$U^\beta$$ is the four velocity of the curve. Writing this in covariant derivative form I believe it should just be,$$\frac{d\phi}{d\tau}=\phi_{;\beta}U^\beta$$ So if a scalar (like a dot product between vectors) is constant along the geodesic then I believe it means that,$$\frac{d\phi}{d\tau}=0.$$ Is this correct?
In the question I am trying to solve, the condition is that $$p^\alpha\epsilon_\alpha=\text{constant}$$ along the geodesic. I am trying to write the condition of what this means to proceed with further calculation.
So I ended up trying to solve the problem, assuming what I stated above is correct. I will state the problem here for reference, Show that if a vector field $$\epsilon^\alpha$$ satisfies Killing’s equation then $$p^\alpha\epsilon_\alpha$$ is constant along the geodesic. So I just took covariant derivative of $$\phi$$ as, $$\frac{d\phi}{d\tau}=U^\beta\phi_{;\beta}$$ and then set it to zero. Here I defined $$\phi = p^\alpha \epsilon_\alpha$$. Then basically expanded the covariant derivative of $$\epsilon$$ and got to the point (using the equation of geodesic $$U^\alpha U^\beta_{;\alpha}=0$$) where you get $$U^\alpha U^\beta \epsilon_{\alpha,\beta}$$. Since $$\epsilon^\alpha$$ is known to be Killing vector field then it satisfies the Killing equation $$\epsilon_{\alpha,\beta}=-\epsilon_{\beta,\alpha}$$ which basically makes it anti-symmetric hence the sum $$U^\alpha U^\beta \epsilon_{\alpha,\beta}$$ is zero which proves that, $$\frac{d(p^\alpha\epsilon_\alpha)}{d\tau}=U^\beta(p^\alpha\epsilon_\alpha)_{;\beta}=0$$.
Hope this made sense. I also checked solution manual (which I only check once I am out of all options to figure it out myself) this is how it is done as well.
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# 8: The Chi-Square Distribution
A chi-squared test is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-square distribution when the null hypothesis is true.
• 8.1: Prelude to The Chi-Square Distribution
You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution.
• 8.2: Facts About the Chi-Square Distribution
he chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population.
• 8.3: Test of Independence
Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test.
• 8.4: Test for Homogeneity
The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.
• 8.5: Comparison of the Chi-Square Tests
You have seen the Chi-square test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which Chi-square test is the appropriate one to use.
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# Classical Mechanics
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### Physics: Springs and Compression
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# Algebra
posted by .
Translate the following statement into an algebraic equaion. Let x represent the number.
6 times a number is 14 more than that number.
I got 6x=x+14
is that right?
• Algebra -
where is the question?
## Similar Questions
1. ### math correction
Translate the following statements into inequalities. Let x represent the number in each case. 2 times a number, increased by 28, is less than or equal to 6 times that number.
2. ### Math correction
(1)Translate the following statement into an algebraic equation. Lex x represent the number. 12 less than 5 times a number is 18 my answer: 5x - 12 = 18 (#2) 4 times a number, increased by 9, is 17. Find the number. My answer: 4x + …
3. ### math
Translate the following statement into an algebraic equation. Let x represent the number. 9 more than 10 times a number is 5 times that same number.
4. ### algebra trying again
Translate the following statement into an algebraic equaion. Let x represent the number. 6 times a number is 14 more than that number. I got 6x=x+14 Is this correct
5. ### algebra 1A
translate into an algebraic expression. use x and y for variables. the sum of three times a number plus five times another number. the translation is. is any one of these correct y+3=8 or 3+8=8 please help. thanks
6. ### algebra
Translate into an algebraic expression Use x and y for any variables. The sum of four time a number plus five times another number
7. ### Algebra
Translate the sentence into an algebraic expression. ”One more than three times a number.”
8. ### Algebra
1. Which is an example of an algebraic expression?
9. ### math
Translate the following into an algebraic espression and set up the related equation. Let y be the unknown number. Some number minus 6 is equal to the number times 17 Translate "the number times 17"
10. ### algebra
translate this expression into a mathematical problem the difference between one more than three times a number and a number is 19 find the smaller number
More Similar Questions
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https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-10-geometry-chapter-summary-review-and-test-review-exercises-page-682/42
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## Thinking Mathematically (6th Edition)
Circumference = $20\pi \approx 62.8$ meters Area = $100\pi \approx 314.2$ square meters
RECALL: In a circle: (1) Circumference = $C = 2\pi{r}$ (2) Area = $A = \pi{r^2}$ (3) Radius = $r = \frac{d}{2}$ where r = radius and d = diameter The given circle has a diameter of 20 meters. Since the radius is equal to half of the diameter, the radius of the circle is: $r = \frac{20}{2} = 10$ meters Solve for the circumference and the area using formulas (1) and (2), respectively, to obtain: Circumference = $2\pi(10) = 20\pi \approx 62.8$ meters Area = $\pi(10^2) = \pi(100) = 100\pi \approx 314.2$ square meters
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# How to Calculate the YTM of a Bond in Excel: Easy Guide
YTM stands for yield to maturity. It is a common term used in the financial industry. Whether you are buying or selling a bond then, you need to calculate YTM. Bonds are mostly sold by companies, financial institutions, and even Governments. As an investor, it is important to calculate the YTM of a bond in order to find out its projected returns. One of the best ways to calculate the yield to maturity of a bond is by using Excel.
In this tutorial, I will guide you on different ways to calculate YTM in Excel.
## Method 1: Use direct Formula to calculate the YTM of a bond
This method we the actual YTM of a bond formula and apply it to Excel. The actual Formula to calculate YTM can be written as Follows.
YTM=(C+(FV-PV)/n)/(FV+PV/2)
Where
C: is the Annual Coupon Amount
FV: is the Face Value
PV: is the Present Value
N: refers to years to maturity
Let’s use the following data to illustrate how to calculate YTM using direct Formula
Face value = \$2000
Present Value = \$2100
Annual Coupon = \$40
Years of Maturity = 15
Yield to maturity =?
Follow these steps to calculate in Excel.
1. Launch your Excel program and create a new worksheet
2. Create all the filled we have created above and fill in the values in the adjacent cells.
3. Click on cell B8 and type in the following formula
=(B6+((B4-B5)/B7))/(B4+B5/2)
4. Press enter, and Excel will compute the Yield to Maturity automatically
5. In case you get an answer like the one below, then you need to format the cell as a percentage
6. Right Click on it and then select format cells
7. On the new pop-up window, select percentage, then click ok
8. Now your answer should appear as shown in the image below
## Method 2: Use the Rate() function to calculate YTM in Excel
The simplest way to calculate Yield to Maturity in Excel is by using the inbuilt Rate function. This function features arguments that are computed to produce YTM. Follow these simple steps:
1. Arrange the following fields in your Excel file
Par Value of Bond: \$3000
Coupons Per Year: 6
Current Bond Price: \$1900
Number of Coupons: 50
Periods: 20
Yield To Maturity (YTM):
2. Type the following Formula on cell 9B
=RATE(B8,B7,-B6,B4)*B5
3. Press the enter button to apply the Rate formula
As you can see from the image above, we have utilized the rate function to calculate YTM
## Method 3: Calculate the YTM of a Bond using the IRR function
We can get the YTM of a bond by multiplying the by the number of coupons per year. In this method, I will show you how to use the IRR function to calculate YTM. The IRR function takes the values that we have provided to calculate the internal Rate of return.
1. Arrange the field and their respective values as follows.
2. Write the following Formula in the IRR section
=IRR(C5:C9)
3. Press enter
4. Write the coupons per year
5. Multiply the IRR by the number of coupons per year to get the YTM
=C10*C11
## Method 4: Use the YIELD function
The inbuilt Excel YIELD() function can help users calculate YTM. To calculate yield to maturity using the YIELD function, you need the following fields.
SD: Settlement Date
MD: Maturity Date
Rate: Coupon Rate
PR: Market value
Redemption: Call Price, also known as Par Value
Frequency: Frequency of payments
Basis: Day Count basis
The Yield function can be written as
= YIELD (SD, MD, Rate, pr, redemption, frequency, [basis])
1. Arrange your fields and values as shown in the image below.
2. Enter the following Formula on cell 11B
=YIELD(B6,B7,B5,B10,B4,B8)
3. Press Enter, and Excel will compute the YTM of your values automatically
Similar Read: How to calculate per capita in Excel
## Method 5: Using VBA code to calculate YTM in Excel
Excel is one of the best finance software thanks to the expandable features using VBA. It stands for Visual Basic Application. You can use it to create Macro programs that run inside Excel Software. Therefore, you can write a VBA code Tailored to calculate the YTM of a bond. Follow these simple steps.
1. Open an existing or create a new Excel File
2. Click on the developer tab, then click on Visual Basic (In case you don’t see the developer tab on your Excel menu. Click on File>Options>Customize Ribbon and Add the Developer tab)
3. After clicking Visual Basic, a new window will pop up. Click on Insert>Module
4. Copy and Paste the VBA code below
```Function YTM(F As Double, P As Double, C As Double, N As Integer, Optional Guess As Double = 0.1) As Double
'This function calculates Yield to Maturity (YTM) using Excel's built-in RATE function.
YTM = WorksheetFunction.Rate(N, C, -P, F, Guess)
End Function
```
5. Head over the back to the worksheet and calculate your BOND using the function that we have just created
For example, if you have a bond with a face value of \$1000, a price of \$950, a coupon rate of 5%, a coupon frequency of semi-annual, and five years to maturity, you can use the following Formula on the Excel cell
==YTM(1000,950,0.05,5)
As you can see from the image above, we have used the VBA code to calculate the YTM of a bond
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JEE > JEE Main 2019 Question Paper with Solutions (11th January - Morning)
# JEE Main 2019 Question Paper with Solutions (11th January - Morning)
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## 90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2019 Question Paper with Solutions (11th January - Morning)
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JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 1
### A body is projected at t = 0 with a velocity 10 ms-1 at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms-2, the value of R is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 1
vx = 10cos60° = 5 m/s
vy = 10cos30° = 5 √3 m/s
velocity after t = 1 sec.
vx = 5 m/s
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 2
### A particle is moving along a circular path with a constant speed of 10 ms-1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60° around the centre of the circle?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 2
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 3
### A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be : (hc = 12500 eV – Å)
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 3
Energy of photon = 12500/980 = 12.75eV
∴ Electron will excite to n= 4
Since 'R' ∝ n2
∴ Radius of atom will be 16a0
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 4
A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 4
Momentum per second carried by liquid per second is ρav2
net force due to reflected liquid =
net force due to stopped liquid =
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 5
An electromagnetic wave of intensity 50 Wm-2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 5
similarly
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 6
An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 x 104)]sin(5.5 x 105t). Here t is in seconds. The sideband frequencies (in kHz) are, [Given π = 22/7]
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 6
Side band frequency are
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 7
The force of interaction between two atoms is given by where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 7
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 8
The charges Q + q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 8
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 9
In the circuit shown, the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time (t0), the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time 't' is given by :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 9
From time t = 0 to t = t0, growth of current takes place and after that decay of current takes place.
most appropriate is (2)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 10
Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured is SI units. The tension in the string is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 10
y = 0.03 sin(450 t – 9x)
⇒ T = 2500 × 5 × 10–3
= 12.5 N
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 11
An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. It the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 11
Suppose M is mass and a is side of larger
triangle, then M/4 and a/2 will be mass and side length of smaller triangle.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 12
There are two long co-axial solenoids of same length l. the inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 12
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 13
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature,. The relation between temperature and volume of this process is TVx = constant, then x is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 13
For adiabatic process : TVgγ-1 = constant
For diatomic process :
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 14
The gas mixture constists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total inernal energy of the system is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 14
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 15
In a Young's double slit experiment, the path different, at a certain point on the screen, between two interfering waves is 1/8th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 15
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 16
If the deBronglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of frequency 6 × 1014 Hz, then the speed of electron is equal to :
(Speed of light = 3 × 108 m/s
Planck's constant = 6.63 × 10–34 J.s
Mass of electron = 9.1 × 10–31 kg)
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 16
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 17
A slab is subjected to two forces of same magnitude F as shown in the figure. Force is in XY-plane while force F1 acts along z-axisat the point . The moment of these forces about point O will be :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 17
Torque for F1 force
Torque for F2 force
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 18
A satellite is revolving in a circular orbit at a height h from the earth surface, such that h << R where R is the radius of the earth.
Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed requried so that the satellite could escape from the gravitational field of earth is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 18
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 19
In an experiment electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied.
[Charge of the electron = 1.6 × 10–19 C Mass of the electron = 9.1 × 10–31 kg]
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 19
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 20
A particle undergoing simple harmonic motion has time dependent displacement given by The ratio of kinetic to potential energy of this particle at t = 210 s will be :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 20
Hence ratio is 3 (most appropriate)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 21
Ice at –20°C is added tp 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to
(Specific heat of water = 4.2 J/g/°C)
Specific heat of Ice = 2.1 J/g/°C
Heat of fusion of water at 0°C = 334 J/g)
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 21
Let amount of ice is m gm.
According to principal of calorimeter
heat taken by ice = heat given by water
∴ 20 × 2.1 × m + (m - 20) × 334
= 50 × 4.2 × 40
376 m = 8400 + 6680
m = 40.1 g
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 22
In the figure shown below, the charge on the left plate of the 10 μF capacitor is -30 μC. The charge on the right plate of the 6 μF capacitor is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 22
6µF & 4µF are in parallel & total charge on this combination is 30 µC
∴ Charge on 6µF capacitor =
= 18 µC
Since charge is asked on right plate therefore is +18µC
Correct answer is (4)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 23
In the given circuit the current through Zener Diode is close to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 23
Since voltage across zener diode must be less than 10V therefore it will not work in breakdown region, & its resistance will be infinite & current through it = 0
∴ correct answer is (4)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 24
The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if Dm is the angle of minimum deviation?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 24
Since Dm = (µ – 1)A
& on increasing the wavelength, µ decreases & hence Dm decreases. Therefore correct answer is (2)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 25
The resistance of the meter bridge AB in the given figure is 4Ω. With a cell of emf ε = 0.5 V and rheostat resistance Rh = 2Ω the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε2 the same null point J is found for Rh = 6 Ω. The emf ε2 is;
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 25
Potential gradient with Rh = 2Ω
Let null point be at ℓ cm
Now with Rh = 6Ω new potential gradient is and at null point
dividing equation (1) by (2) we get
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 26
The given graph shows variation (with distance r from centre) of :
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 27
Two equal resistance when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 27
In series condition, equivalent resistance is 2R thus power consumed is
In parallel condition, equivalent resistance is R/ 2 thus new power is
o r P' = 4P = 240W
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 28
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 28
From lens equation
velocity of image wrt. to lens is given by vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s
= 1.16 × 10–3 m/s towards the lens
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 29
A body of mass 1 kg falls freely from a height of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 x 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms-2, the value of x will be close to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 29
Velocity of 1 kg block just before it collides with 3kg block =
Applying momentum conversation just before and just after collision.
initial compression of spring 1.25 × 106 x0 = 30 ⇒ x0 ≈ 0
applying work energy theorem,Wg + Wsp = ΔKE
solving x ≈ 4 cm
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 30
In a Wheatstone bridge (see fig.), Resistances P and Q are approximately equal. When R = 400 Ω, the bridge is equal. When R = 400 Ω, the bridge is balanced. On inter-changing P and Q, the value of R, for balance, is 405 Ω. The value of X is close to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 30
After interchanging P and Q
From (i) and (ii)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 31
For the cell Zn(s) | Zn2+(aq) || Mx+ (aq) | M(s), different half cells and their standard electrode potentials are given below :
If , which cathode will give a maximum value of E0cell per electron transferred?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 31
We have,
E0cell= E0cathode−E0anode0
For a high value of E0cell the value of SRP of cathode should be high.
Here the highest value is for Au3+/Au
E0cell= 1.4−(−0.76)=2.16V
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 32
The correct match between items-I and II is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 32
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 33
If a reaction follows the Arrhenius equation, the plot lnk vs 1/(RT) gives straight line with agradient (–y) unit. The energy required to activate the reactant is :
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 34
The concentration of dissolved oxygen (DO) in cold water can go upto :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 34
In cold water, dissolved oxygen (DO) can reach a concentration upto 10 ppm
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 35
The major product of the following reaction is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 35
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 36
Th correct statements among (a) to (d) regarding H2 as a fuel are :
(a) It produces less pollutant than petrol
(b) A cylinder of compressed dihydrogen weighs ~ 30times more than a petrol tank producing the same amount of energy
(c) Dihydrogen is stored in tanks of metal alloys like NaNi5
(d) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 37
The major poduct of the following reaction is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 37
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 38
The element that usually does not show variable oxidation states is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 38
Usally Sc(Scandium) does not show variable oxidation states.
Most common oxidation states of :
(i) Sc : +3
(ii) V : +2, +3, +4, +5
(iii) Ti : +2, +3, +4
(iv) Cu : +1, +2
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 39
An organic compound is estimated through Dumus method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 39
Hence, C6H8N2
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 40
The major product of the following reaction is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 40
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 41
Among the following compound which one is found in RNA?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 41
For the given structure 'uracil' is found in RNA
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 42
Which compound(s) out of the following is/are not aromatic?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 42
out of the given options only is aromatic.
Hence (B), (C) and (D) are not aromatic
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 43
The correct match between Item (I) and Item (II) is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 43
(A) Norethindrone – Antifertility
(B) Ofloaxacin – Anti-Biotic
(C) Equanil – Hypertension (traiquilizer)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 44
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose?
[RH = 1 × 105 cm–1, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1]
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 45
Consider the reaction,
N2(g) + 3H2(g) ⇔ 2NH3(g)
The equilibrium constant of the above reaction is KP. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by
(Assume that PNH3 <<Ptotal at equilibrium)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 46
Match the ores(Column A) with the metals (column B) :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 46
Siderite : FeCO3
Kaolinite : Al2(OH)4Si2O5
Malachite : Cu(OH)2.CuCO3
Calamine : ZnCO3
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 47
The correct order of the atomic radii of C, Cs, Al and S is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 47
Atomic radii order : C < S < Al < Cs
Atomic radius of C : 170 pm
Atomic radius of S : 180 pm
Atomic radius of Al : 184 pm
Atomic radius of Cs : 300 pm
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 48
Match the metals (Column I) with the coordination compound(s) / enzyme(s) (Column II)
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 48
(i) Wilkinson catalyst : RhCl(PPh3)3
(ii) Chlorophyll : C55H72O5N4Mg
(iii) Vitamin B12(also known as cyanocobalamin) contain cobalt.
(iv) Carbonic anhydrase contains a zinc ion.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 49
A 10 mg effervescent tablet contianing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol-1]
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 49
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 50
The major product of the following reaction is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 50
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 51
Two blocks of the same metal having same mass and at temperature T1 and T2, respectively. are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, ΔS, for this process is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 51
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 52
The chloride that CANNOT get hydrolysed is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 52
CCl4 cannot get hydrolyzed due to the absence of vacant orbital at carbon atom.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 53
For the ch emical reaction X ⇔ Y, the standard reaction Gibbs energy depends on temperature T (in K) as :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 53
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 54
The freezing point of a diluted milk sample is found to be –0.2°C, while it should have been –0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 54
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 55
A solid having density of 9 × 103 kg m–3 forms face centred cubic crystals of edge length 200√2 pm. What is the molar mass of the solid ?
(Avogadro constant ≌ 6 x 1023 mol-1, π ≌ 3)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 56
The polymer obtained from the following reactions is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 56
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 57
An example of solid sol is :
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 58
Peroxyacetyl nitrate (PAN), an eye irritant is produced by :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 58
Photochemical smog produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate (PAN).
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 59
NaH is an example of :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 59
NaH is an example of ionic hydride which is also known as saline hydride.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 60
The amphoteric hydroxide is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 60
Be(OH)2 is amphoteric in nature while rest all alkaline earth metal hydroxide are basic in nature.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 61
Let It AAT = I3, then |p| is
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 61
A is orthogonal matrix
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 62
The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 :-
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 62
x = 4y – 2 & x2 = 4y
⇒ x2 = x + 2 ⇒ x2 – x – 2 = 0
x = 2, – 1
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 63
The outcome of each of 30 items was observed; 10 items gave an outcome 1/2 -d each, 10 items gave outcome 1/2 each and the remaining 10 items gave outcome 1/2 + d each. If the variance of this outcome data is 4/3 then |d| equals :-
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 63
Variance is independent of origin. So we shift the given data by 1/2.
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 64
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 64
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 65
Let and be coplanar vectors. Then the non-zero vector is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 65
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 66
Let , where x and y are real numbers, then y – x equals :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 66
Hence, y – x = 198 – 107 = 91
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 67
Let and g(x) = |f(x)| + f (|x|). Then, in the interval (–2, 2), g is :-
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 67
and f(|x|) = x2 - 1, x ∈ [-2, 2]
It is not differentiable at x = 1
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 68
Let f : R → R be defined by x ∈ R. Then the range of f is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 68
f(0) = 0 & f(x) is odd.
Further, if x > 0 then
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 69
The sum of the real values of x for which the middle term in the binomial expansion of equals 5670 is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 69
⇒ 70x8 = 5670
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 70
The value of r for which 20Cr 20C0 + 20Cr–1 20C1 + 20Cr–2 20C2 + .... 20C0 20Cr is maximum, is
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 70
Given sum = coefficient of xr in the expansion of (1 + x)20(1 + x)20,
which is equal to 40Cr
It is maximum when r = 20
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 71
Let a1, a2, ....., a10 be a G.P. If a3/a1 = 25, then a9/a5 equals.
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 71
a1, a2, ....., a10 are in G.P.,
Let the common ratio be r
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 72
If for a suitable chosen integer m and a function A(x), where C is a constant of integration then (A(x))m equals :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 72
Case-II x ≤ 0
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 73
In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 - c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 73
Given a + b = x and ab = y
If x2 – c2 = y ⇒ (a + b)2 – c2 = ab
⇒ a2 + b2 – c2 = –ab
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 74
The value of the integral (where [x] denotes the greatest integer less than 20Cr or equal to x) is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 74
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 75
If the system of linear equations
2x + 2y + 3z = a
3x - y + 5z = b
x - 3y + 2z = c
where a, b, c are non-zero real numbers, has more then one solution, then :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 75
P1 : 2x + 2y + 3z = a
P2 : 3x – y + 5z = b
P3 : x – 3y + 2z = c
We find
P1 + P3 = P2 ⇒ a + c = b
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 76
A square is inscribed inthe circle x2 + y2 – 6x + 8y – 103 = 0 with its sides parallel to the corrdinate axes. Then the distance of the vertex of this square which is nearest to the origin is :-
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 76
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 77
Let fk(x) = 1/k (sinx + cosk x) for k = 1,2,3,....... Then for all x ∈ R, the value of f4(x) – f6(x) is equal to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 77
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 78
Let [x] denote the greatest integer less than or equal to x. Then :-
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 78
R.H.L. ≠ L.H.L.
*Multiple options can be correct
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 79
The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an anlge π/4 with the plane y–z+5=0 are:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 79
Let the equation of plane be a(x – 0) + b(y + 1) + c(z – 0) = 0 It passes through (0,0,1) then
b + c = 0 ...(1)
⇒ a2 = –2bc and b = –c
we get a2 = 2c2
⇒ direction ratio (a, b, c) = (√2, -1, 1) or (√2, 1, -1)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 80
If xloge(loge x) – x2 + y2 = 4(y > 0), then dy/dx at x = e is equal to :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 80
Differentiating with respect to x,
at x = e we get
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 81
The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 81
Equation of circle
Equation of tangent of origin is 2x + y = 0
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 82
If q is false and p ∧ q⇔r is true, then which one of the following statements is a tautology?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 82
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 83
If y(x) is the solution of the differential equation where, then :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 83
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 84
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = {x ∈ R : x2 + 30 ≤ 11x} is:
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 84
S = {x ∈ R, x2 + 30 – 11x ≤ 0}
= {x ∈ R, 5 ≤ x ≤ 6}
Now f(x) = 3x3 – 18x2 + 27x – 40
⇒ f'(x) = 9(x – 1)(x – 3),
which is positive in [5, 6]
⇒ f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 85
If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 85
81x2 + kx + 256 = 0 ; x = α, α3
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 86
Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 86
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 87
Equation of a common tangent to the parabola y2 = 4x and the hyperbole xy = 2 is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 87
Let the equation of tangent to parabola
y2 = 4x be y = mx + 1/m
It is also a tangent to hyperbola xy = 2
So tangent is 2y + x + 4 = 0
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 88
The plane containing the line and also containing its projection on the plane 2x + 3y – z = 5, contains which one of the following points ?
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 88
The normal vector of required plane
So, direction ratio of normal is (–1, 1, 1) So required plane is
–(x – 3) + (y + 2) + (z – 1) = 0
⇒ –x + y + z + 4 = 0
Which is satisfied by (2, 0, –2)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 89
If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted betwen the coordinate axes lie on the curve :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 89
Equation of general tangent on ellipse
Let the midpoint be (h, k)
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 90
Two integers are selected at random from the set {1, 2,...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :
Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 90
Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space
= 5C2 + 6C2
so required probability =
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# Momentum, Impulse, and Collisions. Momentum (P) A quantity that expresses the motion of a body and its resistance to slowing down. P = mv m = mass (kg)
## Presentation on theme: "Momentum, Impulse, and Collisions. Momentum (P) A quantity that expresses the motion of a body and its resistance to slowing down. P = mv m = mass (kg)"— Presentation transcript:
Momentum, Impulse, and Collisions
Momentum (P) A quantity that expresses the motion of a body and its resistance to slowing down. P = mv m = mass (kg) v = velocity (m/s)
Try it… A toy car has a mass of 5kg. If it is moving at a velocity of 5m/s, what is its momentum? 25N·s What is the momentum of a 25kg boy moving at a speed of 1m/s? 25N·s
Impulse measures a change in momentum; arrived at by multiplying the average force acting on a body by the length of time it acts. Impulse = F(t) F = force (N or Newtons) t = time (s)
Impulse
Another one… An 8N force acts upon an object for 5 seconds. What impulse is given this object? 40N·s
Putting it all together… F=ma (Newton’s 2 nd Law) a=Δv/t Insert “Δv/t” for “a” in “F=ma” This gives us “F=mΔv/t” Multiply by “t” to get- F(t) = m(v f -v i ) Notice impulse, on the left side, is equal to the change in momentum of an object.
Try it… A ball changes velocity from 20m/s to 30m/s. If it has a mass of 5kg, what impulse was necessary to cause this change? 50N·s How long would it take a 5N force to change the velocity of the ball? (Hint: use Impulse=F(t)) 10s
The Law of Conservation of Momentum In an isolated system, the total momentum does not change. The total momentum remains constant in any type of collision.
Two Types of Collisions…. Elastic The two objects collide and then separate. Inelastic When two objects collide and move together as one mass.
There are 2 types of collisions Elastic m 1i v 1i + m 2i v 2i = m 1f v 1f + m 2f v 2f Where 1 implies object 1, 2 implies object 2 Where i implies initial & f implies final Inelastic m 1i v 1i + m 2i v 2i = m T v T Where T implies total Use + and – velocities to show direction.
Try these… A 2kg ball strikes another ball head on at an initial velocity of +2m/s. If the second ball had a mass of 3kg and was initially moving at -1m/s, what is the final velocity of the 2kg ball if the v f of the 3kg ball is now +1.33m/s? Set it up- m 1i v 1i + m 2i v 2i = m 1f v 1f + m 2f v 2f 2kg(2m/s)+3kg(-1m/s) = 2kg(v 1f )+3kg(1.33m/s) v 1f = -1.5m/s
Last one… A 500kg car is driving at a velocity of 10m/s. Another car (700kg) hits it from behind at a velocity of 25m/s and the two cars interlock bumpers. What is the resulting velocity of the two cars? Set it up- m 1i v 1i + m 2i v 2i = m T v T 500kg(10m/s)+700kg(25m/s) = 1,200kg(v T ) v T = 18.75 m/s
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# PSYCH 2300
### Card Set Information
Author: xx.chelsii ID: 225636 Filename: PSYCH 2300 Updated: 2013-07-02 12:30:10 Tags: Chapter 15 Folders: Description: Correlation and Regression Show Answers:
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1. Correlation
Statistical technique that is used to measure and describe a relationship between two variables
2. Positive Correlation
The two variables tend to change in the same direction.
As the value of X increases, the Y also increases. Vice Versa
3. Negative Correlation
• Two variables go in opposite directions.
• As X increases, Y decreases - an inverse relationship
4. Perfect Correlation
Identified by a correlation of 1.00 and indicates a perfectly consistent relationship
5. Pearson Correlation
Measures the degree to and direction of the linear relationship between two variables
6. Sum of Products or SP
Measure the amount of co-variability between TWO variables
7. Restricted Range
A correlation that is computed from scores that do not represent the full range of possible values, should be cautious when interpreting correlation.
8. Coefficient of determination =
• Measures the proportion of variability in one variable that can be determined from the relationship with the other variable.
• A correlation of r=0.80 (or -0.80), means that =0.64 (or 64%) of the variability in the Y scores can be predicted from the relationship with X
9. Linear Relationship
• A linear relationship between two variables can be expressed by the equation:
• Where a and b are fixed constants
10. Linear Equation
11. Slope
Slope determines how much the Y variable changes when X is increased by 1 point
12. Y-Intercept
• Determines the value of Y when X=0
• a value
13. Regression
Statistical technique for finding the best fitting straight line for a set of data
14. Regression Line
The resulting straight line
15. Least-squared-error
• The best fitting line has the smallest total squared error
16. Regression Equation for Y
17. Standard Error of Estimate
Gives a measure of the standard distance between a regression line and the actual data points
18. Predicted Variability (SSregression)
A statistical technique used to measure the amount of variance in a data set that is not explained by the regression model
19. Unpredicted Variability (SSresidual)
residual of an observed value is the difference between the observed value and theestimated function value
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# Numpy dot Product
Posted in /
Vinay Khatri
Last updated on October 6, 2022
Numpy is one of the Powerful Python Data Science Libraries. It comes with a built-in robust Array data structure that can be used for many mathematical operations. The numpy library supports many methods and ``` numpy.dot() ``` is one of those. Using the numpy dot() method, we can calculate the dot product of two arrays. The numpy ``` dot(array1,array2) ``` method accepts two arrays as a parameter and returns their dot product or matrix multiplication.
## Numpy dot() syntax
``numpy.dot(array1, array2, out=None)``
#### Parameters
arrray1 and array2 represent the array-like structure. The out parameter represents the output argument. By default, its value is ``` None, ``` and if specified explicitly, it needs to be the exact kind of return output of the dot() method.
#### Return value
The dot() product returns a ``` ndarray. ```
### Python numpy dot() method examples
Example1:
Python dot() product if both array1 and array2 are 1-D arrays.
``````>>> import numpy as np
>>> array1 = [1,2,3]
>>> array2 = [4,5,6]
>>> print(np.dot(array1, array2))
32``````
If both the arrays are 1D, the ``` dot() ``` method performs the inner product between the arrays and returns the output as a number.
``````>>>1*4 + 2*5 + 3*6
>>>4+10+18
32``````
Example 2: Python dot() product if both array1 and array2 are 2-D arrays
``````>>> import numpy as np
>>> array1 = [[4,0], [1,-9]]
>>> array2 = [[8,0],[2,-18]]
>>> print(np.dot(array1, array2))
[[ 32 0]
[-10 162]]``````
If both arrays are 2D, the dot will perform the matrix multiplication between them.
``````>>>[[4*8 + 0*2, 4*0 + 0*-18 ]
[1*8 + -9*2, 1*0 + -9*-18 ]
>>>[[32, 0]
[-10, 162]]``````
Note: As a matrix multiplication, the row size of ``` array1 ``` must be equal to the column size of ``` array2 ``` else, the dot() method throws a ValueError.
Example3:
Python dot() product if either of array1 or array2 is a 0-D(scalar) array
``````>>> import numpy as np
>>> array1 = 10
>>> array2 = [[8,0],[2,-18]]
>>> print(np.dot(array1, array2))
[[ 80 0]
[ 20 -180]]``````
If one array of either is a 0-D array, the ``` dot() ``` multiply the 0-D array with the other array.
## Summary
The numpy ``` dot() ``` method finds out the product of two arrays based on their shape. Here are some important facts about ``` dot(array1, array2) ``` method of how it computes the product for different array shapes.
• If ``` array1, ``` and ``` array2 ``` are 1-D array, the dot() method performs inner product between both the arrays.
• If ``` array1 ``` and ``` array2 ``` are 2D arrays, the numpy dot() method performs matrix multiplication between them.
• If anyone between the ``` array1 ``` or ``` array2 ``` is a scalar or 1d array, the numpy ``` dot() ``` method will multiply that 1d or scalar number with another array.
• If ``` array1 ``` is an N-D array and ``` array2 ``` is a 1-D array, then the numpy ``` dot() ``` method will calculate the sum-product over the last axis of ``` array1 ``` and ``` array2. ```
• If ``` array1 ``` is an N-D array and ``` array2 ``` is an M-D array (M>=2 ), then the ``` dot() ``` method will calculate the sum-product over the last axis of ``` array1 ``` and second to the last axis of ``` array2. ```
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# How To Find Quadratic Equations Without Graphing Ideas
How To Find Quadratic Equations Without Graphing. ( 2) + 2 > 0; ( a, b, and c can have any value, except that a can’t be 0.) try changing a, b and c to see what the graph looks like.
A = − ( w − 1 2) 2 + 1 4 4. A parabola can open up or down.
### 42A Graphing Quadratic Equations In Vertex Form
A x 2 + b x + c = 0, w h e r e a ≠ 0. Also see the roots (the solutions to the equation).
### How To Find Quadratic Equations Without Graphing
By applying x = 3 in (1), we get.Continuing from @calculus’ comments, consider the function.Domain f x x 3.Each method also provides information about the corresponding quadratic graph.
F ‘(x) = 6x + 12.F ( x) = x 2 + 2 x − 3.F (x) = ax 2 + bx + c where a, b, and c are real numbers and a is not equal to zero.F(x) = 0.5(x−1) 2 + 1.
Find the intersection point of the straight lines.Find the values of y for which the values of x, obtained from x = g (y) are real and its domain of f.For a ≠ 0 we have a x 2 + b x + c = a g ( x) + d where g ( x) = ( x + b 2 a) 2 and d = c − b 2 4 a.For more information and detailed examples about changing from one quadratic form to another, check out our review article on the three forms of quadratic equations.
Get the equation in the form y = ax 2 + bx + c.Graphing quadratic functions y = ax 2 + bx + c 2.I can graph quadratic functions in vertex form (using basic transformations).I can rewrite quadratic equations from standard to vertex and vice.
I will explain these steps in following examples.If a = 0, then the equation is linear, not quadratic, as there is no a x 2 {\displaystyle ax^{2}} term.If a > 0 the range of a g ( x) is { a y:If the parabola is defined by y = f (x) = 3×2 + 12x +7.
If the parabola opens up, the lowest point is called the vertex.In algebra, a quadratic equation is any equation that can be rearranged in standard form as a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0} where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0.In order to do that, you will need to convert both equations of a problem into the y=mx+b format.In this form, the quadratic equation is written as:
Learn how with this free video lesson.Let x = g (y) step 3 :Now that the quadratic is in vertex form:Now you have all you need to graph.
On the other hand, f ( 0) = − 2 and f ″ ( x) = 2 x log 2.Once you have done this, you will be analyzing the m and b values.Put y = f (x) step 2 :Quadratic function has the form $f(x) = ax^2 + bx + c$ where a, b and c are numbers.
Quadratic functions the graph of a quadratic function is a parabola.Range of a quadratic function.Set $$y=0$$ and solve for $$x\text{.}$$ find the vertex:So the point of intersection of the given straight lines is (3, 0).
So the range of a x 2 + b x + c is { z + d:So there is another root.Solve quadratic equations by factorising, using formulae and completing the square.Solve the equation y = f (x) for x in terms of y.
Substitute the value you obtained for x back into the expression for the parabola to get the y component of the vertex coordinate.The easiest way to identify the range of other functions such as root and fraction functions is to draw the graph of the function using a graphing calculator.The equation they’ve given me to solve is:The numbers a, b, and c are the coefficients of the equation and may be distinguished.
The picture they’ve given me shows the graph of the related quadratic function:The range of a function is the set of all real values of y that you can get by plugging real numbers into x.The range of g ( x) is [ 0, ∞).The set of values of y obtained in step 3 is the range of the given function.
The x value of the axis of symmetry is 2 since 2 is equidistant from (that is, in the middle of) x=1 and x=3.Then read more about the quadratic equation.There are a few rules to follow.This may not be the correct equation for the data, but it’s a good model and the best we can come up with.
This tutorial looks at how to describe a linear system without actually graphing it.To graph a quadratic equation.Using inspection f ( − 1) = − 3 2, f ( − 2) =.We find the vertex of a quadratic equation with the following steps:
Writing equations of quadratic functions 8.Y = 3( − 2)2 +12( − 2) + 7.Y ∈ [ 0, ∞) } = [ 0, ∞).You can sketch quadratic function in 4 steps.
You know by now how to solve a quadratic equation using factoring.Z ∈ [ 0, ∞) } = [ d, ∞).
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# Position vector
Two points and their position vectors
Position vectors (here denoted by and ) in the Cartesian coordinate system${\ displaystyle {\ vec {r}} _ {P}}$${\ displaystyle {\ vec {r}} _ {Q}}$
In mathematics and in physics, a position vector (also radius vector or position vector ) of a point is a vector that points from a fixed reference point to this point (location). In elementary and synthetic geometry , these vectors can be defined as classes of arrows with the same displacement or as parallel displacements .
Position vectors make it possible to use vector calculation for the description of points , sets of points and images . If a Cartesian coordinate system is used as a basis, the origin of the coordinates is generally chosen as the reference point for the position vectors of the points. In this case the coordinates of a point with respect to this coordinate system coincide with the coordinates of its position vector.
In analytical geometry , position vectors are used to describe images of an affine or Euclidean space and to describe sets of points (such as straight lines and planes ) by means of equations and parametric representations .
In physics , position vectors are used to describe the location of a body in Euclidean space. In coordinate transformations, position vectors show a different transformation behavior than covariant vectors .
## Spellings
In geometry, the reference point (origin) is usually referred to as (for Latin origo ). The notation for the position vector of a point is then: ${\ displaystyle O}$${\ displaystyle P}$
${\ displaystyle {\ overrightarrow {OP}}}$
Occasionally the lowercase letters with vector arrows are also used, which correspond to the uppercase letters with which the points are designated, for example:
${\ displaystyle {\ vec {p}} = {\ overrightarrow {OP}}, \ {\ vec {q}} = {\ overrightarrow {OQ}}, \ {\ vec {a}} = {\ overrightarrow {OA }}, \ {\ vec {b}} = {\ overrightarrow {OB}}, \ \ dots, \ {\ vec {x}} = {\ overrightarrow {OX}}}$
The notation that the capital letter denoting the point is provided with a vector arrow is also common:
${\ displaystyle {\ vec {P}} = {\ overrightarrow {OP}}, \ {\ vec {Q}} = {\ overrightarrow {OQ}}, \ {\ vec {A}} = {\ overrightarrow {OA }}, \ {\ vec {B}} = {\ overrightarrow {OB}}, \ \ dots, \ {\ vec {X}} = {\ overrightarrow {OX}}}$
In physics in particular, the position vector is also called the radius vector and is written with a vector arrow as or (especially in theoretical physics) in semi-bold as . ${\ displaystyle {\ vec {r}}}$${\ displaystyle \ mathbf {r}}$
## Examples and applications in geometry
### Connection vector
For the connection vector of two points and with the position vectors and the following applies: ${\ displaystyle {\ overrightarrow {PQ}}}$${\ displaystyle P}$${\ displaystyle Q}$${\ displaystyle {\ vec {p}} = {\ overrightarrow {OP}}}$${\ displaystyle {\ vec {q}} = {\ overrightarrow {OQ}}}$
${\ displaystyle {\ overrightarrow {PQ}} = {\ overrightarrow {OQ}} - {\ overrightarrow {OP}} = {\ vec {q}} - {\ vec {p}}}$
### Cartesian coordinates
The following applies to the coordinates of the position vector of the point with the coordinates : ${\ displaystyle {\ overrightarrow {OP}}}$${\ displaystyle P}$${\ displaystyle (p_ {1}, p_ {2}, p_ {3})}$
${\ displaystyle {\ overrightarrow {OP}} = {\ begin {pmatrix} p_ {1} \\ p_ {2} \\ p_ {3} \ end {pmatrix}}}$
### shift
A shift around the vector maps the point onto the point . Then the following applies to the position vectors: ${\ displaystyle {\ vec {v}}}$${\ displaystyle X}$${\ displaystyle X ^ {\ prime}}$
${\ displaystyle {\ overrightarrow {OX '}} = {\ overrightarrow {OX}} + {\ vec {v}}}$
${\ displaystyle {\ vec {x}} '= {\ vec {x}} + {\ vec {v}}}$
### Rotation around the origin
A rotation in the plane with the center of rotation to the angle counter -clockwise , with the aid of a as follows in Cartesian coordinate rotation matrix will be described: If the position vector of a point and the position vector of the pixel , then: ${\ displaystyle O}$ ${\ displaystyle \ varphi}$${\ displaystyle {\ vec {x}} = {\ tbinom {x_ {1}} {x_ {2}}} = {\ overrightarrow {OX}}}$${\ displaystyle X}$${\ displaystyle {\ vec {x}} '= {\ tbinom {x_ {1}'} {x_ {2} '}} = {\ overrightarrow {OX'}}}$${\ displaystyle X '}$
${\ displaystyle {\ begin {pmatrix} x_ {1} '\\ x_ {2}' \ end {pmatrix}} = {\ begin {pmatrix} \ cos \ varphi & - \ sin \ varphi \\\ sin \ varphi & \ cos \ varphi \ end {pmatrix}} {\ begin {pmatrix} x_ {1} \\ x_ {2} \ end {pmatrix}}}$
### Affine figure
A general affine mapping that maps the point to the point can be represented with position vectors as follows: ${\ displaystyle X}$${\ displaystyle X '}$
${\ displaystyle {\ vec {x}} '= L ({\ vec {x}}) + {\ vec {v}}}$
Here, the position vector von , the position vector von , is a linear mapping and a vector that describes a displacement. In Cartesian coordinates, the linear mapping can be represented by a matrix and the following applies: ${\ displaystyle {\ vec {x}}}$${\ displaystyle X}$${\ displaystyle {\ vec {x}} '}$${\ displaystyle X '}$${\ displaystyle L}$${\ displaystyle {\ vec {v}}}$${\ displaystyle L}$${\ displaystyle A}$
${\ displaystyle {\ vec {x}} = A \ cdot {\ vec {x}} + {\ vec {v}}}$
In three-dimensional space this results in:
${\ displaystyle {\ begin {pmatrix} x_ {1} '\\ x_ {2}' \\ x_ {3} '\ end {pmatrix}} = {\ begin {pmatrix} a_ {11} & a_ {12} & a_ {13} \\ a_ {21} & a_ {22} & a_ {23} \\ a_ {31} & a_ {32} & a_ {33} \ end {pmatrix}} \, {\ begin {pmatrix} x_ {1} \ \ x_ {2} \\ x_ {3} \ end {pmatrix}} + {\ begin {pmatrix} v_ {1} \\ v_ {2} \\ v_ {3} \ end {pmatrix}}}$
Corresponding representations are also available for other dimensions.
### Parametric representation of a straight line
The straight line through the points and contains exactly those points whose position vector is the representation ${\ displaystyle P}$${\ displaystyle Q}$${\ displaystyle X}$${\ displaystyle {\ vec {x}}}$
${\ displaystyle {\ vec {x}} = {\ overrightarrow {OP}} + t \, {\ overrightarrow {PQ}}}$ With ${\ displaystyle t \ in \ mathbb {R}}$
owns. One speaks here of the parametric form of a straight line equation .
### Normal form of the plane equation
The plane through the point (support point) with normal vector contains exactly those points whose position vector corresponds to the normal equation${\ displaystyle P}$ ${\ displaystyle {\ vec {n}}}$${\ displaystyle X}$${\ displaystyle {\ vec {x}}}$
${\ displaystyle {\ vec {x}} \ cdot {\ vec {n}} = {\ vec {p}} \ cdot {\ vec {n}}}$
Fulfills. It is the position vector ( support vector ) of the support point and the Malpunkt denotes the scalar product . ${\ displaystyle {\ vec {p}}}$${\ displaystyle P}$
## Position vector in different coordinate systems
Cartesian coordinate system
The point described by a position vector can be expressed by the coordinates of a coordinate system, the reference point of the position vector usually being placed in the coordinate origin .
### Cartesian coordinates
Usually the position vector is in Cartesian coordinates in the form
${\ displaystyle {\ vec {r}} = {\ vec {r}} \, (x, y, z) = {\ begin {pmatrix} x \\ y \\ z \ end {pmatrix}}}$
Are defined. Therefore the Cartesian coordinates are also the components of the position vector.
### Cylindrical coordinates
The position vector as a function of cylinder coordinates is obtained by converting the cylinder coordinates into the corresponding Cartesian coordinates
${\ displaystyle {\ vec {r}} = {\ vec {r}} \, (\ rho, \ varphi, z) = {\ begin {pmatrix} \ rho \, \ cos \ varphi \\\ rho \, \ sin \ varphi \\ z \ end {pmatrix}}.}$
Here denotes the distance of the point from the -axis, the angle is counted from the -axis in the direction of the -axis. and are therefore the polar coordinates of the point projected orthogonally onto the - plane. ${\ displaystyle \ rho}$${\ displaystyle z}$${\ displaystyle \ phi}$${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle \ rho}$${\ displaystyle \ varphi}$${\ displaystyle x}$${\ displaystyle y}$
From a mathematical point of view, the mapping (function) that assigns the Cartesian coordinates of the position vector to the cylinder coordinates is considered here . ${\ displaystyle (\ rho, \ varphi, z)}$${\ displaystyle (x, y, z)}$
### Spherical coordinates
The position vector as a function of spherical coordinates is obtained by converting the spherical coordinates into the corresponding Cartesian coordinates
${\ displaystyle {\ vec {r}} = {\ vec {r}} \, (r, \ theta, \ varphi) = {\ begin {pmatrix} r \, \ sin \ theta \, \ cos \ varphi \ \ r \, \ sin \ theta \, \ sin \ varphi \\ r \, \ cos \ theta \ end {pmatrix}}.}$
Here denotes the distance of the point from the origin (i.e. the length of the position vector), the angle is measured in the - -plane from the -axis in the direction of the -axis, the angle is the angle between the -axis and the position vector. ${\ displaystyle r}$${\ displaystyle \ varphi}$${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle \ theta}$${\ displaystyle z}$
## physics
### Celestial mechanics
In order to indicate the position of a celestial body moving on an orbit around a center of gravity , this center of gravity is selected as the origin of the location or radius vector in celestial mechanics . The radius vector then always lies in the direction of the gravitational line . The path of the position vector is called the driving beam . The driving beam plays a central role in Kepler's second law (area theorem) .
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# What is the ratio called, when Universal gas constant (R) is divided by Avogadro’s number (N)?
By Ritesh|Updated : November 1st, 2022
1. Plank’s constant
2. Boltzmann’s constant
3. Rydberg’s constant
4. Van der Waal’s constant
When the Universal gas constant (R) is divided by Avogadro’s number (N) is called Boltzmann’s constant. The ratio of the Avogadro number to the Universal Gas Constant is known as the Boltzmann constant. It establishes a relationship between the temperature of a gas and the average kinetic energy of its particles.
### Applications of Boltzmann Constant
Numerous branches of physics make use of the Boltzmann constant. They include some of the following:
• The Boltzmann Constant is used to indicate the equipartition of an atom's energy in traditional statistical mechanics.
• This is how the Boltzmann factor is expressed.
• It has a significant impact on how entropy is defined statistically.
• It is a term used to express thermal voltage in semiconductor physics.
Universal Gas Constant (R)
When one mole of particles at a specific temperature is taken into account, the gas constant is the proportionality constant used to relate the energy scale to the temperature scale in physics. The intersection of Boyle's law, Avogadro's number, Charles' law, and Gay-law Lussac's yields the ideal gas constant. The gas constant R can be expressed as follows:–
Gas constant R = 8.3144598(48) J⋅mol−1⋅K−1
Avogadro's number or Avogadro's constant refers to the number of units contained in one mole of any substance. It is equal to 6.022140857×1023. Depending on the nature of the reaction and the substance, the units may be electrons, ions, atoms, or molecules.
Summary:
## What is the ratio called, when Universal gas constant (R) is divided by Avogadro’s number (N)? (a)Plank’s constant (b)Boltzmann’s constant (c)Rydberg’s constant (d)Van der Waal’s constant
Boltzmann's constant is what results when the universal gas constant (R) is divided by Avogadro's number (N). Max Planck is credited with introducing the Boltzmann constant, which bears Ludwig Boltzmann's name. It is a physical constant that is calculated by dividing the gas constant by the Avogadro number.
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# Minimum product subset of an array
• Difficulty Level : Easy
• Last Updated : 16 Feb, 2023
### INTRODUCTION:
The minimum product subset of an array refers to a subset of elements from the array such that the product of the elements in the subset is minimized. To find the minimum product subset, various algorithms can be used, such as greedy algorithms, dynamic programming, and branch and bound. The choice of algorithm depends on the specific constraints and requirements of the problem.
1. One common algorithm used to find the minimum product subset of an array is the greedy algorithm. The basic idea of this algorithm is to start with the first element of the array and add the next element to the subset only if it will result in a smaller product. The advantage of this algorithm is its simplicity and ease of implementation. However, the greedy algorithm may not always produce the optimal solution and can be very slow for large arrays.
2. Another algorithm used for this problem is dynamic programming. The dynamic programming algorithm divides the problem into subproblems and solves each subproblem only once, using the solutions of smaller subproblems to find the solution for larger ones. This can lead to significant time and space savings. The advantage of dynamic programming is that it always provides the optimal solution, but it can be more complex to implement compared to the greedy algorithm.
3. Branch and bound is another algorithm that can be used to find the minimum product subset of an array. This algorithm involves searching for a solution by branching into multiple possibilities and bounding the search to only consider valid solutions. The advantage of this algorithm is that it provides the optimal solution and can be faster than other algorithms for certain cases. However, it can also be more complex to implement and may require more time and space compared to other algorithms.
In conclusion, the choice of algorithm depends on the specific constraints and requirements of the problem, such as the size of the array, the required solution accuracy, and the available computational resources.
Given array a, we have to find the minimum product possible with the subset of elements present in the array. The minimum product can be a single element also.
Examples:
Input : a[] = { -1, -1, -2, 4, 3 }
Output : -24
Explanation : Minimum product will be ( -2 * -1 * -1 * 4 * 3 ) = -24
Input : a[] = { -1, 0 }
Output : -1
Explanation : -1(single element) is minimum product possible
Input : a[] = { 0, 0, 0 }
Output : 0
A simple solution is to generate all subsets, find the product of every subset and return the minimum product.
A better solution is to use the below facts.
1. If there are even number of negative numbers and no zeros, the result is the product of all except the largest valued negative number.
2. If there are an odd number of negative numbers and no zeros, the result is simply the product of all.
3. If there are zeros and positive, no negative, the result is 0. The exceptional case is when there is no negative number and all other elements positive then our result should be the first minimum positive number.
Implementation:
## C++
`// CPP program to find maximum product of` `// a subset.` `#include ` `using` `namespace` `std;` `int` `minProductSubset(``int` `a[], ``int` `n)` `{` ` ``if` `(n == 1)` ` ``return` `a[0];` ` ``// Find count of negative numbers, count of zeros,` ` ``// maximum valued negative number, minimum valued` ` ``// positive number and product of non-zero numbers` ` ``int` `max_neg = INT_MIN, min_pos = INT_MAX, count_neg = 0,` ` ``count_zero = 0, prod = 1;` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// If number is 0, we don't multiply it with` ` ``// product.` ` ``if` `(a[i] == 0) {` ` ``count_zero++;` ` ``continue``;` ` ``}` ` ``// Count negatives and keep track of maximum valued` ` ``// negative.` ` ``if` `(a[i] < 0) {` ` ``count_neg++;` ` ``max_neg = max(max_neg, a[i]);` ` ``}` ` ``// Track minimum positive number of array` ` ``if` `(a[i] > 0)` ` ``min_pos = min(min_pos, a[i]);` ` ``prod = prod * a[i];` ` ``}` ` ``// If there are all zeros or no negative number present` ` ``if` `(count_zero == n || (count_neg == 0 && count_zero > 0))` ` ``return` `0;` ` ``// If there are all positive` ` ``if` `(count_neg == 0)` ` ``return` `min_pos;` ` ``// If there are even number of negative numbers and` ` ``// count_neg not 0` ` ``if` `(!(count_neg & 1) && count_neg != 0)` ` ``// Otherwise result is product of all non-zeros` ` ``// divided by maximum valued negative.` ` ``prod = prod / max_neg;` ` ``return` `prod;` `}` `int` `main()` `{` ` ``int` `a[] = { -1, -1, -2, 4, 3 };` ` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` ` ``cout << minProductSubset(a, n);` ` ``return` `0;` `}` `// This code is contributed by Sania Kumari Gupta`
## C
`// C program to find maximum product of` `// a subset.` `#include ` `#include ` `// Find maximum between two numbers.` `int` `max(``int` `num1, ``int` `num2)` `{` ` ``return` `(num1 > num2) ? num1 : num2;` `}` `// Find minimum between two numbers.` `int` `min(``int` `num1, ``int` `num2)` `{` ` ``return` `(num1 > num2) ? num2 : num1;` `}` `int` `minProductSubset(``int` `a[], ``int` `n)` `{` ` ``if` `(n == 1)` ` ``return` `a[0];` ` ``// Find count of negative numbers, count of zeros,` ` ``// maximum valued negative number, minimum valued` ` ``// positive number and product of non-zero numbers` ` ``int` `max_neg = INT_MIN, min_pos = INT_MAX, count_neg = 0,` ` ``count_zero = 0, prod = 1;` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// If number is 0, we don't multiply it with` ` ``// product.` ` ``if` `(a[i] == 0) {` ` ``count_zero++;` ` ``continue``;` ` ``}` ` ``// Count negatives and keep track of maximum valued` ` ``// negative.` ` ``if` `(a[i] < 0) {` ` ``count_neg++;` ` ``max_neg = max(max_neg, a[i]);` ` ``}` ` ``// Track minimum positive number of array` ` ``if` `(a[i] > 0)` ` ``min_pos = min(min_pos, a[i]);` ` ``prod = prod * a[i];` ` ``}` ` ``// If there are all zeros or no negative number present` ` ``if` `(count_zero == n || (count_neg == 0 && count_zero > 0))` ` ``return` `0;` ` ``// If there are all positive` ` ``if` `(count_neg == 0)` ` ``return` `min_pos;` ` ``// If there are even number of negative numbers and` ` ``// count_neg not 0` ` ``if` `(!(count_neg & 1) && count_neg != 0)` ` ``// Otherwise result is product of all non-zeros` ` ``// divided by maximum valued negative.` ` ``prod = prod / max_neg;` ` ``return` `prod;` `}` `int` `main()` `{` ` ``int` `a[] = { -1, -1, -2, 4, 3 };` ` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` ` ``printf``(``"%d"``, minProductSubset(a, n));` ` ``return` `0;` `}` `// This code is contributed by Sania Kumari Gupta`
## Java
`// Java program to find maximum product of` `// a subset.` `class` `GFG {` ` ``static` `int` `minProductSubset(``int` `a[], ``int` `n)` ` ``{` ` ``if` `(n == ``1``)` ` ``return` `a[``0``];` ` ``// Find count of negative numbers,` ` ``// count of zeros, maximum valued` ` ``// negative number, minimum valued` ` ``// positive number and product of` ` ``// non-zero numbers` ` ``int` `negmax = Integer.MIN_VALUE;` ` ``int` `posmin = Integer.MAX_VALUE;` ` ``int` `count_neg = ``0``, count_zero = ``0``;` ` ``int` `product = ``1``;` ` ``for` `(``int` `i = ``0``; i < n; i++) {` ` ``// if number is zero,count it` ` ``// but dont multiply` ` ``if` `(a[i] == ``0``) {` ` ``count_zero++;` ` ``continue``;` ` ``}` ` ``// count the negative numbers` ` ``// and find the max negative number` ` ``if` `(a[i] < ``0``) {` ` ``count_neg++;` ` ``negmax = Math.max(negmax, a[i]);` ` ``}` ` ``// find the minimum positive number` ` ``if` `(a[i] > ``0` `&& a[i] < posmin)` ` ``posmin = a[i];` ` ``product *= a[i];` ` ``}` ` ``// if there are all zeroes` ` ``// or zero is present but no` ` ``// negative number is present` ` ``if` `(count_zero == n` ` ``|| (count_neg == ``0` `&& count_zero > ``0``))` ` ``return` `0``;` ` ``// If there are all positive` ` ``if` `(count_neg == ``0``)` ` ``return` `posmin;` ` ``// If there are even number except` ` ``// zero of negative numbers` ` ``if` `(count_neg % ``2` `== ``0` `&& count_neg != ``0``) {` ` ``// Otherwise result is product of` ` ``// all non-zeros divided by maximum` ` ``// valued negative.` ` ``product = product / negmax;` ` ``}` ` ``return` `product;` ` ``}` ` ``// main function` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `a[] = { -``1``, -``1``, -``2``, ``4``, ``3` `};` ` ``int` `n = ``5``;` ` ``System.out.println(minProductSubset(a, n));` ` ``}` `}` `// This code is contributed by Arnab Kundu.`
## Python3
`# Python3 program to find maximum` `# product of a subset.` `# def to find maximum` `# product of a subset` `def` `minProductSubset(a, n):` ` ``if` `(n ``=``=` `1``):` ` ``return` `a[``0``]` ` ``# Find count of negative numbers,` ` ``# count of zeros, maximum valued` ` ``# negative number, minimum valued` ` ``# positive number and product` ` ``# of non-zero numbers` ` ``max_neg ``=` `float``(``'-inf'``)` ` ``min_pos ``=` `float``(``'inf'``)` ` ``count_neg ``=` `0` ` ``count_zero ``=` `0` ` ``prod ``=` `1` ` ``for` `i ``in` `range``(``0``, n):` ` ``# If number is 0, we don't` ` ``# multiply it with product.` ` ``if` `(a[i] ``=``=` `0``):` ` ``count_zero ``=` `count_zero ``+` `1` ` ``continue` ` ``# Count negatives and keep` ` ``# track of maximum valued` ` ``# negative.` ` ``if` `(a[i] < ``0``):` ` ``count_neg ``=` `count_neg ``+` `1` ` ``max_neg ``=` `max``(max_neg, a[i])` ` ``# Track minimum positive` ` ``# number of array` ` ``if` `(a[i] > ``0``):` ` ``min_pos ``=` `min``(min_pos, a[i])` ` ``prod ``=` `prod ``*` `a[i]` ` ``# If there are all zeros` ` ``# or no negative number` ` ``# present` ` ``if` `(count_zero ``=``=` `n ``or` `(count_neg ``=``=` `0` ` ``and` `count_zero > ``0``)):` ` ``return` `0` ` ``# If there are all positive` ` ``if` `(count_neg ``=``=` `0``):` ` ``return` `min_pos` ` ``# If there are even number of` ` ``# negative numbers and count_neg` ` ``# not 0` ` ``if` `((count_neg & ``1``) ``=``=` `0` `and` ` ``count_neg !``=` `0``):` ` ``# Otherwise result is product of` ` ``# all non-zeros divided by` ` ``# maximum valued negative.` ` ``prod ``=` `int``(prod ``/` `max_neg)` ` ``return` `prod` `# Driver code` `a ``=` `[``-``1``, ``-``1``, ``-``2``, ``4``, ``3``]` `n ``=` `len``(a)` `print``(minProductSubset(a, n))` `# This code is contributed by` `# Manish Shaw (manishshaw1)`
## C#
`// C# program to find maximum product of` `// a subset.` `using` `System;` `public` `class` `GFG {` ` ``static` `int` `minProductSubset(``int``[] a, ``int` `n)` ` ``{` ` ``if` `(n == 1)` ` ``return` `a[0];` ` ``// Find count of negative numbers,` ` ``// count of zeros, maximum valued` ` ``// negative number, minimum valued` ` ``// positive number and product of` ` ``// non-zero numbers` ` ``int` `negmax = ``int``.MinValue;` ` ``int` `posmin = ``int``.MinValue;` ` ``int` `count_neg = 0, count_zero = 0;` ` ``int` `product = 1;` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// if number is zero, count it` ` ``// but dont multiply` ` ``if` `(a[i] == 0) {` ` ``count_zero++;` ` ``continue``;` ` ``}` ` ``// count the negative numbers` ` ``// and find the max negative number` ` ``if` `(a[i] < 0) {` ` ``count_neg++;` ` ``negmax = Math.Max(negmax, a[i]);` ` ``}` ` ``// find the minimum positive number` ` ``if` `(a[i] > 0 && a[i] < posmin) {` ` ``posmin = a[i];` ` ``}` ` ``product *= a[i];` ` ``}` ` ``// if there are all zeroes` ` ``// or zero is present but no` ` ``// negative number is present` ` ``if` `(count_zero == n` ` ``|| (count_neg == 0 && count_zero > 0))` ` ``return` `0;` ` ``// If there are all positive` ` ``if` `(count_neg == 0)` ` ``return` `posmin;` ` ``// If there are even number except` ` ``// zero of negative numbers` ` ``if` `(count_neg % 2 == 0 && count_neg != 0) {` ` ``// Otherwise result is product of` ` ``// all non-zeros divided by maximum` ` ``// valued negative.` ` ``product = product / negmax;` ` ``}` ` ``return` `product;` ` ``}` ` ``// main function` ` ``public` `static` `void` `Main()` ` ``{` ` ``int``[] a = ``new` `int``[] { -1, -1, -2, 4, 3 };` ` ``int` `n = 5;` ` ``Console.WriteLine(minProductSubset(a, n));` ` ``}` `}` `// This code is contributed by Ajit.`
## PHP
` 0) ` ` ``\$min_pos` `= min(``\$min_pos``, ``\$a``[``\$i``]); ` ` ``\$prod` `= ``\$prod` `* ``\$a``[``\$i``];` ` ``}` ` ``// If there are all zeros` ` ``// or no negative number` ` ``// present` ` ``if` `(``\$count_zero` `== ``\$n` `|| ` ` ``(``\$count_neg` `== 0 && ` ` ``\$count_zero` `> 0))` ` ``return` `0;` ` ``// If there are all positive` ` ``if` `(``\$count_neg` `== 0)` ` ``return` `\$min_pos``;` ` ``// If there are even number of` ` ``// negative numbers and count_neg` ` ``// not 0` ` ``if` `(!(``\$count_neg` `& 1) && ` ` ``\$count_neg` `!= 0)` ` ``{` ` ``// Otherwise result is product of` ` ``// all non-zeros divided by maximum` ` ``// valued negative.` ` ``\$prod` `= ``\$prod` `/ ``\$max_neg``;` ` ``}` ` ``return` `\$prod``;` `}` `// Driver code` `\$a` `= ``array``( -1, -1, -2, 4, 3 );` `\$n` `= sizeof(``\$a``);` `echo``(minProductSubset(``\$a``, ``\$n``));` `// This code is contributed by Ajit.` `?>`
## Javascript
``
Output
`-24`
Complexity Analysis:
• Time Complexity: O(n)
• Auxiliary Space: O(1)
My Personal Notes arrow_drop_up
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# Adding Fractions is “Painless”, When You Know the Rules!
Adding fractions is a pretty easy and straightforward operation, if you follow the rules. Actually, the “key” is to make sure the denominators are the same. That way, our point of reference is the same and we can now work the problem.
#### Here’s the Rule…
1. If needed, build each fraction so that the denominators are the same.
3. The denominator of your answer will be the new denominator of the built-up fractions.
4. Reduce or simplify the answer, if required.
If the fractions have the same denominator, simply omit Step#1.
### 1/3 + 1/3 =(1 + 1)/3 = 2/3
When working with different denominators, we do all the steps.
## Want to practice adding fractions?
### Related Topics
If you’ve mastered adding fractions, why not try your hand at one of the following: Subtracting Fractions, Multiplying Fractions or Dividing Fractions.
### Fraction Calculator
Learn how to solve fraction problems, then check your work with our online fraction calculator. Click here >
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If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Grade 5 math (FL B.E.S.T.)>Unit 1
Lesson 3: Decimals in expanded form
# Write decimals in expanded form
To write a decimal in expanded form, we need to break down each digit according to its place value. Start with the whole number portion, identifying the hundreds, tens, and ones places. Then, move on to the tenths, hundredths, and thousandths places. Keep in mind the order of operations when combining the expanded terms. Created by Sal Khan.
## Want to join the conversation?
• Why did you say point on this video instead of saying and?The right way to say a decimal is and.
• Either way is correct. Generally when you write decimals out in words you will say "and", and when you write them in numbers you will say "point".
• Why the left side of decimal is once and other right side is tenths
• because there are such a thing as a ones digit (1,2,3,4,5,6,7,8,9), but there is no such thing as a oneths digit. .1 = 1/10 (one tenth) .2 = 2/10 (two tenths) etc.
• how do you even make a project?
• Why when you did the seven hundredth you put it in a fraction. Why is that
• Because decimals are a fraction. In expanded form, we write each digit with its corresponding place value. All the decimal digits have place values that are fractions.
• i'm finding this complicated with all those numbers, is there an easier way to remember or to understand it?
• Absolutely. If you don't understand the words break them down into smaller words and if you don't understand the math you could watch the video a few times and you will probably get it. If you don't reach out again and someone will be able to help you
• I do not get what you are talking about because it is confusing brake it down easyer
• is a decimal point the same thing as the word "and"?
• can a decimal go forever
• Yes, it can. They are called non-terminating decimals. An example is pi, mathematicians are still trying to find the end, if there is one. Also, if there is a number that is like 3.66666666666666666666, you can put a bar notation over the first 6.
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## CS302 Lecture Notes - Uniform, Exponential and Weibull Random Numbers
• September 22, 2009
• James S. Plank
• Directory: /home/plank/cs302/Notes/Random
When we generate a sequence of random numbers, there need to be parameters involved to say what the properties that sequence has. We usually assume a "uniform" distribution between two numbers. For example, drand48() returns a number uniformly distributed between 0 and 1. This means that all numbers between 0 and 1 have an equal chance of being generated. When we roll a die, we are generating a random integer uniformly distributed between 1 and 6, meaning that each of those values has an equal chance of being generated.
A probability distribution defines the characteristics of generating random numbers. One way to characterize this is with a cumulative distribution function (CDF), defined as: F(x) = the probability that a random number is less than or equal to x for all possible x.
Think about a uniform distribution whose minimum value is a and whose maximum value is b. What is F(b)? It is one, since all values are less than or equal to b. The mean of a and b is (a+b)/2. So what about F((b+a)/2)? It is 0.5, since half of the numbers will be below and mean and half will be above the mean.
We plot CDF's with the x values on the x-axis, and the F(x) values on the Y axis. Note, the Y axis therefore only goes from zero to one, and the CDF has to be a monotonically increasing function. Here is the CDF for the uniform distribution whose minimum value is a and whose maximum value is b:
Mathematically, this is:
F(x) = (x-a)/(b-a)
We can use a CDF to generate random numbers using drand48(). Let y be a number generated with drand48(). What we do to generate the number according to a CDF is to find the value of x for which F(x) = y, and x is our random number.
Try it for the uniform distribution above. Suppose we generate y = drand48(). We need to find the value of x for which:
y = F(x) = (x-a)/(b-a)
Using algebra we can massage that properly:
y = (x-a)/(b-a)
(b-a)y = x-a
(b-a)y + a = x.
So when we want to emit random numbers, we do:
``` number = drand48() * (b - a) + a;
```
If you think about it, that would be how you would generate uniformly distributed random numbers between a and b, even if there were no math or CDF's to guide you. It's nice when the theory matches what you would do anyway.
### Exponential Distributions
Exponential Distributions come up in real-life scenarios quite a bit. Examples are arrival events, like people showing up to a store, or cars on a highway, and failure events, such as light bulbs failing. The uniform distribution above was defined by a maximum and a minimum. An exponential distribution is parameterized by a variable λ, which is the mean arrival rate/failure rate. For example, light bulbs may fail at a rate of 1 every 1000 hours, or people may enter a store at a rate of 30 people an hour.
The CDF for an exponential distribution is defined for x ≥ 0, and is:
F(x,λ) = 1- e-λx
Here are CDF's for three values of λ:
Note a few things -- the function is asymptotic, meaning any value of x, from one to a gazillion can occur. It's just that the gazillion has a really low probability of occurring.
The mean value is 1/λ. However, in looking at the graph, you will see that well over half of the values generated will be below the mean - roughly 60 percent. This means that when you buy a light bulb that is rated to last for 1000 hours, chances are 60 percent that it will fail before 1000 hours. However, there's also a chance that the light bulb will last for 1000 years. It's just a really small chance.
We can generate values from an exponential distribution by using the CDF, just as we did above:
y = F(x,&lambda)
y = 1- e-λx
y - 1 = -e-λx
1 - y = e-λx
ln(1 - y) = ln(e-λx)
ln(1 - y) = -λx
-ln(1 - y)/λ = x
So, to generate random numbers according to an exponential, we do:
``` number = -1.0 * log(1.0 - drand48()) / lambda;
```
How about we try this out? Look at expon.cpp:
```#include #include #include #include #include #include using namespace std; main(int argc, char **argv) { double lambda; int iterations; int i; if (argc != 3 || sscanf(argv[1], "%lf", &lambda) != 1 || lambda <= 0 || sscanf(argv[2], "%d", &iterations) != 1) { cerr << "usage: expon lambda iterations\n"; exit(1); } srand48(time(0)); for (i = 0; i < iterations; i++) { cout << log(1-drand48())/lambda*-1 << endl; } } ```
Let's try it out on a simulation scenario. Suppose we buy a crate of 1000 light bulbs, each of which has an average lifetime of 1000 hours. We can simulate the failure time of each bulb by choosing 1000 numbers from an exponential distribution where λ = 1/1000 = 0.001:
```UNIX> expon 0.001 1000 > bulb.txt
286.928
120.251
1234.99
185.829
4899.49
2493.85
3980.23
1912.57
697.718
687.042
UNIX> sort -n bulb.txt | head -n 5
1.06986
1.20739
2.45384
2.62195
2.73788
UNIX> sort -n bulb.txt | tail -n 5
5608.55
5671.58
5792.11
7189.54
8034.5
UNIX>
```
You can see from the first head statement that the random numbers are indeed all over the place. The two sort statements show you the five bulbs that fail the soonest, and the five that last the longest. It's amazing, no? One unlucky bulb fails after a little over an hour, while one lasts almost a year: 8035 hours = 335 days.
Suppose we sort bulb.txt, and plot the numbers as a graph: the y-axis will be the line number of the file, and the x-axis will have the value on that line:
Think to yourself -- how is that related to the CDF for an exponential? Well, divide the values on the y-axis by 1000 and what do you get? You should get an approximation for the CDF for λ = 1000. To show that, the graph below graphs two lines: the previous graph with the y-axis divided by 1000, and the CDF function y = 1 - e-0.001 x:
Awesome, no?
### Weibull Distributions
A Weibull distribution is an enriched exponential that is often used to model physical components that wear out over time. It is charactertized by three variables:
• η is the lifetime -- sometimes called a shape. Wikipedia calls it λ, but I find that confusing because it is not the same λ as an exponential, so I'm using an alternate form that I've seen in other papers.
• β is a shape which describes whether failures/events increase or decrease over time:
• If β equals one, then the Weibull is the same as an exponential with λ = 1 / η. This makes η the expected value, which makes sense.
• If β is greater than one, then failures/events increase over time.
• If β is less than one, then failures/events decrease over time.
• γ is a minimum value.
The CDF for a Weibull is defined for x ≥ γ:
F(x,β,η,&gamma) = 1- e-((x-γ)/η)β
So, to generate random numbers, you do:
y = 1- e-((x-γ)/η)β
1 -y = e-(x-γ)/η)β
ln(1 -y) = -((x-γ)/η)β
(-ln(1 -y)) = ((x-γ)/η)β
(-ln(1 -y))1/&beta = (x-γ)/η
η(-ln(1 -y))1/&beta = x-γ
η(-ln(1 -y))1/&beta + γ = x
Yuck. How do you compute that 1/&beta? Use the pow() math routine (man pow). That allows us to compute a Weibull:
``` tmp1 = -log(1.0 - drand48());
tmp2 = pow(tmp1, 1.0/beta);
return eta * tmp2 + gamma;
```
Use this exact generator in your labs.
You can test it in weibull.cpp
```#include #include #include #include #include #include using namespace std; main(int argc, char **argv) { double eta, beta, tmp1, tmp2, gamma; int iterations; int i; if (argc != 5 || sscanf(argv[1], "%lf", &beta) != 1 || beta <= 0 || sscanf(argv[2], "%lf", &eta) != 1 || eta <= 0 || sscanf(argv[3], "%lf", &gamma) != 1 || gamma < 0 || sscanf(argv[4], "%d", &iterations) != 1) { cerr << "usage: weibull beta eta gamma iterations\n"; exit(1); } srand48(time(0)); for (i = 0; i < iterations; i++) { tmp1 = -log(1.0 - drand48()); tmp2 = pow(tmp1, 1.0/beta); cout << tmp2 * eta + gamma << endl; } } ```
### What do Wiebull's Look Like?
Since the lifetime and minimum value parameters (&eta and &gamma) are straightforward, you may ask "so, exactly what does this shape parameter do?" We'll explore that below. While CDF's are nice for generating random numbers, I find that looking at histograms of probability distributions gives a more intuitive feeling about them.
Rather than generate these histograms mathematically, as we would do in a course on probability, we're going to generate them empirically. The program histogram.cpp takes numbers on standard input and turns them into histograms of a given granularity. We use this program to take a look at a random number generator that generates numbers uniformly between zero and 2000. The command used to generate the graph below is:
```UNIX> uniform 0 2000 1000000 | histogram 100
```
The histogram breaks up the random numbers into "bins" of size 100. Thus, the numbers 0-100 are in the first bin, 100-200 are in the second bin, etc. The Y-axis plots the fraction of all numbers generated in each bin. Since there are 20 bins in the above graph, and numbers are generated uniformly, we would expect each bin to hold roughly 1/20 = 0.05 of the numbers, and indeed, the histogram shows that.
Let's now turn our attention to the Exponential, which is equivalent to the Weibull with β equal to one. We use the following command:
```UNIX> weibull 1 1000 0 1000000 | histogram 100
```
to generate the following histogram:
See how fundamentally different the exponential is from the uniform number generator? The bin with the most values is the smallest -- nearly 1/10th of the values are between zero and 100, even though the mean is 1000.
Let's increase β to 1.3 and see what it does to the histogram:
It shifts the values toward the mean. In other words, more of the values generated are closer to the mean. There are fewer values in the 0-100 bin, and more in the bins closer to 1000. If we increase β to 3, we see something approximating a bell curve:
Increasing β to 50, we see that the values become completely concentrated around the mean:
If we view Weibulls as generating failure data, we say that increasing β "increases the failure rate", because as time gets closer to the mean, if you have not had a failure yet, your probability increases as β increases.
What about decreasing β? We plot three histograms below. First, β = .8:
Then β = .3:
And finally β = .05:
Each one concentrates values further towards zero. However, remember that the mean must still be 1000. This means that there are some really big values generated (or more precisely that the really big bins probably have higher frequencies than their counterparts for larger β). This is sometimes called a "decreasing" failure rate.
### Can we look at CDF's?
Sure -- the program cdf.cpp takes input data and emits X/Y values for a CDF. Using that, we plot the CDF's of all the Weibulls above:
Does that give you a better intuitive feel than the histograms? I don't think so, although all of the conclusions that we drew from the histograms are present in the CDF's. Most are just harder to see. One that is easier to see is when β equals 0.05. Just above, we said that the really big values have higher frequencies than their counterparts for larger β. That is clear from the CDF, and less clear from the histogram.
I didn't expect the CDF(1000) to be independent of β, but I'm sure if you look at the math, you'll see that it is. Interesting.
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in
# How To Find Asymptotes Of Hyperbola
It is the continuous diagonals of the central rectangle that intersect at c. Every hyperbola has two asymptotes.
When a hyperbola shifts around the coordinate plant, use
### The asymptotes pass through the center of the hyperbola (h, k) and intersect the vertices of a rectangle with side lengths of 2a and 2b.
How to find asymptotes of hyperbola. Find the equations of the asymptotes of the hyperbola. Before discussing rectangular hyperbolas, we must first understand what asymptotes are. By using this website, you agree to our cookie policy.
If the hyperbola is vertical, the asymptotes have the equation. If the parabola is given as mx2+ny2 = l, by defining. Asymptotes of a hyperbola are the lines that pass through center of the hyperbola.
Click to see full answer also know, do parabolas have asymptotes? To find the asymptotes of a hyperbola, use a simple manipulation of the equation of the parabola. ( 3 x 2 + 18 x) + ( − 2 y 2) + 15 = 0.
Find the asymptotes of the curve 2 x 2 + 5 x y + 2 y 2 + 4 x + 5 y = 0, and find the general equation of all hyperbolas having the same asymptotes. This can be factored into two linear equations, corresponding to two lines. The eccentricity of the hyperbola whose asymptotes are 3 x + 4 y = 2 and 4 x.
Hyperbola, foci ( 8 , 3 ) and ( 2 , 3 ) , asymptotes y + 3 =. Learn how with this free video lesson. How do you find the asymptotes of a rectangular hyperbola?
First bring the equation of the parabola to above given form. How to find the asymptotes of a hyperbola. Therefore, parabolas don't have asymptotes.when asked to find the equation of the asymptotes, your answer depends on whether the hyperbola is horizontal or vertical.
Basis for drawing the hyperbola. Additionally, why do hyperbolas have asymptotes? When asked to find the equation of the asymptotes, your answer depends on whether the hyperbola is horizontal or vertical.
Every hyperbola has two asymptotes. Learning how to do both may help you understand the concept. And, thanks to the internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next.
As gets larger and larger without bound goes to 0, and so we can say. There are two different approaches you can use to find the asymptotes. ( 3 x 2 + 18 x) + ( − 2 y 2) = − 15.
Find an equation of the conic satisfying the given conditions. It is a useful tool for graphing the hyperbola and its asymptotes. Looking at the denominators, i see that a 2 = 25 and b 2 = 144, so a = 5 and b = 12.
If this sounds confusing, you can think of an asymptote as follows: The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. Need instruction on how to find the equation of a hyperbola using an asymptote?
To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. Then, factor the left side of the equation into 2 products, set each equal to 0, and solve them both for “y” to get the equations for the asymptotes. Includes full solutions and score reporting.
An asymptote to a curve is a straight line, to which the tangent to the curve tends as the point of contact goes to infinity. To find the equations of the asymptotes of a hyperbola, start by writing down the equation in standard form, but setting it equal to 0 instead of 1. An asymptote to a curve is a straight line such that the perpendicular distance of a point \(p(x,\,y.
A hyperbola has two asymptotes as shown in figure 1: Find the center, vertices, foci, eccentricity, and asymptotes of the hyperbola with the given equation, and sketch: The asymptotes of rectangular hyperbola are y = ± x.
If the hyperbola is horizontal, the asymptotes are given by the line with the equation. The hyperbola gets closer and closer to the asymptotes, but can never reach them. Even though parabolas and hyperbolas look very similar, parabolas are formed by the distance from a point and the distance to a line being the same.
The unit hyperbola is the set of points (x,y) in the
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# Descriptive Statistics Calculations and Practical Application Part 2 1.
## Presentation on theme: "Descriptive Statistics Calculations and Practical Application Part 2 1."— Presentation transcript:
Descriptive Statistics Calculations and Practical Application Part 2 1
Content Normal Z distribution Z Score Calculation and Application Z and t distributions 95% confidence interval of the mean Friendly Introductory Statistics Help (FISH) More Descriptive Graphics Test for Normal Distribution Hand Calculations 2
3
http://davidmlane.com/hyperstat/z_table.html Visual of Normal Curve Z Distribution Below plus Above = 100% 4
Z Score A Z score takes a raw score and converts it to a number that expresses how far that value is from the mean in standard deviation units A Z score can be positive, above the mean, negative, below the mean, and 0 equal to the mean A Z score can represent a percentile and probability value The following is a formula to calculate a Z score where X = raw score, X bar = mean and S = standard deviation Z scores explained z score calculation 5
Z Score to Percent Probability http://www.measuringusability.com/pcalcz.php % Below% Above % Tails % Within 6
Example Answers Example from excel spreadsheet on descriptive statistics with filled in values for calculations. Note the percentages are calculated for you. % below = percentile rank for calculated Z score 7
What is Standard Error of the Mean: (error) standard error of the mean is an estimate of the amount that an obtained mean may be expected to differ by chance from the true population mean. http://medical- dictionary.thefreedictionary.com/standard+error+of+the+mean http://medical- dictionary.thefreedictionary.com/standard+error+of+the+mean The larger the n the smaller the SE M. The smaller the Std Dev, the smaller the SE M
Confidence Interval of the Mean for Statistical Inference About Population Using Z score CI 95% = Mean ± 1.96 x (standard error mean) CI 95% = 86.8 ± 1.96 x (0.467) CI 95% = 86.8 ±.915 CI 95% = 85.89 to 87.2 Using t score CI 95% = Mean ± t value x (standard error mean) CI 95% = 86.8 ± 2.2621 x (0.467) CI 95% = 86.8 ± 1.06 CI 95% = 85.74 to 87.862
Z and t distributions at 95%; Statistical Inference Z 5% n-1 2.5% + 2.5% = 5% t distribution
t value approximates Z when sample size is large D egrees of freedom (df) for single group = n-1) http://statpages.org/pdfs.html t = Z
Example using Friendly Introductory Statistics Help (FISH), enter data STEP 1 and perform STEPS 2-10 to check your calculations.
95% Confidence Interval for the Mean using the Z distribution 13 http://www.mccallum-layton.co.uk/stats/ConfidenceIntervalCalc.aspxwww.mccallum-layton.co.uk/stats/ConfidenceIntervalCalc.aspx
FISH with 95% Confidence interval with t distribution
Normal distribution should have density in the middle central values like the distribution shown in this table 15
Positive skewed not normal; the median or mode may better represent this group Not skewed normal; the mean would represent this group Negative skewed not normal; the median or mode may better represent this group 16 Positive Negative
Stem plot 17
Fit of Normal Distribution Mean is a good representation of scores because mean, median, and mode (as shown by the yellow arrow) are at center of the distribution within a distribution that is reasonably bell shaped. Mean is not a good representation of scores because mean in green is pulled negatively towards the outliers to the left. In this case the median in red better represents the density of the distribution. Mean is not a good representation of scores because mean in green is pulled positively towards the outliers to the left. In this case the median in red better represents the density of the distribution. http://bcs.whfreeman.com/ips4e/pages/bcs-main.asp?s=00010&n=99000&i=99010.01&v=category&o=&ns=0&t=&uid=0&rau=0 18 Positive Skewness Negative Skewness
Too high and skinny not normal; positive value Too short and wide not normal; negative value Kurtosis 19
Measures for Normality Skewness (If skewness divided by its error is greater than +1.96 or less than – 1.96 then skewness could cause data to fail normal distribution) 0 if mean and median equal Positive if mean is greater than median Negative if mean is less than median Kurtosis (If kurtosis divided by its error is greater than +1.96 or less than – 1.96 then kurtosis could cause data to fail normal distribution) Mesokurtic is normal =0 Leptokurtic is high and skinny = positive value Platykurtic is short and wide = negative value Normality test : Shapiro-Wilk test Significance level: alpha > 0.05 Inference: Null Hypothesis Retained: hypothesis that data does not differ from the theoretical normal distribution is supported if the significance level is greater than.05; data are normally distributed; it is therefore OK to use the mean as representative of a given group or time. 20
If p value is greater than level of significance then accept that your data are normally distributed. That is, the data do not significantly differ from the normal distribution. 21 Normal Distribution Calculator Calculated using, mean, median, SD, range, skewness and kurtosis
Excel Descriptive Statistics 22 Variable Name Knee ROM Mean86.80 Standard Error0.47 Median87.00 Mode87.00 Standard Deviation1.48 Sample Variance2.18 Kurtosis0.26 Skewness-0.61 Range5.00 Minimum84.00 Maximum89.00 Sum868.00 Count10 Tabled t value 2.26 Confidence Level (95.0%)1.06 Upper Hinge 75%88 Lower Hinge 25%86 Interquartile Range2
Calculator for SEM, tabled value and CI
Proceed to Excel Workbooks to develop your understanding and application of this content 24
Use FISH on Part B Excel to Check Assignment Calculations 25
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+1 vote
95 views
A company produces five types of shirts - A, B, C, D, E - using cloth of three qualities - High, Medium and Low - using dyes of three qualities - High, Medium, and Low. The following tables give, respectively: 1. The number of shirts (of each category) produced, in thousands. 2. The percentage distribution of cloth quality in each type of shirt, and 3. The percentage distribution of dye quality in each type of shirt.
Note: Each shirt requires 1.5 metres of cloth
What is the ratio of low-quality dye used for C-shirts to that used for D-shirts?
1. 3:2
2. 2:1
3. 1:2
4. 2:3
edited | 95 views
+1 vote
Edit : In the figure : Its Distribution of "Dye" not "Day"
$\frac{30k \ * \ 0.4}{10k \ * \ 0.6 } = \frac{12}{6} =2:1[option B]$
answered by (870 points) 1 2 17
+1 vote
Number of shirts where low-quality dyes are used in C-type shirt = 40% * 30,000 = 12,000
Number of shirts where low-quality dyes are used in D-type shirt = 60% * 10,000 = 6,000
Ratio of low-quality dye used for C-shirts to that used for D-shirts = 12,000 : 6,000
= 2:1
answered by (5.7k points) 7 30 142
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# User Forum
Subject :IMO Class : Class 4
## Ans 1:
Class : Class 10
Thanks for the solution
Class : Class 6
## Ans 3:
Class : Class 10
Thank you for pointing out the mistake. We have corrected the sequence now.
Water in the jug = 1/12
Water added in the jug = 5/6
Total water = 1/2 + 5/6 = 11/12
Water overflowed = 1/4
Capacity of jug = 11/12 - 1/4 = 8/12 = 2/3 L
Subject :IMO Class : Class 4
## Ans 1:
Class : Class 7
The 2nd smallest odd 5 digit number is 10003. So difference between the ones place and thousands is 3-0=3. Ans. 3. which is option A
## Ans 2:
Class : Class 5
Answer: 1, Because smallest 5 digit odd number is 10001. Therefore difference between thousandth place digit and ones place digit is 1-0=1. Answer: (C)
Class : Class 4
It's A only
## Ans 4:
Class : Class 6
oh thanks by mistake instead of 10003 i thought 11113
Subject :IMO Class : Class 4
## Ans 1:
Class : Class 10
Solution
The weight of 9 balls = 500 + 500 + 800 = 1800 g
Weight of 1 ball = 200 g
Weight of 5 balls = 1000g
PS: The figure may not be clear. We are replacing it with a more clear figure.
Class : Class 9
[D] 1000g
## Ans 3:
Class : Class 5
The figure is not clear on right hand side it is actually 800+500+500= 1800 kg left side = 9 balls so 1 ball weight is 1800 /9 = 200 kg 5 balls = 1000 kg answer D
## Ans 4:
Class : Class 5
figure at the top was not clear it should be clear so that students can gain marks in a justified way
Subject :IMO Class : Class 5
## Ans 1:
Class : Class 6
640 you idiot
Subject :IMO Class : Class 7
## Ans 1:
Class : Class 9
Subject :IMO Class : Class 7
Class : Class 7
Class : Class 7
## Ans 3:
Class : Class 7
In question the phrase 'is added to itself' should be ommited. Then the answer matches with the question
Subject :IMO Class : Class 10
Class : Class 1
Subject :IMO Class : Class 10
Class : Class 10
## Ans 2:
Class : Class 10
Subject :IMO Class : Class 8
## Ans 1:
Class : Class 4
from where the 12 come?
Class : Class 3
Class : Class 8
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# Numerical integration of double integral with two variables
Posted 7 months ago
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I am trying to numerically integrate the following double integral in MATHEMATICA. This question has also been asked at Mathematica Stackexchangewhere $Im$ is the imaginary part of the expression, $i$ is the imaginary number, $x$ and $y$ are variables while $a$, $b$, $c$ and $Q$ are constants greater than $0$.Here is my attempt to solve this. a = 3 b = 0.0137 c = 0.0023 Q = 6 NIntegrate[Im[Exp[-I x c + b (i x y/(y^a - I x))]]/x, {x, 0, \[Infinity]}, {y, 0, Q}, AccuracyGoal -> 10] Is this the correct way of applying numerical integration with more than one variable? I am getting an error when I run this expression which reads as "evaluated to non-numerical values for all sampling points in the \region with boundaries {{[Infinity],0.},{0,6}}.Can anyone please guide me to correct the implementation of the expression given above.
6 Replies
Sort By:
Posted 7 months ago
Since x and y are Real, ComplexExpand may be able to help a=3;b=0.0137;c=0.0023;Q=6; NIntegrate[ComplexExpand[Im[Exp[-I x c+b (I x y/(y^a-I x))]]/x], {x,0,Infinity},{y,0,Q},AccuracyGoal->10] returns -8.79547 but warns it was not able to obtain 10 digits of accuracy near x==0.Dividing the integral into two parts a=3;b=0.0137;c=0.0023;Q=6; NIntegrate[ComplexExpand[Im[Exp[-I x c+b(I x y/(y^a-I x))]]/x],{x,1/10,Infinity},{y,0,Q},AccuracyGoal->10]+ NIntegrate[Im[Exp[-I x c+b(I x y/(y^a-I x))]]/x,{x,0,1/10},{y,0,Q},AccuracyGoal->10] returns -8.79548+0. I with no warning, but I would study the accuracy of that carefully
Posted 7 months ago
exp should be Exp and i should be I. Mathematica is case sensitive. Also make sure there is a space between two variables which are being multiplied.
Posted 7 months ago
With the code change above, I do not get the non-numerical error. However, the NIntegrate reports the integral as not converging. Dealing with convergence of multidimensional integrals is tricky. It could be that the integrand is oscillatory or contains singularities, or that it in fact does not converge.
Posted 7 months ago
When you simultaneously cross-post the same question (https://mathematica.stackexchange.com/questions/229514/numerical-integration-of-double-integral-with-two-variables), you should mention that in both posts.
Thank you for your response, I have incorporated the changes recommended. However, I still am not getting an error when I run this script which reads as "evaluated to non-numerical values for all sampling points in the \region with boundaries {{[Infinity],0.},{0,6}}. a = 3 b = 0.0137 c = 0.0023 Q = 6 NIntegrate[ Im[Exp[-I x c + b (I x y/(y^a - I x))]]/x, {x, 0, \[Infinity]}, {y, 0, Q}, AccuracyGoal -> 10]
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# Queries to find number of connected grid components of given sizes in a Matrix
• Difficulty Level : Medium
• Last Updated : 10 Sep, 2021
Given a matrix mat[][] containing only of 0s and 1s, and an array queries[], the task is for each query, say k, is to find the number of connected grid components (cells consisting of 1s) of size k
Note: Two cells are connected if they share an edge in the direction up, down, left, and right not diagonal.
Example:
Input: mat[][] = [[1 1 1 1 1 1],
[1 1 0 0 0 0],
[0 0 0 1 1 1],
[0 0 0 1 1 1],
[0 0 1 0 0 0],
[1 0 0 0 0 0]]
queries[] = [6, 1, 8, 2]
Output: [1, 2, 1, 0]
Explanation: There are 4 connected components of sizes 8, 6, 1, 1 respectively hence the output the queries array is [1, 2, 1, 0]. We can see that the number of connected components of different sizes are marked down in the image:
Input: matrix[][] = [[1 1 0 0 0],
[1 0 0 1 0],
[0 0 1 1 0],
[1 1 0 0 0]]
queries[]= [3, 1, 2, 4]
Output: [2, 0, 1, 0]
Explanation: The number of connected components of sizes 3, 2 are 2, 1 all other sizes are zero hence the output array is [2, 0, 1, 0]
Approach: The idea is to count and store frequency of the number of connected components of ones of all sizes in a unordered map using BFS/DFS in the grid, then we can iterate over the queries array and assign the count of connected component for each given size in the array.
Following are the steps to solve the problem:
• Iterate through matrix and perform BFS from the unvisited cell which contains “1”
• Check if the unvisited valid adjacent cells contains 1 and push these cells in the queue.
• Repeat the above two steps for all unvisited cells having 1 in them.
• Print the array having the number of connected components for each given size in the queries.
Below is the implementation of the above approach:
## C++
`// C++ implementation for the above approach` `#include ` `using` `namespace` `std;` `const` `int` `n = 6;` `const` `int` `m = 6;` `const` `int` `dx[] = { 0, 1, -1, 0 };` `const` `int` `dy[] = { 1, 0, 0, -1 };` `// stores information about which cell` `// are already visited in a particular BFS` `int` `visited[n][m];` `// Stores the count of cells in` `// the largest connected component` `int` `COUNT;` `// Function checks if a cell is valid, i.e.` `// it is inside the grid and equal to 1` `bool` `is_valid(``int` `x, ``int` `y, ``int` `matrix[n][m])` `{` ` ``if` `(x < n && y < m && x >= 0 && y >= 0) {` ` ``if` `(visited[x][y] == ``false` ` ``&& matrix[x][y] == 1)` ` ``return` `true``;` ` ``else` ` ``return` `false``;` ` ``}` ` ``else` ` ``return` `false``;` `}` `// Map to count the frequency of` `// each connected component` `map<``int``, ``int``> mp;` `// function to calculate the` `// largest connected component` `void` `findComponentSize(``int` `matrix[n][m])` `{` ` ``// Iterate over every cell` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `j = 0; j < m; j++) {` ` ``if` `(!visited[i][j] && matrix[i][j] == 1) {` ` ``COUNT = 0;` ` ``// Stores the indices of the matrix cells` ` ``queue > q;` ` ``// Mark the starting cell as visited` ` ``// and push it into the queue` ` ``q.push({ i, j });` ` ``visited[i][j] = ``true``;` ` ``// Iterate while the queue` ` ``// is not empty` ` ``while` `(!q.empty()) {` ` ``pair<``int``, ``int``> p = q.front();` ` ``q.pop();` ` ``int` `x = p.first, y = p.second;` ` ``COUNT++;` ` ``// Go to the adjacent cells` ` ``for` `(``int` `i = 0; i < 4; i++) {` ` ``int` `newX = x + dx[i];` ` ``int` `newY = y + dy[i];` ` ``if` `(is_valid(newX, newY, matrix)) {` ` ``q.push({ newX, newY });` ` ``visited[newX][newY] = ``true``;` ` ``}` ` ``}` ` ``}` ` ``mp[COUNT]++;` ` ``}` ` ``}` ` ``}` `}` `// Drivers Code` `int` `main()` `{` ` ``// Given input array of 1s and 0s` ` ``int` `matrix[n][m]` ` ``= { { 1, 1, 1, 1, 1, 1 }, { 1, 1, 0, 0, 0, 0 }, ` ` ``{ 0, 0, 0, 1, 1, 1 }, { 0, 0, 0, 1, 1, 1 }, ` ` ``{ 0, 0, 1, 0, 0, 0 }, { 1, 0, 0, 0, 0, 0 } };` ` ``// queries array` ` ``int` `queries[] = { 6, 1, 8, 2 };` ` ``// sizeof queries array` ` ``int` `N = ``sizeof``(queries) / ``sizeof``(queries[0]);` ` ``// Initialize all cells unvisited` ` ``memset``(visited, ``false``, ``sizeof` `visited);` ` ``// Count the frequency of each connected component` ` ``findComponentSize(matrix);` ` ``// Iterate over the given queries array and` ` ``// And answer the queries` ` ``for` `(``int` `i = 0; i < N; i++)` ` ``cout << mp[queries[i]] << ``" "``;` ` ``return` `0;` `}`
## Java
`// Java implementation for the above approach` `import` `java.util.*;` `class` `GFG{` ` ``static` `class` `pair` ` ``{ ` ` ``int` `first, second; ` ` ``public` `pair(``int` `first, ``int` `second) ` ` ``{ ` ` ``this``.first = first; ` ` ``this``.second = second; ` ` ``} ` ` ``} ` `static` `int` `n = ``6``;` `static` `int` `m = ``6``;` `static` `int` `dx[] = { ``0``, ``1``, -``1``, ``0` `};` `static` `int` `dy[] = { ``1``, ``0``, ``0``, -``1` `};` `// stores information about which cell` `// are already visited in a particular BFS` `static` `int` `[][]visited = ``new` `int``[n][m];` `// Stores the final result grid` `static` `int``[][] result = ``new` `int``[n][m];` `// Stores the count of cells in` `// the largest connected component` `static` `int` `COUNT;` `// Function checks if a cell is valid, i.e.` `// it is inside the grid and equal to 1` `static` `boolean` `is_valid(``int` `x, ``int` `y, ``int` `matrix[][])` `{` ` ``if` `(x < n && y < m && x >= ``0` `&& y >= ``0``) {` ` ``if` `(visited[x][y] == ``0` ` ``&& matrix[x][y] == ``1``)` ` ``return` `true``;` ` ``else` ` ``return` `false``;` ` ``}` ` ``else` ` ``return` `false``;` `}` `// Map to count the frequency of` `// each connected component` `static` `HashMap mp = ``new` `HashMap();` `// function to calculate the` `// largest connected component` `static` `void` `findComponentSize(``int` `matrix[][])` `{` ` ``// Iterate over every cell` ` ``for` `(``int` `i = ``0``; i < n; i++) {` ` ``for` `(``int` `j = ``0``; j < m; j++) {` ` ``if` `(visited[i][j]==``0` `&& matrix[i][j] == ``1``) {` ` ``COUNT = ``0``;` ` ``// Stores the indices of the matrix cells` ` ``Queue q = ``new` `LinkedList<>();` ` ``// Mark the starting cell as visited` ` ``// and push it into the queue` ` ``q.add(``new` `pair( i, j ));` ` ``visited[i][j] = ``1``;` ` ``// Iterate while the queue` ` ``// is not empty` ` ``while` `(!q.isEmpty()) {` ` ``pair p = q.peek();` ` ``q.remove();` ` ``int` `x = p.first, y = p.second;` ` ``COUNT++;` ` ``// Go to the adjacent cells` ` ``for` `(``int` `k = ``0``; k < ``4``; k++) {` ` ``int` `newX = x + dx[k];` ` ``int` `newY = y + dy[k];` ` ``if` `(is_valid(newX, newY, matrix)) {` ` ``q.add(``new` `pair(newX, newY ));` ` ``visited[newX][newY] = ``1``;` ` ``}` ` ``}` ` ``}` ` ``if``(mp.containsKey(COUNT)){` ` ``mp.put(COUNT, mp.get(COUNT)+``1``);` ` ``}` ` ``else``{` ` ``mp.put(COUNT, ``1``);` ` ``}` ` ``}` ` ``}` ` ``}` `}` `// Drivers Code` `public` `static` `void` `main(String[] args)` `{` ` ``// Given input array of 1s and 0s` ` ``int` `matrix[][]` ` ``= { { ``1``, ``1``, ``1``, ``1``, ``1``, ``1` `}, { ``1``, ``1``, ``0``, ``0``, ``0``, ``0` `}, ` ` ``{ ``0``, ``0``, ``0``, ``1``, ``1``, ``1` `}, { ``0``, ``0``, ``0``, ``1``, ``1``, ``1` `}, ` ` ``{ ``0``, ``0``, ``1``, ``0``, ``0``, ``0` `}, { ``1``, ``0``, ``0``, ``0``, ``0``, ``0` `} };` ` ``// queries array` ` ``int` `queries[] = { ``6``, ``1``, ``8``, ``2` `};` ` ``// sizeof queries array` ` ``int` `N = queries.length;` ` ` ` ``// Count the frequency of each connected component` ` ``findComponentSize(matrix);` ` ``// Iterate over the given queries array and` ` ``// And answer the queries` ` ``for` `(``int` `i = ``0``; i < N; i++)` ` ``System.out.print((mp.get(queries[i])!=``null``?mp.get(queries[i]):``0``)+ ``" "``);` `}` `}` `// This code is contributed by 29AjayKumar`
## Python3
`# Python 3 implementation for the above approach` `n ``=` `6` `m ``=` `6` `dx ``=` `[``0``, ``1``, ``-``1``, ``0``]` `dy ``=` `[``1``, ``0``, ``0``, ``-``1``]` `# stores information about which cell` `# are already visited in a particular BFS` `visited ``=` `[[``False` `for` `i ``in` `range``(m)] ``for` `j ``in` `range``(n)]` `# Stores the final result grid` `result ``=` `[[``0` `for` `i ``in` `range``(m)] ``for` `j ``in` `range``(n)]` `# Stores the count of cells in` `# the largest connected component` `COUNT ``=` `0` `# Function checks if a cell is valid, i.e.` `# it is inside the grid and equal to 1` `def` `is_valid(x, y, matrix):` ` ``if` `(x < n ``and` `y < m ``and` `x >``=` `0` `and` `y >``=` `0``):` ` ``if` `(visited[x][y] ``=``=` `False` `and` `matrix[x][y] ``=``=` `1``):` ` ``return` `True` ` ``else``:` ` ``return` `False` ` ``else``:` ` ``return` `False` `# Map to count the frequency of` `# each connected component` `mp ``=` `{}` `# function to calculate the` `# largest connected component` `def` `findComponentSize(matrix):` ` ``# Iterate over every cell` ` ``for` `i ``in` `range``(n):` ` ``for` `j ``in` `range``(m):` ` ``if``(visited[i][j]``=``=` `False` `and` `matrix[i][j] ``=``=` `1``):` ` ``COUNT ``=` `0` ` ``# Stores the indices of the matrix cells` ` ``q ``=` `[]` ` ``# Mark the starting cell as visited` ` ``# and push it into the queue` ` ``q.append([i, j])` ` ``visited[i][j] ``=` `True` ` ``# Iterate while the queue` ` ``# is not empty` ` ``while` `(``len``(q)!``=``0``):` ` ``p ``=` `q[``0``]` ` ``q ``=` `q[``1``:]` ` ``x ``=` `p[``0``]` ` ``y ``=` `p[``1``]` ` ``COUNT ``+``=` `1` ` ``# Go to the adjacent cells` ` ``for` `i ``in` `range``(``4``):` ` ``newX ``=` `x ``+` `dx[i]` ` ``newY ``=` `y ``+` `dy[i]` ` ``if` `(is_valid(newX, newY, matrix)):` ` ``q.append([newX, newY])` ` ``visited[newX][newY] ``=` `True` ` ``if` `COUNT ``in` `mp:` ` ``mp[COUNT] ``+``=` `1` ` ``else``:` ` ``mp[COUNT] ``=` `1` `# Drivers Code` `if` `__name__ ``=``=` `'__main__'``:` ` ``# Given input array of 1s and 0s` ` ``matrix ``=` `[[``1``, ``1``, ``1``, ``1``, ``1``, ``1``],` ` ``[``1``, ``1``, ``0``, ``0``, ``0``, ``0``],` ` ``[``0``, ``0``, ``0``, ``1``, ``1``, ``1``],` ` ``[``0``, ``0``, ``0``, ``1``, ``1``, ``1``],` ` ``[``0``, ``0``, ``1``, ``0``, ``0``, ``0``],` ` ``[``1``, ``0``, ``0``, ``0``, ``0``, ``0``]]` ` ``# queries array` ` ``queries ``=` `[``6``, ``1``, ``8``, ``2``]` ` ``# sizeof queries array` ` ``N ``=` `len``(queries)` ` ``# Count the frequency of each connected component` ` ``findComponentSize(matrix)` ` ``# Iterate over the given queries array and` ` ``# And answer the queries` ` ``for` `i ``in` `range``(N):` ` ``if` `queries[i] ``in` `mp:` ` ``print``(mp[queries[i]],end ``=` `" "``)` ` ``else``:` ` ``print``(``0``,end ``=` `" "``)` ` ``# This code is contributed by SURENDRA_GANGWAR.`
## C#
`// C# implementation for the above approach` `using` `System;` `using` `System.Collections.Generic;` `public` `class` `GFG{` ` ``class` `pair` ` ``{ ` ` ``public` `int` `first, second; ` ` ``public` `pair(``int` `first, ``int` `second) ` ` ``{ ` ` ``this``.first = first; ` ` ``this``.second = second; ` ` ``} ` ` ``} ` `static` `int` `n = 6;` `static` `int` `m = 6;` `static` `int` `[]dx = { 0, 1, -1, 0 };` `static` `int` `[]dy = { 1, 0, 0, -1 };` `// stores information about which cell` `// are already visited in a particular BFS` `static` `int` `[,]visited = ``new` `int``[n,m];` `// Stores the count of cells in` `// the largest connected component` `static` `int` `COUNT;` `// Function checks if a cell is valid, i.e.` `// it is inside the grid and equal to 1` `static` `bool` `is_valid(``int` `x, ``int` `y, ``int` `[,]matrix)` `{` ` ``if` `(x < n && y < m && x >= 0 && y >= 0) {` ` ``if` `(visited[x,y] == 0` ` ``&& matrix[x,y] == 1)` ` ``return` `true``;` ` ``else` ` ``return` `false``;` ` ``}` ` ``else` ` ``return` `false``;` `}` `// Map to count the frequency of` `// each connected component` `static` `Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();` `// function to calculate the` `// largest connected component` `static` `void` `findComponentSize(``int` `[,]matrix)` `{` ` ``// Iterate over every cell` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `j = 0; j < m; j++) {` ` ``if` `(visited[i,j]==0 && matrix[i,j] == 1) {` ` ``COUNT = 0;` ` ``// Stores the indices of the matrix cells` ` ``List q = ``new` `List();` ` ``// Mark the starting cell as visited` ` ``// and push it into the queue` ` ``q.Add(``new` `pair( i, j ));` ` ``visited[i,j] = 1;` ` ``// Iterate while the queue` ` ``// is not empty` ` ``while` `(q.Count>0) {` ` ``pair p = q[0];` ` ``q.RemoveAt();` ` ``int` `x = p.first, y = p.second;` ` ``COUNT++;` ` ``// Go to the adjacent cells` ` ``for` `(``int` `k = 0; k < 4; k++) {` ` ``int` `newX = x + dx[k];` ` ``int` `newY = y + dy[k];` ` ``if` `(is_valid(newX, newY, matrix)) {` ` ``q.Add(``new` `pair(newX, newY ));` ` ``visited[newX,newY] = 1;` ` ``}` ` ``}` ` ``}` ` ``if``(mp.ContainsKey(COUNT)){` ` ``mp[COUNT] += 1;` ` ``}` ` ``else``{` ` ``mp.Add(COUNT, 1);` ` ``}` ` ``}` ` ``}` ` ``}` `}` `// Drivers Code` `public` `static` `void` `Main()` `{` ` ``// Given input array of 1s and 0s` ` ``int` `[,]matrix` ` ``= ``new` `int``[,]{ { 1, 1, 1, 1, 1, 1 }, { 1, 1, 0, 0, 0, 0 }, ` ` ``{ 0, 0, 0, 1, 1, 1 }, { 0, 0, 0, 1, 1, 1 }, ` ` ``{ 0, 0, 1, 0, 0, 0 }, { 1, 0, 0, 0, 0, 0 } };` ` ``// queries array` ` ``int` `[]queries = { 6, 1, 8, 2 };` ` ``// sizeof queries array` ` ``int` `N = queries.Length;` ` ``for``(``int` `i=0;i
Output:
`1 2 1 0`
Time Complexity: O(n*m)
Space Complexity: O(n*m)
My Personal Notes arrow_drop_up
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# Excel calculates the number of bills
Source: Internet
Author: User
Here's a very practical example for you to explain, that is, in Excel, give a sum of the amount, how to automatically calculate 100 yuan, 50 yuan, 20 yuan, 10 yuan, 5 yuan, 2 yuan, 1 yuan, 5, 2, 1 cents, 5 points, 2 points, 1 points each have a few!
Let's look at the table below first.
In the table above, A2 gives a number of amounts, and then, in column B, gives a different currency for the amount, the C column calculates each by how many Zhang hundred Yuan, Wu picks the yuan, the double picks up the yuan, picks the yuan, the Wu Yuan, the Double yuan, one yuan, Woocher, the angle, a corner, Wu cent, the minute, a cent composition.
Although the use of the formula is not complicated, but it is not so simple, for the convenience of everyone's learning, the above figure has added an auxiliary column, listed in the C column of the formula, please carefully look, in addition, for the convenience of your study, and then write the formula to the following:
IF (int (A2)/100) >=1,int (int (A2)/100), "0")
INT ((a2-(c2*100))/50)
INT ((a2-c2*100-c3*50)/20)
INT ((a2-c2*100-c3*50-c4*20)/10)
INT ((A2-C2*100-C3*50-C4*20-C5*10)/5)
INT ((a2-c2*100-c3*50-c4*20-c5*10-c6*5)/2)
INT ((a2-c2*100-c3*50-c4*20-c5*10-c6*5)-C7*2/1)
INT ((A2-int (A2))/0.5)
INT ((A2-int (A2)-c9*0.5)/0.2)
INT ((A2-int (A2)-c9*0.5-c10*0.2)/0.1)
INT ((A2-int (A2)-c9*0.5-c10*0.2-c11*0.1)/0.05)
INT ((A2-int (A2)-c9*0.5-c10*0.2-c11*0.1-c12*0.05)/0.02)
INT ((A2-int (A2)-c9*0.5-c10*0.2-c11*0.1-c12*0.05)/0.02)
Let's look at the formula for C4 cell, and hope you have some insight.
Computer Tutorials
Finally, let's look at how to calculate the formula for the angle: see the picture.
Related Keywords:
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Scheme in 33 Miniatures
This weekend I was trying to wake my old brain up by looking and the clever and mind-bending problems from my graduate school. I have really gotten excited about Madhur Tulsi‘s Mathematical Toolkit class — as a “data scientist”/”machine learning engineer” (I sometimes wonder what these designations mean — heck I just am grateful to have been able to work on A.I. with all its challenges and contradictions for a couple of decades) I find myself treasuring the notes and exercises.
One gem that I found is the text Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra by Jiřì Matoušek. This is full of memorable tricks and insights that leverage linear algebra. While the text is mostly about proofs, I used some of the algorithms presented discussions do some Lisp programming this weekend.
I started off with the first two miniatures on Fibonacci numbers — the first presenting a fast method to get the nth Fibonacci number and the second which is probably not so fast is yet elegant in that it presents a function to compute Fibonacci numbers.
So first, I wanted to re-awaken my Lisp programming brain. Racket is definitely one of the more innovative Lisp languages to have emerged over the last 10 years. You can think of Racket as a package that provides a collection of Lisp-languages, among them standard Scheme — a paired down Lisp dialect.
The first implementation achieves $O(logn)$ running time by using exponentiation by squaring. This is achieved by first realizing that the Fibonacci recurrence can be written in terms of a matrix polynomial $\begin{pmatrix} F_(n+1)\\ F_(n)\end{pmatrix} = M^n \begin{pmatrix} 1\\ 0\end{pmatrix}$ where $M = \begin{pmatrix} 1 & 1\\ 1 & 0\end{pmatrix}$. I hadn’t known this but $n$ is a power of 2 we can compute the $M^n$ by repeatedly squaring $M$ and then the case where $n$ is odd is taken care of by a multiply. I’m guessing that the core exponentiation routine would use this (or a faster method) anyway, but still interesting to consider.
So first to see if a quick exponentiation by squaring does against the method provided in the racket math/matrix package. Here’s my quick recursive method, based on the tail recursive method in the wikipedia article
;;identity
(define one-matrix (matrix [[1 0] [0 1]]))
;;recursive exponentiation by squaring
(define (exp-by-squaring x n)
(letrec
(
[exp-by-squaring-aux
(lambda (y x n)
(cond
[(= n 0) y]
[(= n 1) (matrix* y x)]
[(= (modulo n 2) 0)
(exp-by-squaring-aux y (matrix* x x) (/ n 2))]
[else
(exp-by-squaring-aux
(matrix* y x) (matrix* x x) (/ (- n 1) 2))]))])
(exp-by-squaring-aux one-matrix x n)))
And here is a quick running time analysis
> (time (dotimes (x 1000) (exp-by-squaring (matrix [[1 1 ] [1 0]]) 100)))
cpu time: 793 real time: 802 gc time: 175
> (time (dotimes (x 1000) (matrix-expt (matrix [[1 1 ] [1 0]]) 100)))
cpu time: 143 real time: 144 gc time: 36
>
Looks like I’m better off with the library — although curious to know what the library implementation is.
So sticking with the matrix-expt library function:
;;;-- Miniature 1: Fibonacci Numbers, Quickly
(define one-zero-matrix (matrix [[1 0]]))
(define fib-matrix (matrix [[1 1 ] [1 0]]))
(define (fast-fib n)
;;; Fast method of computing nth fibonnaci number
(let* (
[m-expt (matrix-expt fib-matrix n)]
[idx0 0]
[idx1 1]
[prod (matrix* one-zero-matrix m-expt )]
)
(matrix-ref prod idx0 idx1 )))
Now here’s an old-school tail recursive implementation
;;; Old school fib
(define (os-fib n)
(letrec ([os-fib-helper
(lambda (n a b)
(cond
[(= n 0) b]
[else
(os-fib-helper (- n 1) b (+ a b) )]))])
(cond
[(= n 0) 0]
[(= n 1) 1]
[else
(os-fib-helper (- n 2) 1 1)])))
Now let’s look at how they compare.
> (time (dotimes (x 10000) (fast-fib 100)))
cpu time: 2554 real time: 2612 gc time: 617
> (time (dotimes (x 10000) (os-fib 100)))
cpu time: 41 real time: 41 gc time: 11
Wow. Maybe we’ll have to go to type checked racket to see any gains there!
Typed racket claims to get more of a boost in performance by using type hints. Here is the typed version of the fast Fibonacci
;;; -- Miniature 1: Fibonacci Numbers, Quickly (with type hints)
(: one-zero-matrix (Matrix Integer))
(define one-zero-matrix (matrix [[1 0]]))
(: fib-matrix (Matrix Integer))
(define fib-matrix (matrix [[1 1 ] [1 0]]))
(: fast-fib (-> Integer Integer))
(define (fast-fib n)
;;; Fast method of computing nth fibonnaci number
(let*
(
[m-expt (matrix-expt fib-matrix n)]
[idx0 : Integer 0]
[idx1 : Integer 1]
[prod : (Matrix Any) (matrix* one-zero-matrix m-expt )]
)
(cast (matrix-ref prod idx0 idx1 ) Integer)))
;;Hacky runner to allow my bootleg benchmark
(: run-fast-fib-100 (-> Integer Integer))
(define (run-fast-fib-100 n)
(dotimes (x n)
(fast-fib 100))
n)
> (time (run-fast-fib-100 10000))
cpu time: 689 real time: 728 gc time: 251
- : Integer
10000
> (time (run-fast-fib-100 10000))
cpu time: 672 real time: 733 gc time: 244
- : Integer
10000
> (time (run-fast-fib-100 10000))
cpu time: 704 real time: 750 gc time: 267
- : Integer
10000
> (time (dotimes (x 10000) (os-fib 100)))
cpu time: 28 real time: 28 gc time: 0
> (time (dotimes (x 10000) (os-fib 100)))
cpu time: 31 real time: 30 gc time: 0
> (time (dotimes (x 10000) (os-fib 100)))
cpu time: 35 real time: 35 gc time: 0
>
About an order of magnitude better, but still order of magnitude worse than the iterative version though. Better off sticking with the simple iteration.
Now in the second miniature, Matoušek derives a numeric expression for the $nth$ Fibonacci number by solving a system of linear equations that follow from the recurrence. This is $F_n = 1/\sqrt{5} \left[ \right( (1 + \sqrt{5})/2 \left) ^n - \right( (1 - \sqrt{5})/2 \left) ^n \right]$
;;; -- Miniature 2: Fibonacci Numbers, the Formula(: root-five Real)
(define root-five (sqrt 5.0))
(define inv-root-five (/ 1.0 root-five))
(define first-tau-root (/ (+ 1.0 root-five) 2.0))
(define second-tau-root (/ (- 1.0 root-five) 2.0))
(define (fib-root n)
;;; Not so fast but fun ?
(exact-round
(* inv-root-five
(- (expt first-tau-root n) (expt second-tau-root n))))
How does compare against an “old school” iterative version?
> (time (dotimes (x 10000) (fib-root 100)))
cpu time: 3 real time: 4 gc time: 0
> (time (dotimes (x 10000) (os-fib 100)))
cpu time: 27 real time: 28 gc time: 0
Well, that seems to best it. At least until we get further out in the sequence, where overflow gets to be an issue
> (time (dotimes (x 10000) (fib-root 1200)))
cpu time: 4 real time: 4 gc time: 0
> (time (dotimes (x 10000) (os-fib 1200)))
cpu time: 1271 real time: 1419 gc time: 293
> > (time (dotimes (x 10000) (fib-root 2000)))
. . exact-round: contract violation
expected: rational?
given: +inf.0
> (time (dotimes (x 10000) (os-fib 2000)))
cpu time: 2619 real time: 2893 gc time: 725
So we’ve got constant time with the analytic solution, we might loose the better performance once we get better precision. Interesting. I’ll hold it there for now and contemplate. It’s been a nice Scheme appetizer — a feel my inner Scheme brain re-awakening.
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A215798 Numbers n such that 2^n-1 can be written in the form a^2 + 3*b^2. 4
2, 3, 5, 6, 7, 9, 13, 14, 15, 17, 18, 19, 21, 25, 26, 27, 31, 37, 38, 39, 42, 45, 49, 51, 54, 57, 61, 62, 63, 65, 67, 74, 75, 78, 81, 85, 89, 93, 98, 101, 103, 107, 111, 114, 117, 122, 125, 126, 127, 133, 134, 135, 139, 147, 153, 162, 171, 183, 186, 189, 195, 201, 217, 221, 222, 225, 234, 243, 254, 255, 257, 259, 267, 269, 271, 278, 279, 281, 293, 294 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS These 2^n-1 numbers have no prime factors of the form 2 (mod 3) to an odd power. LINKS V. Raman, Table of n, a(n) for n = 1..159 Samuel S. Wagstaff, Jr., The Cunningham Project, Factorizations of 2^n-1, for odd n's < 1200. EXAMPLE 2^67-1 = 10106743618^2+3*3891344499^2 = 9845359982^2+3*4108642899^2 PROG (PARI) for(i=2, 100, a=factorint(2^i-1)~; has=0; for(j=1, #a, if(a[1, j]%3==2&&a[2, j]%2==1, has=1; break)); if(has==0, print(i" -\t"a[1, ]))) CROSSREFS Cf. A000043, A000225, A215799, A215806, A215807. Sequence in context: A153763 A050747 A261514 * A074780 A292938 A056900 Adjacent sequences: A215795 A215796 A215797 * A215799 A215800 A215801 KEYWORD nonn AUTHOR V. Raman, Aug 23 2012 EXTENSIONS Added 14 more terms - V. Raman, Aug 29 2012 STATUS approved
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Last modified August 7 11:46 EDT 2020. Contains 336276 sequences. (Running on oeis4.)
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# Oxidation Number Worksheet
Do you aspire to transform young learners to "Mini Mathematicians"? Then, this is the place to be! Introduce the elementary math concepts for kindergarten through fifth graders with this set of number sense worksheets featuring topics like counting objects, skip counting, performing the four arithmetic operations like addition, subtraction, multiplication and division, real-life word problems and exclusive place value drills. Additionally, learn the types of numbers like prime and composite, odd and even, Roman numerals and much more. Included here are rounding off and estimation practices as well.
These worksheets will generate nine times tables target circles drills as selected by the user. The user may select from times tables ranging from 1 to 15, and the range for the table to be between 0 and 12. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, and 3rd Grade. These multiplication worksheets are configured to produce problems in the range of 0 thru 12 in a vertical format. The numbers for each factor may be individually varied to generate different sets of Multiplication problems. If you select the number 5 in the one group and all of the numbers 0 through 12 in the other group, then you will produce a multiplication worksheet that generates problems for the 5 times tables. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
These multiplication worksheets help teach multiplication by learning how to draw and determine the size of arrays. The student will be given a description of an array and then asked to both draw the array and determine the number of units in the array. You may select the range of rows and columns used for the arrays, as well as the description given to draw the array. These multiplication worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade. These multiplication worksheets help teach multiplication by learning how to draw and determine the size of arrays based off a given word problem. The student will be given a word problem in which they must draw an array and write a multiplication equation to describe and solve the word problem. You may select the range of rows and columns used for the arrays. These multiplication worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
These multiplication worksheets may be configured for 1 or 2 Digits on the right of the decimal and up to 2 digits on the left of the decimal. You may vary the numbers of multiplication problems on the multiplication worksheets from 12 to 25. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. These multiplication worksheets may be configured for either single or multiple digit horizontal problems. The factors may be selected to be positive, negative or mixed numbers for these multiplication worksheets. You may vary the numbers of multiplication problems on the multiplication worksheets from 12 to 30. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
These multiplication worksheets may be configured for 2, 3, or 4 digit multiplicands being multiplied by 1, 2, or 3 digit multipliers. You may select between 12 and 25 multiplication problems to be displayed on the multiplication worksheets. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. These multiplication worksheets may be configured for either single or multiple digit horizontal problems with 2 factors. You may select between 12 and 30 multiplication problems to be displayed on the multiplication worksheets. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
These multiplication worksheets may be configured for up to 3 digits on the left of the decimal. The currency symbol may be selected from Dollar, Pound, Euro, and Yen. You may vary the numbers of problems for each worksheet from 12, 16 or 20. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. These multiplication worksheets use arrays to help teach multiplication and how to write out multiplication equations. The student will be given an array and asked to write out the numbers of rows and columns in the array, as well as a multiplication equation to describe the array. You may select the range of rows and coumns used for the arrays. These multiplication worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
These multiplication times table worksheets are colorful and a great resource for teaching kids their multiplication times tables. A complete set of free printable multiplication times tables for 1 to 12. These multiplication times table worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. This multiplication times table charts is a great resource for teaching kids their multiplication times tables. The chart is sized based off the magnitude of the multiplied number. This multiplication chart is great for a visual representation of the multiplication times tables. This multiplication chart is appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
## Triangle Angle Sum Worksheet
### Worksheet Periodic Trends Answer Key
#### Rate This oxidation number worksheet
90/100 by 719 users
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## Calculus (3rd Edition)
$$x=\sqrt[3] 2.$$
The zeros of $f(x)=2- x^3$ can be obtained by putting $$2-x^3=0\Longrightarrow x=\sqrt[3] 2.$$ Since $f(-x)\neq -f(x)$ nor $f(-x)\neq f(x)$ then the function is not symmetric about the origin nor about the y-axis and it is decreasing on the interval $(-\infty,\infty)$. See the figure below.
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17 terms
# Honors Geometry - Chapter 5
#### Terms in this set (...)
Perpendicular Bisector
Goes through midpoint and is perpendicular to a side.
Circumcenter
The point of intersection of the perpendicular bisectors.
Circumcenter Theorem
This point of a triangle is equidistant from the vertices of the triangle.
Median
Starts at the midpoint and goes to a vertex.
Centroid
The point of intersection of the medians.
Centroid Theorem
The special point involved with this theorem is located 2/3 of the distance from a vertex to the midpoint of the side opposite the vertex on a median.
Altitude
Starts at a vertex and goes perpendicular to a side.
Orthocenter
The point of intersection of the altitudes.
Angle Bisector
A segment that cuts an angle in half equally.
Incenter
The point of intersection of the angle bisectors.
Incenter Theorem
The special point involved with this theorem is equidistant from each side of the triangle.
Exterior Angle Theorem
If an angle is an exterior angle of a triangle, then its measure is greater than the measure of either of its corresponding remote interior angles.
Opposite Angle Theorem
If one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.
Opposite Side Theorem
The bigger angle is opposite the longer side and the smaller angle is opposite the smaller side.
Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
SAS Inequality / Hinge Theorem
If two sides of one triangle are congruent to two sides of another triangle, then the bigger included angle is opposite the longer side and the smaller included angle is opposite the shorter side.
SSS Inequality Theorem
If two sides of one triangle are congruent to two sides of another triangle, then the third side that is longer is opposite the bigger angle and the third side that is shorter is opposite the smaller angle.
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# What does the ABC stand for in standard form Ax+By=C?
What does the ABC stand for in standard form Ax+By=C?
Ax+By+C is a generalized form (in fact the standard generalized form) for a linear equation
where A, B, and C are placeholders for constants (the x and y are variables).
The generalized form, Ax+By+C, includes equations such as
\$color(white)(XXXXX)\$\$7x+3y=98\$
\$color(white)(XXXXX)\$\$(1)x+(-5)y=14\$
\$color(white)(XXXXX)\$\$9261x+83y=-7491\$
\$color(white)(XXXXX)\$\$color(white)(XXXXX)\$any infinitely many others
By normal usage A should be an integer value which is not negative.
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## Conversion formula
The conversion factor from kilometers to decimeters is 10000, which means that 1 kilometer is equal to 10000 decimeters:
1 km = 10000 dm
To convert 22.7 kilometers into decimeters we have to multiply 22.7 by the conversion factor in order to get the length amount from kilometers to decimeters. We can also form a simple proportion to calculate the result:
1 km → 10000 dm
22.7 km → L(dm)
Solve the above proportion to obtain the length L in decimeters:
L(dm) = 22.7 km × 10000 dm
L(dm) = 227000 dm
The final result is:
22.7 km → 227000 dm
We conclude that 22.7 kilometers is equivalent to 227000 decimeters:
22.7 kilometers = 227000 decimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 decimeter is equal to 4.4052863436123E-6 × 22.7 kilometers.
Another way is saying that 22.7 kilometers is equal to 1 ÷ 4.4052863436123E-6 decimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-two point seven kilometers is approximately two hundred twenty-seven thousand decimeters:
22.7 km ≅ 227000 dm
An alternative is also that one decimeter is approximately zero times twenty-two point seven kilometers.
## Conversion table
### kilometers to decimeters chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers to decimeters
kilometers (km) decimeters (dm)
23.7 kilometers 237000 decimeters
24.7 kilometers 247000 decimeters
25.7 kilometers 257000 decimeters
26.7 kilometers 267000 decimeters
27.7 kilometers 277000 decimeters
28.7 kilometers 287000 decimeters
29.7 kilometers 297000 decimeters
30.7 kilometers 307000 decimeters
31.7 kilometers 317000 decimeters
32.7 kilometers 327000 decimeters
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cum
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Search
# CSEC Mathematics: Appreciation and Depreciation
In this lesson, we will cover:
1. what appreciation and depreciation are
2. how to calculate and solve problems involving appreciation or depreciation
Appreciation is the increase of the value of something over time. Depreciation, on the other hand, is the opposite- it is the decrease of value of something over time.
You may be familiar with appreciation in terms of the value of houses or assets over time, since a house or property will appreciate the longer you own it. A car, however, depreciates over time when bough from new due to wear and tear.
Problems involving appreciation or depreciation are almost exactly the same as problems asking about compound interest. In fact, the formula for calculating it is the same as the formula we discussed in a previous post on interest.
Appreciation: V = I x (1 + r)^n
Depreciation: V = I x (1 - r)^n
V= Final value
I = Initial value
r = rate of increase or decrease of value over time
n = number of periods (years, months, weeks, depending on the question)
Example 1: A villa is purchased in 2004 for \$16.5 million. What is the value of the villa in 2009 if it appreciates in value by 7% each year?
This is a problem of appreciation, so we add the rate of change of value.
V = I x (1 + r)^n
= \$16,500,000 x (1 + 0.07)^5
= \$23,142,103.56
Example 2: A Toyota Corolla purchased in 1992 for \$1.2 million depreciated in value yearly by 1.5%. What is the car's value in 2008?
This is a problem of depreciation, so we subtract the rate of change of value.
V = I x (1 - r)^n
= \$1,200,000 x (1-0.015)^16
= \$942,238.69
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What Would The Net Force Be // arena-sever.ru
# What is Net Force? - Definition, Magnitude & Equations.
A net force is defined as the sum of all the forces acting on an object. The equation below is the sum of N forces acting on an object. The equation below is the sum of N forces acting on an object. Since forces act on an object in opposite directions, we calculate the net force, which is the the vector sum of all forces acting on an object. The net force is the difference between the two forces, just like your net pay is the difference between your gross pay and the deductions that come out of your paycheck. Add these forces together.Find the sum of these relevant forces by applying their force, in Newtons, to your diagram. Forces moving toward the right in the free-body diagram are considered positive forces, and those moving toward the left are negative. Add the forces together. The result is.
Net force is when two or more forces act on an object at the same time and the forces combine. Balanced forces are forces on an object that are equal in size and opposite in direction. Mar 21, 2017 · The force is given by the formula pAv² P density of the liquid A- area of the hole V- velocity of efflux Velocity of efflux is V=√2gh So, F=pAv² = pA2gh =2pAgh Since the two holes are in opposite directions the force will be subtracted. Net for.
Q. You and your friend, are playing tug of war. Your friend pulls with a force of 55 N, to the rights. You pull with a force of 65 N, to the left. What is the net force on the rope? Sep 26, 2019 · Net force is the total amount of force acting on an object when you take into account both magnitude and direction. An object with a net force of zero is stationary. An unbalanced force, or net force of a magnitude greater than or less than zero, leads to acceleration of the object. Net force synonyms, Net force pronunciation, Net force translation, English dictionary definition of Net force. a force which is the result of two or more forces acting conjointly, or a motion which is the result of two or more motions combined.
The net force of an object if balanced is zero, this is called static equilibrium, if the net force is unbalanced then it's in motion. Oct 20, 2009 · Net force is the sum of all forces that act on an object. An unbalanced force means that net force is not equal to zero. If that is the case, the object will accelerate. Net force is the sum of all forces that act on an object. Oct 18, 2014 · Think about the forces that you get from theand - stationary charges. What will the forces be on the positive test charge? Where will it feel zero net force? So once you know about where it will be, then you can more intuitively write out the sum of the 2 force equations = 0. Definition of net force in thedictionary. Meaning of net force. What does net force mean? Information and translations of net force in the most comprehensive dictionary definitions resource on the web. TheForce.Net, Your Daily Dose of Star Wars, get up to the minute updates on Star Wars Movies, Star Wars Television, Star Wars Literature, Star Wars Games, Star Wars Fandom, and so much more!
• The Net Force takes into account both how strong the forces are and in what direction they act • The Net Force determines the acceleration of the object net force Example: Net force = 0 Net force = 0 • An object can have many forces acting on it at the same time. • If all the forces oppose each other exactly then the net force = 0 and the object will.
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# Find the domain and range of each of the relations given below:
Question:
Find the domain and range of each of the relations given below:
(i) $R=\{(-1,1),(1,1),(-2,4),(2,4),(2,4),(3,9)\}$
(ii) $\mathrm{R}=\left\{\left(\mathrm{x}, \frac{1}{\mathrm{x}}\right): \mathrm{x}\right.$ is an interger, $\left.0<\mathrm{x}<5\right\}$
(iii) $R=\{(x, y): x+2 y=8$ and $x, y \in N\}$
(iv) $R=\{(x, y),: y=|x-1|, x \in Z$ and $|x| \leq 3\}$
Solution:
(i) Given: $R=\{(-1,1),(1,1),(-2,4),(2,4),(2,4),(3,9)\}$
$\operatorname{Dom}(R)=\{x:(x, y) \in R\}=\{-2,-1,1,2,3\}$
Range $(R)=\{y:(x, y) \in R\}=\{1,4,9\}$
(ii) Given:
$R=\left\{\left(x, \frac{1}{x}\right): x\right.$ is an interger, $\left.0 That means,$R=\left\{(1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\right\}\operatorname{Dom}(R)=\left\{x:(x, y)^{\in} R\right\}=\{1,2,3,4\}$Range$(R)=\left\{y:(x, y) \in_{R}\right\}=\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$(iii) Given:$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}+2 \mathrm{y}=8$and$\mathrm{x}, \mathrm{y} \in \mathrm{N}\}$That means,$R=\{(2,3),(4,2),(6,1)\}\operatorname{Dom}(R)=\{x:(x, y) \in R\}=\{2,4,6\}$Range$(R)=\left\{y:(x, y)^{\in} R\right\}=\{1,2,3\}$(iv) Given:$R=\{(x, y): y=|x-1|, x \in Z$and$|x| \leq 3\}\operatorname{Dom}(R)=\left\{x:(x, y)^{\in} R\right\}=\{-3,-2,-1,0,1,2,3\}$Range$(R)=\left\{y:(x, y)^{\in} R\right\}=\{0,1,2,3,4\}\$
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# Precalculus with Limits (2010)
## Educators
### Problem 1
A rectangular array of real numbers that can be used to solve a system of linear equations is called a ________.
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### Problem 2
A matrix is ________ if the number of rows equals the number of columns.
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### Problem 3
For a square matrix, the entries $a_{11}, a_{22}, a_{33}, \ldots, a_{n n}$ are the________,_______entries.
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### Problem 4
A matrix with only one row is called a ________ matrix, and a matrix with only one column is called a ________ matrix.
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### Problem 5
The matrix derived from a system of linear equations is called the ________ matrix of the system.
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### Problem 6
The matrix derived from the coefficients of a system of linear equations is called the ________ matrix of the system.
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### Problem 7
Two matrices are called ________ if one of the matrices can be obtained from the other by a sequence of elementary row operations.
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### Problem 8
A matrix in row-echelon form is in ________ ________ ________ if every column that has a leading 1 has zeros in every position above and below its leading 1.
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### Problem 9
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{ll}{7} & {0}\end{array}\right]$$
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### Problem 10
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{llll}{5} & {-3} & {8} & {7}\end{array}\right]$$
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### Problem 11
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{r}{2} \\ {36} \\ {3}\end{array}\right]$$
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### Problem 12
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{rrrr}{-3} & {7} & {15} & {0} \\ {0} & {0} & {3} & {3} \\ {1} & {1} & {6} & {7}\end{array}\right]$$
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### Problem 13
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{cc}{33} & {45} \\ {-9} & {20}\end{array}\right]$$
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### Problem 14
In Exercises 9–14, determine the order of the matrix.
$$\left[\begin{array}{rrr}{-7} & {6} & {4} \\ {0} & {-5} & {1}\end{array}\right]$$
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### Problem 15
In Exercises 15–20, write the augmented matrix for the system of linear equations.
$$\left\{\begin{array}{l}{4 x-3 y=-5} \\ {-x+3 y=12}\end{array}\right.$$
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### Problem 16
In Exercises 15–20, write the augmented matrix for the system of linear equations.
$$\left\{\begin{array}{l}{7 x+4 y=22} \\ {5 x-9 y=15}\end{array}\right.$$
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### Problem 17
In Exercises 15–20, write the augmented matrix for the system of linear equations.
\left\{\begin{aligned} x+10 y-2 z &=2 \\ 5 x-3 y+4 z &=0 \\ 2 x+y &=6 \end{aligned}\right.
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### Problem 18
In Exercises 15–20, write the augmented matrix for the system of linear equations.
$$\left\{\begin{array}{rr}{-x-8 y+5 z=} & {8} \\ {-7 x-15 z=}& {-38} \\ {3 x-y+8 z=} & {20}\end{array}\right.$$
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### Problem 19
In Exercises 15–20, write the augmented matrix for the system of linear equations.
\left\{\begin{aligned} 7 x-5 y+z =13 \\ 19 x \quad -8 z=10 \end{aligned}\right.
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### Problem 20
In Exercises 15–20, write the augmented matrix for the system of linear equations.
\left\{\begin{aligned} 9 x+2 y-3 z &=20 \\-25 y+11 z &=-5 \end{aligned}\right.
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### Problem 21
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{rrrr}{1} & {2} & {\vdots} & {7} \\ {2} & {-3} & {\vdots} & {4}\end{array}\right]$$
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### Problem 22
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{rrrr}{7} & {-5} & {\vdots} & {0} \\ {8} & {3} & {\vdots} & {-2}\end{array}\right]$$
Check back soon!
### Problem 23
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{ccccc}{2} & {0} & {5} & {\vdots} & {-12} \\ {0} & {1} & {-2} & {\vdots} & {7} \\ {6} & {3} & {0} & {\vdots} & {2}\end{array}\right]$$
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### Problem 24
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{rrrrr}{4} & {-5} & {-1} & {\vdots} & {18} \\ {-11} & {0} & {6} & {\vdots} & {25} \\ {3} & {8} & {0} & {\vdots} & {-29}\end{array}\right]$$
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### Problem 25
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{rrrrrr}{9} & {12} & {3} & {0} & {\vdots} & {0} \\ {-2} & {18} & {5} & {2} & {\vdots} & {10} \\ {1} & {7} & {-8} & {0} & {\vdots} & {-4} \\ {3} & {0} & {2} & {0} & {\vdots} & {-10}\end{array}\right]$$
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### Problem 26
In Exercises $21-26,$ write the system of linear equations represented by the augmented matrix. (Use variables $x, y, z$ and $w,$ if applicable.)
$$\left[\begin{array}{rrrrrr}{6} & {2} & {-1} & {-5} & {\vdots} & {-25} \\ {-1} & {0} & {7} & {3} & {\vdots} & {7} \\ {4} & {-1} & {-10} & {6} & {\vdots} & {23} \\ {0} & {8} & {1} & {-11} & {\vdots} & {-21}\end{array}\right]$$
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### Problem 27
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\begin{array}{rrr}{\left[\begin{array}{ccc}{1} & {4} & {3} \\ {2} & {10} & {5}\end{array}\right]} \\ {\left[\begin{array}{rr}{1} & {4} & {3} \\ {0} & {} & {-1} \end{array}\right]}\end{array}$$
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### Problem 28
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrr}{3} & {6} & {8} \\ {4} & {-3} & {6}\end{array}\right]\\\left[\begin{array}{rrr}{1} & {} & {\frac{8}{3}} \\ {4} & {-3} & {6}\end{array}\right]$$
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### Problem 29
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrr}{1} & {1} & {1} \\ {5} & {-2} & {4}\end{array}\right]\\\left[\begin{array}{rr}{1} & {1} & {1} \\ {0} &{}& {-1}\end{array}\right]$$
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### Problem 30
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrr}{-3} & {3} & {12} \\ {18} & {-8} & {4}\end{array}\right]\\\left[\begin{array}{rrr}{1} & {-1} & {} \\ {18} & {-8} & {4}\end{array}\right]$$
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### Problem 31
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrrr}{1} & {5} & {4} & {-1} \\ {0} & {1} & {-2} & {2} \\ {0} & {0} & {1} & {-7}\end{array}\right]\\\left[\begin{array}{cccr}{1} & {0} \\ {0} & {1} & {-2} & {2} \\ {0} & {0} & {1} & {-7}\end{array}\right]$$
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### Problem 32
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrrr}{1} & {0} & {6} & {1} \\ {0} & {-1} & {0} & {7} \\ {0} & {0} & {-1} & {3}\end{array}\right]\\\left[\begin{array}{cccc}{1} & {0} & {6} & {1} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$$
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### Problem 33
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrrr}{1} & {1} & {4} & {-1} \\ {3} & {8} & {10} & {3} \\ {-2} & {1} & {12} & {6}\end{array}\right]\\\left[\begin{array}{cccc}{1} & {1} & {4} & {-1} \\ {0} & {5} & {} & {} \\ {0} & {3} & {} & {}\end{array}\right]\\\left[\begin{array}{rrrr}{1} & {1} & {4} & {-1} \\ {0} & {1} & {-\frac{2}{5}} & {\frac{6}{5}} \\ {0} & {3} & {}\end{array}\right]$$
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### Problem 34
In Exercises 27–34, fill in the blank(s) using elementary row operations to form a row-equivalent matrix.
$$\left[\begin{array}{rrrr}{2} & {4} & {8} & {3} \\ {1} & {-1} & {-3} & {2} \\ {2} & {6} & {4} & {9}\end{array}\right]\\\left[\begin{array}{rrrr}{1} \\ {1} & {-1} & {-3} & {2} \\ {2} & {6} & {4} & {9}\end{array}\right]\\\left[\begin{array}{cccc}{1} & {2} & {4} & {\frac{3}{2}} \\ {0} & {} & {-7} & {\frac{1}{2}} \\ {0} & {2} & {}\end{array}\right]$$
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### Problem 35
In Exercises 35–38, identify the elementary row operation(s) being performed to obtain the new row-equivalent matrix.
$$\begin{array}{ll}{\text { Original Matrix }} & {\text { New Row-Equivalent Matrix }} \\ {\left[\begin{array}{rrr}{-2} & {5} & {1} \\ {3} & {-1} & {-8}\end{array}\right]} & {\left[\begin{array}{rrr}{13} & {0} & {-39} \\ {3} & {-1} & {-8}\end{array}\right]}\end{array}$$
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### Problem 36
In Exercises 35–38, identify the elementary row operation(s) being performed to obtain the new row-equivalent matrix.
$$\begin{array}{ll}{\text { Original Matrix }} & {\text { New Row-Equivalent Matrix }} \\ {\left[\begin{array}{rrr}{3} & {-1} & {-4} \\ {-4} & {3} & {7}\end{array}\right]} & {\left[\begin{array}{rrr}{3} & {-1} & {-4} \\ {5} & {0} & {-5}\end{array}\right]}\end{array}$$
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### Problem 37
In Exercises 35–38, identify the elementary row operation(s) being performed to obtain the new row-equivalent matrix.
$$\begin{array}{ll}{\text { Original Matrix }} & {\text { New Row-Equivalent Matrix }} \\ {\left[\begin{array}{rrr}{0} & {-1} & {-5} & {5} \\ {-1} & {3} & {-7} & {6} \\ {4} & {-5} & {1} & {3}\end{array}\right]} & {\left[\begin{array}{rrr}{-1} & {3} & {-7} & {6} \\ {0} & {-1} & {-5} & {5} \\ {0} & {7} & {-27} & {27}\end{array}\right]}\end{array}$$
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### Problem 38
In Exercises 35–38, identify the elementary row operation(s) being performed to obtain the new row-equivalent matrix.
$$\begin{array}{ll}{\text {Original Matrix}} & {\text {New Row-Equivalent Matrix}} \\ {\left[\begin{array}{rrrr}{-1} & {-2} & {3} & {-2} \\ {2} & {-5} & {1} & {-7} \\ {5} & {4} & {-7} & {6}\end{array}\right]} & {\left[\begin{array}{rrrr}{-1} & {-2} & {3} & {-2} \\ {0} & {-9} & {7} & {-11} \\ {0} & {-6} & {8} & {-4}\end{array}\right]}\end{array}$$
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### Problem 39
Perform the sequence of row operations on the matrix. What did the operations accomplish?
$$\left[\begin{array}{rrr}{1} & {2} & {3} \\ {2} & {-1} & {-4} \\ {3} & {1} & {-1}\end{array}\right]$$
(a) Add $-2$ times $R_{1}$ to $R_{2}.$
(b) Add $-3$ times $R_{1}$ to $R_{3}.$
(c) Add $-1$ times $R_{2}$ to $R_{3}.$
(d) Multiply $R_{2}$ by $-\frac{1}{5.}$
(e) Add $-2$ times $R_{2}$ to $R_{1}.$
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### Problem 40
Perform the sequence of row operations on the matrix. What did the operations accomplish?
$$\left[\begin{array}{rr}{7} & {1} \\ {0} & {2} \\ {-3} & {4} \\ {4} & {1}\end{array}\right]$$
(a) Add $R_{3}$ to $R_{4}.$
(b) Interchange $R_{1}$ and $R_{4}$ .
(c) Add 3 times $R_{1}$ to $R_{3}.$
(d) Add $-7$ times $R_{1}$ to $R_{4}.$
(e) Multiply $R_{2}$ by $\frac{1}{2}.$
(f) Add the appropriate multiples of $R_{2}$ to $R_{1}, R_{3},$ and $R_{4}$ .
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### Problem 41
In Exercises 41–44, determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form.
$$\left[\begin{array}{llll}{1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {5} \\ {0} & {0} & {0} & {0}\end{array}\right]$$
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### Problem 42
In Exercises 41–44, determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form.
$$\left[\begin{array}{llll}{1} & {3} & {0} & {0} \\ {0} & {0} & {1} & {8} \\ {0} & {0} & {0} & {0}\end{array}\right]$$
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### Problem 43
In Exercises 41–44, determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form.
$$\left[\begin{array}{llll}{1} & {0} & {0} & {1} \\ {0} & {1} & {0} & {-1} \\ {0} & {0} & {0} & {2}\end{array}\right]$$
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### Problem 44
In Exercises 41–44, determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form.
$$\left[\begin{array}{llll}{1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {2} \\ {0} & {0} & {1} & {0}\end{array}\right]$$
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### Problem 45
In Exercises 45–48, write the matrix in row-echelon form. (Remember that the row-echelon form of a matrix is not unique.)
$$\left[\begin{array}{rrrr}{1} & {1} & {0} & {5} \\ {-2} & {-1} & {2} & {-10} \\ {3} & {6} & {7} & {14}\end{array}\right]$$
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### Problem 46
In Exercises 45–48, write the matrix in row-echelon form. (Remember that the row-echelon form of a matrix is not unique.)
$$\left[\begin{array}{rrrr}{1} & {2} & {-1} & {3} \\ {3} & {7} & {-5} & {14} \\ {-2} & {-1} & {-3} & {8}\end{array}\right]$$
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### Problem 47
In Exercises 45–48, write the matrix in row-echelon form. (Remember that the row-echelon form of a matrix is not unique.)
$$\left[\begin{array}{rrrr}{1} & {-1} & {-1} & {1} \\ {5} & {-4} & {1} & {8} \\ {-6} & {8} & {18} & {0}\end{array}\right]$$
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### Problem 48
In Exercises 45–48, write the matrix in row-echelon form. (Remember that the row-echelon form of a matrix is not unique.)
$$\left[\begin{array}{rrrr}{1} & {-3} & {0} & {-7} \\ {-3} & {10} & {1} & {23} \\ {4} & {-10} & {2} & {-24}\end{array}\right]$$
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### Problem 49
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{rrr}{3} & {3} & {3} \\ {-1} & {0} & {-4} \\ {2} & {4} & {-2}\end{array}\right]$$
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### Problem 50
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{ccc}{1} & {3} & {2} \\ {5} & {15} & {9} \\ {2} & {6} & {10}\end{array}\right]$$
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### Problem 51
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{rrrr}{1} & {2} & {3} & {-5} \\ {1} & {2} & {4} & {-9} \\ {-2} & {-4} & {-4} & {3} \\ {4} & {8} & {11} & {-14}\end{array}\right]$$
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### Problem 52
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{rrrr}{-2} & {3} & {-1} & {-2} \\ {4} & {-2} & {5} & {8} \\ {1} & {5} & {-2} & {0} \\ {3} & {8} & {-10} & {-30}\end{array}\right]$$
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### Problem 53
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{rrrr}{-3} & {5} & {1} & {12} \\ {1} & {-1} & {1} & {4}\end{array}\right]$$
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### Problem 54
In Exercises 49–54, use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form.
$$\left[\begin{array}{rrrr}{5} & {1} & {2} & {4} \\ {-1} & {5} & {10} & {-32}\end{array}\right]$$
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### Problem 55
In Exercises $55-58$ , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables $x, y,$ and $z,$ if applicable.)
$$\left[\begin{array}{rrrr}{1} & {-2} & {\vdots} & {4} \\ {0} & {1} & {\vdots} & {-3}\end{array}\right]$$
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### Problem 56
In Exercises $55-58$ , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables $x, y,$ and $z,$ if applicable.)
$$\left[\begin{array}{llll}{1} & {5} & {\vdots} & {0} \\ {0} & {1} & {\vdots} & {-1}\end{array}\right]$$
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### Problem 57
In Exercises $55-58$ , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables $x, y,$ and $z,$ if applicable.)
$$\left[\begin{array}{rrrrr}{1} & {-1} & {2} & {\vdots} & {4} \\ {0} & {1} & {-1} & {\vdots} & {2} \\ {0} & {0} & {1} & {\vdots} & {-2}\end{array}\right]$$
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### Problem 58
In Exercises $55-58$ , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables $x, y,$ and $z,$ if applicable.)
$$\left[\begin{array}{rrrrr}{1} & {2} & {-2} & {\vdots} & {-1} \\ {0} & {1} & {1} & {\vdots} & {9} \\ {0} & {0} & {1} & {\vdots} & {-3}\end{array}\right]$$
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### Problem 59
In Exercises $59-62,$ an augmented matrix that represents a system of linear equations (in variables $x, y,$ and $z,$ if applicable) has been reduced using Gauss.Jordan elimination. Write the solution represented by the augmented matrix.
$$\left[\begin{array}{llll}{1} & {0} & {\vdots} & {3} \\ {0} & {1} & {\vdots} & {-4}\end{array}\right]$$
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### Problem 60
In Exercises $59-62,$ an augmented matrix that represents a system of linear equations (in variables $x, y,$ and $z,$ if applicable) has been reduced using Gauss.Jordan elimination. Write the solution represented by the augmented matrix.
$$\left[\begin{array}{llll}{1} & {0} & {\vdots} & {-6} \\ {0} & {1} & {\vdots} & {10}\end{array}\right]$$
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### Problem 61
In Exercises $59-62,$ an augmented matrix that represents a system of linear equations (in variables $x, y,$ and $z,$ if applicable) has been reduced using Gauss.Jordan elimination. Write the solution represented by the augmented matrix.
$$\left[\begin{array}{ccccc}{1} & {0} & {0} & {\vdots} & {-4} \\ {0} & {1} & {0} & {\vdots} & {-10} \\ {0} & {0} & {1} & {\vdots} & {4}\end{array}\right]$$
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### Problem 62
In Exercises $59-62,$ an augmented matrix that represents a system of linear equations (in variables $x, y,$ and $z,$ if applicable) has been reduced using Gauss.Jordan elimination. Write the solution represented by the augmented matrix.
$$\left[\begin{array}{ccccc}{1} & {0} & {0} & {\vdots} & {5} \\ {0} & {1} & {0} & {\vdots} & {-3} \\ {0} & {0} & {1} & {\vdots} & {0}\end{array}\right]$$
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### Problem 63
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y &=7 \\ 2 x+y &=8 \end{aligned}\right.
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### Problem 64
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\left\{\begin{array}{l}{2 x+6 y=16} \\ {2 x+3 y=7}\end{array}\right.$$
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### Problem 65
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 3 x-2 y &=-27 \\ x+3 y &=13 \end{aligned}\right.
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### Problem 66
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\left\{\begin{array}{rr}{-x+y=} & {4} \\ {2 x-4 y=} & {-34}\end{array}\right.$$
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### Problem 67
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned}-2 x+6 y &=-22 \\ x+2 y &=-9 \end{aligned}\right.
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### Problem 68
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 5 x-5 y &=-5 \\-2 x-3 y &=7 \end{aligned}\right.
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### Problem 69
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\left\{\begin{array}{l}{8 x-4 y=7} \\ {5 x+2 y=1}\end{array}\right.$$
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### Problem 70
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x-3 y &=5 \\-2 x+6 y &=-10 \end{aligned}\right.
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### Problem 71
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x -3 z=&-2 \\ 3 x+y-2 z &=5 \\ 2 x+2 y+z &=4 \end{aligned}\right.
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### Problem 72
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 2 x-y+3 z &=24 \\ 2 y-z &=14 \\ {7 x-5 y}\qquad &=6 \end{aligned}\right.
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### Problem 73
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned}-x+y-z=-14 \\ 2 x-y+z= 21 \\ 3 x+2 y+z= 19 \end{aligned}\right.
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### Problem 74
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 2 x+2 y-z &=2 \\ x-3 y+z &=-28 \\-x+y\qquad &=14 \end{aligned}\right.
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### Problem 75
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y-3 z=&-28 \\ 4 y+2 z=& 0 \\-x+y-z=&-5 \end{aligned}\right.
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### Problem 76
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 3 x-2 y+z=& 15 \\-x+y+2 z=&-10 \\ x-y-4 z=& 14 \end{aligned}\right.
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### Problem 77
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y &=0 \\-x-y &=0 \end{aligned}\right.
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### Problem 78
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y &=0 \\ 2 x+4 y &=0 \end{aligned}\right.
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### Problem 79
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y+z &=8 \\ 3 x+7 y+6 z &=26 \end{aligned}\right.
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### Problem 80
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+y+4 z &=5 \\ 2 x+y-z &=9 \end{aligned}\right.
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### Problem 81
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned}-x+y =-22 \\ 3 x+4 y =4 \\ 4 x-8 y =32 \end{aligned}\right.
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### Problem 82
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x+2 y &=0 \\ x+y &=6 \\ 3 x-2 y &=8 \end{aligned}\right.
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### Problem 83
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} 3 x+2 y-z+w &=0 \\ x-y+4 z+2 w &=25 \\-2 x+y+2 z-w &=2 \\ x+y+z+w &=6 \end{aligned}\right.
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### Problem 84
In Exercises 63–84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\left\{\begin{aligned} x-4 y+3 z-2 w &=9 \\ 3 x-2 y+z-4 w &=-13 \\-4 x+3 y-2 z+w &=-4 \\-2 x+y-4 z+3 w &=-10 \end{aligned}\right.
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### Problem 85
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} 3 x+3 y+12 z=& 6 \\ x+y+4 z=& 2 \\ 2 x+5 y+20 z=& 10 \\-x+2 y+8 z=& 4 \end{aligned}\right.
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### Problem 86
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} 2 x\quad+10 y+2 z= 6 \\ x\quad+5 y+2 z= 6 \\ x\qquad+5 y+z= 3 \\-3 x-15 y-3 z=-9 \end{aligned}\right.
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### Problem 87
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} 2 x+y-z+2 w =-6 \\ 3 x+4 y \quad\quad\quad+w =1 \\ x+5 y+2 z+6 w =-3 \\ 5 x+2 y-z-w =3 \end{aligned}\right.
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### Problem 88
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} x+2 y+2 z+4 w &=11 \\ 3 x+6 y+5 z+12 w &=30 \\ x+3 y-3 z+2 w &=-5 \\ 6 x-y-z+w &=-9 \end{aligned}\right.
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### Problem 89
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} x+y+z+w &=0 \\ 2 x+3 y+z-2 w &=0\\ 3 x+5 y\quad\quad+z&=0 \end{aligned}\right.
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### Problem 90
In Exercises 85–90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system.
\left\{\begin{aligned} x+2 y+z+3 w =0 \\ x-y \qquad +w=0 \\ y-z+2 w =0 \end{aligned}\right.
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### Problem 91
In Exercises 91–94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices.
$$(a) \left\{\begin{array}{r}{x-2 y+z=-6} \\ {y-5 z=16} \\ {z=-3}\end{array}\right. \quad(b) \left\{\begin{array}{r}{x+y-2 z=6} \\ {y+3 z=-8} \\ {z=-3}\end{array}\right.$$
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### Problem 92
In Exercises 91–94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices.
$$(a) \left\{\begin{array}{r}{x-3 y+4 z=-11} \\ {y-z=-4} \\ {z=2}\end{array}\right. \quad(b) \left\{\begin{array}{r} {x+4 y =-11} \\ {y+3 z =4} \\ {z =2} \end{array}\right.$$
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### Problem 93
In Exercises 91–94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices.
$$(a)\left\{\begin{array}{r}{x-4 y+5 z=27} \\ {y-7 z=-54} \\ {z=8}\end{array}\right. \quad(b) \left\{\begin{array}{r}{x-6 y+z=15} \\ {y+5 z=42} \\ {z=8}\end{array}\right.$$
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### Problem 94
In Exercises 91–94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices.
$$(a)\left\{\begin{array}{r}{x+3 y-z=19} \\ {y+6 z=-18} \\ {z=-4}\end{array}\right.\quad(b) \left\{\begin{array}{r}{x-y+3 z=-15} \\ {y-2 z=14} \\ {z=-4}\end{array}\right.$$
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### Problem 95
In Exercises $95-98,$ use a system of equations to find the quadratic function $f(x)=a x^{2}+b x+c$ that satisfies the equations. Solve the system using matrices.
$$f(1)=1, f(2)=-1, f(3)=-5$$
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### Problem 96
In Exercises $95-98,$ use a system of equations to find the quadratic function $f(x)=a x^{2}+b x+c$ that satisfies the equations. Solve the system using matrices.
$$f(1)=2, f(2)=9, f(3)=20$$
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### Problem 97
In Exercises $95-98,$ use a system of equations to find the quadratic function $f(x)=a x^{2}+b x+c$ that satisfies the equations. Solve the system using matrices.
$$f(-2)=-15, f(-1)=7, f(1)=-3$$
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### Problem 98
In Exercises $95-98,$ use a system of equations to find the quadratic function $f(x)=a x^{2}+b x+c$ that satisfies the equations. Solve the system using matrices.
$$f(-2)=-3, f(1)=-3, f(2)=-11$$
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### Problem 99
In Exercises $99-102,$ use a system of equations to find the cubic function $f(x)=a x^{3}+b x^{2}+c x+d$ that satisfies the equations. Solve the system using matrices.
$$\begin{array}{l}{f(-1)=-5} \\ {f(1)=-1} \\ {f(2)=1} \\ {f(3)=11}\end{array}$$
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### Problem 100
In Exercises $99-102,$ use a system of equations to find the cubic function $f(x)=a x^{3}+b x^{2}+c x+d$ that satisfies the equations. Solve the system using matrices.
$$\begin{array}{l}{f(-1)=4} \\ {f(1)=4} \\ {f(2)=16} \\ {f(3)=44}\end{array}$$
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### Problem 101
In Exercises $99-102,$ use a system of equations to find the cubic function $f(x)=a x^{3}+b x^{2}+c x+d$ that satisfies the equations. Solve the system using matrices.
$$\begin{array}{l}{f(-2)=-7} \\ {f(-1)=2} \\ {f(1)=-4} \\ {f(2)=-7}\end{array}$$
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### Problem 102
In Exercises $99-102,$ use a system of equations to find the cubic function $f(x)=a x^{3}+b x^{2}+c x+d$ that satisfies the equations. Solve the system using matrices.
$$\begin{array}{l}{f(-2)=-17} \\ {f(-1)=-5} \\ {f(1)=1} \\ {f(2)=7}\end{array}$$
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Use the system
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# Recent problems solved by 'ntnk'
logarithm/592472: evaluate the limit! Lim (e^4t) -1/t x->0 i thought e^0=1 t=0 so 0/0 =0 but it said it was wrong i also tried infinity but it was wrong1 solutions Answer 376072 by ntnk(54) on 2012-03-28 15:40:50 (Show Source): You can put this solution on YOUR website!I think your parentheses are misplaced, so based on the work you have shown, my guess is that your question is the following: http://latex.codecogs.com/gif.latex?%5Clim_%7Bt%5Cto%200%7D%5Cfrac%7Be%5E%7B4t%7D-1%7D%7Bt%7D First I would like to point out and emphasize the fact that 0 / 0 is not the same as 0. (However, you are correct that .) The easiest way to solve this is by using L'Hopital's rule. We can take the derivative of the numerator and the derivative of the denominator. The limit above is therefore given by the following http://latex.codecogs.com/gif.latex?%5Clim_%7Bt%5Cto%200%7D%5Cfrac%7B4e%5E%7B4t%7D%7D%7B1%7D Now we can plug in to find out that the limit is 4.
Triangles/592346: a right triangle gas an angle whose measure is 35. what is the measurment of the third angle?1 solutions Answer 376062 by ntnk(54) on 2012-03-28 15:20:07 (Show Source): You can put this solution on YOUR website!There are two very important facts that you must know to do these types of questions. 1. In ANY triangle, the measurements of the three angles must add up to 180. 2. In a RIGHT triangle, one of the angles measures 90 degrees. We have a right triangle with an angle measure of 35. The fact that the triangle is right tells us that one of the angles is 90. We need to find the measure of the third angle, which we will call x. Since the measures of the angles must add up to 180, we have 35 + 90 + x = 180 and we must solve for x. Simplifying a bit on the left hand side gives 125 + x = 180 Now subtract 125 from both sides to get x = 180 - 125 = 55 so the third angle is 55 degrees. I hope this helps!
Age_Word_Problems/592782: A man is twice as old as his son.10years ago he was three times as old.how old will his son be in 5years time?1 solutions Answer 376061 by ntnk(54) on 2012-03-28 15:16:09 (Show Source): You can put this solution on YOUR website!Let's turn the statements you give above into equations. Let's call the man's age M and the son's age S. (1) A man is twice as old as his son: M = 2 * S. Since the man's age today is M and the son's age is S, this means this means that 10 years ago, the man's age was M-10 and the son's age was S-10. (2) 10years ago he was three times as old: M-10 = 3 * (S-10). Now we can solve for the man's age an his son's age by substitution. We can plug the first equation (M=2*S) into the second one so that M - 10 = 3 * (S-10) becomes 2*S - 10 = 3 * (S-10). Now we can solve for S. First use the distributive property on the right-hand side to get 2*S - 10 = 3*S - 30 Now add 30 to both sides to get 2*S + 20 = 3*S and subtract 2*S from both sides to get 20 = S. This is the age of the son today. The question asks how old the son will be in 5 years, so the answer must be 20+5 = 25 years old. Now let's check. We think the son is 20 years old today. Since the man is twice as old, the man must be 40. Ten years ago, the man was 30 and the son was 10. Therefore the man was 3 times as old as the son ten years ago, so everything checks out. I hope this helps!
Age_Word_Problems/592692: My Age: 39 Elijah: 12 At some point , my age will be 3 times Elijah's age. Is this correct: 3x= 3(x-27) How do I solve? Thanks!1 solutions Answer 376020 by ntnk(54) on 2012-03-28 10:26:44 (Show Source): You can put this solution on YOUR website!The equation you have written is not quite correct. Let's see how to solve it before getting to the age question. First we use the distributive property on the right-hand side to get Then we subtract 3x from both sides to get which is obviously false, so that means the equation above does not have a solution. Let's think about what the problem is telling us. Right now you are 39 and Elijah is 12. Next year, you will be 40 and Elijah will be 13. The year after that, you will be 41 and Elijah will be 14. More generally, in X years, your age will be 39 + X and Elijah's age will be 12 + X. (For example, if we plug in X=1 or X=2, we get the statements from the previous paragraph.) We want to know when your age will be three times Elijah's age. So we take three times Elijah's age: 3 * (12 + X) and set it equal to your age: (39 + X). This gives the equation We use the distributive property on the left hand side to get Now subtract X from both sides to get Then subtract 36 from both sides to get and finally divide by 2 to get Now we can check the final answer. In 1.5 years, you will be 39+1.5 = 40.5 years old, and Elijah will be 12+1.5 = 13.5 years old. Indeed, in 1.5 years you will be 3 times older than Elijah since 13.5 * 3 = 40.5
Age_Word_Problems/313306: A man is three times as old as his son, and his daughter is four years younger than his son. If the combined ages of all three is 81 years, find the present age of the father. Ps: Thank you, please demonstrate the work progress so I will know how to do the question next time please and thank you :D1 solutions Answer 224040 by ntnk(54) on 2010-06-10 23:23:11 (Show Source): You can put this solution on YOUR website!So for these you have to turn the statements in the word problem into equations. A man is three times as old as his son: his daughter is four years younger than his son: combined ages of all three is 81 years: Now you can just substitute: Add 4 to both sides to get: Divide by 5 on both sides to get: Now check: and , so , so we do indeed get the combined age as 81, which means our son's age of 17 is correct! I hope that helps! -ntnk
Matrices-and-determiminant/150001: I need some help on this equation. I need to use the gauss-jordan method solving by matrices. thanks 4x - 2y = 3 -2x + 3y = 11 solutions Answer 110104 by ntnk(54) on 2008-07-28 13:01:03 (Show Source): You can put this solution on YOUR website!I don't know how to make matrices here, but I'll try. [ 4 -2 | 3 ] [-2 3 | 1 ] <-- That's your augmented matrix representing the system of equations. Now row one column one has to be a pivot, so row two column one must become zero. Add 1/2*row one to row two. [4 -2 | 3] [0 2 | 5/2] So you have it in the proper inverted triangle form now. Equate the components. Substitute this back into: So there is the solution: (11/8, 5,4) -ntnk
Quadratic_Equations/131612: y=-3x^2-6x+41 solutions Answer 96242 by ntnk(54) on 2008-03-12 21:27:18 (Show Source):
Linear-systems/131815: This question is from textbook Solving Linear Equations: Use linear combinations to solve the system of linear equations. t + r = 1 2r - t = 21 solutions Answer 96241 by ntnk(54) on 2008-03-12 21:22:59 (Show Source): You can put this solution on YOUR website!In any equation, solve for one variable in terms of the other and substitute into the other equation. If , then . If and , then Distributing: Combining like terms: Subtracting 2 from both sides: Dividing both sides by -3: . Substitute this back into either of the original equations to find 'r'. Therefore, , so .
Miscellaneous_Word_Problems/131825: The problem reads: A rectangle is 5 feet longer than it is wide. The perimeter of the rectangle is 34 feet. What is the length of the rectangle? any help would be appreiciated1 solutions Answer 96240 by ntnk(54) on 2008-03-12 21:19:25 (Show Source): You can put this solution on YOUR website!Start by actually writing out every equation that you can deduce from the problem statement. "A rectangle is 5 feet longer than it is wide." "The perimeter of the rectangle is 34 feet." What is the length of the rectangle? L = ? Substitute the first equation into the second equation, obtaining: Distributing: Combining like terms: Subtracting 10 from both sides: Dividing both sides by 4: If W = 6, then L = 11 since as stated earlier. -ntnk
Sequences-and-series/65739: Please help me figure out this word problem. Considering yourself, your parents, your grandparents, and so on, back to your grandparents with the workd "great" used in front 40 times. What is the total number of people that you are considering. Thanks for your help.1 solutions Answer 46451 by ntnk(54) on 2006-12-28 16:49:17 (Show Source): You can put this solution on YOUR website!Each level is 2^x (because each person has two parents and so forth.) Therefore, YOU are 2^0 = 1 (person); PARENTS are 2^1= 2 (people); GRANDPARENTS are 2^2= 4 (people); GREAT GRANDPARENTS are 2^3=8; GREAT GREAT GRANDPARENTS are 2^4=16; GREAT GREAT GREAT GRANDPARENTS are 2^5=32; and so forth. But notice that three "greats" implies 2^5. This means that x "greats" implies x+2. Therefore, 40 "greats" implies 2^42, which is the total number of people that you are considering. Also, the question is a bit unclear as to if you want to know the total number of people including yourself and parents, and grandparents, and so forth until great*40 grandparents, so if you include ALL of those people, then it will be 2^42+2^41+2^40+...+2^20+2^19+...+2^3+2^2+2^1+2^0 and so forth. and so forth ntnk
logarithm/64600: 1. Rewrite in logarithmic form. Any help would be much appreciated1 solutions Answer 46448 by ntnk(54) on 2006-12-28 16:39:45 (Show Source): You can put this solution on YOUR website!If a=b^c, then log(base b)a=c, therefore, if , then log(base (1/2))8=-3.
Rational-functions/64647: This question is from textbook McDougal Littell Algebra 2 how do I graph the given function y=f(x-4)+11 solutions Answer 46446 by ntnk(54) on 2006-12-28 16:32:52 (Show Source): You can put this solution on YOUR website! There is a horizontal shift to the right by four and a vertical shift up by 1. ntnk
Exponential-and-logarithmic-functions/63300: solve algebraically: 1 solutions Answer 46445 by ntnk(54) on 2006-12-28 16:25:31 (Show Source): You can put this solution on YOUR website! If x = y, then log x = log y, therefore, find the log(base3) of both sides. Log(base3)3^(2x+1)=log(base3)(1/243) This simplifies to because log(base3)(1/243)=-5 ntnk
Expressions-with-variables/54990: This question is from textbook
(-4)(10x+2)=-4
I am don't know how you would even get this one out of (). Can you please help me and show me the steps? Thanks
1 solutions
Answer 37187 by ntnk(54) on 2006-10-06 18:01:01 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: This is an equation! Solutions: x=-0.1. Graphical form: Equation was fully solved.Text form: -4*(10x+2)=-4 simplifies to 0=0Cartoon (animation) form: For tutors: `simplify_cartoon( -4*(10x+2)=-4 )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
Moved these terms to the left
It becomes .
Look at .
Expanded term by using associative property on
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
It becomes .
Look at .
It becomes .
Look at .
Solved linear equation equivalent to -40*x-4 =0
It becomes .
Result:
This is an equation! Solutions: x=-0.1.
### Universal Simplifier and Solver
Done!
At the end when you have -40x-4=0, you can add four to both sides, making it -40x=4, then divide both sides by -40 to get x alone and or
real-numbers/54989: This question is from textbook Algebra 1
I am doing a multi-step problem and can't seem to grasp the point here
is the problem:
please show how you did it so I can understand better!!!
Thank you so much!
1 solutions
Answer 37185 by ntnk(54) on 2006-10-06 17:54:27 (Show Source):
You can put this solution on YOUR website!
Read the "detailed explanation" section for the steps, but disregard everything after where its says 0=0, thats an error. It is reexplained below.
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: This is an equation! Solutions: x=12. Graphical form: Equation was fully solved.Text form: x/2+7=x/3+9 simplifies to 0=0Cartoon (animation) form: For tutors: `simplify_cartoon( x/2+7=x/3+9 )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
Moved these terms to the left ,
It becomes .
Look at .
It becomes .
Look at .
Moved to the right of expression
It becomes .
Look at .
It becomes .
Look at .
Eliminated similar terms , replacing them with
It becomes .
Look at .
It becomes .
Look at .
Remove unneeded parentheses around factor ,
It becomes .
Look at .
Remove extraneous '1' from product
It becomes .
Look at .
Solved linear equation equivalent to 0.166666666666667*x-2 =0
It becomes .
Result:
This is an equation! Solutions: x=12.
### Universal Simplifier and Solver
Done!
Sorry, the solver messed up a bit at the end. The correct process is when you have , you add two to both sides to get . Then multiply both sides by six to get the final answer.
Distributive-associative-commutative-properties/51733: please help me factor this equation: 1 solutions Answer 37184 by ntnk(54) on 2006-10-06 17:47:29 (Show Source): You can put this solution on YOUR website!Both monomials are divisible by a common factor; this means that we can "factor" it out so that it can later be distributed back. You can divide both parts by and by , therefore you can factor out . The solution thus becomes because when you use the distributive property, you get the original equation back.
Equations/49139: FIND VALUES OF X1 solutions Answer 33784 by ntnk(54) on 2006-09-04 14:30:56 (Show Source): You can put this solution on YOUR website! The domain of this function is X IS GREATER THAN OR EQUAL TO FIVE.
Polynomials-and-rational-expressions/40711: Please evaluate. Thanks!1 solutions Answer 26099 by ntnk(54) on 2006-06-04 17:37:35 (Show Source):
Quadratic_Equations/33706: If we apply the quadratic formula and find that the value of b2 - 4ac equals zero, what can we conclude about the solutions? A) The equation has no real number solutions. B) The equation has exactly one irrational solution. C) The equation has two different rational solutions D) The equation has exactly one rational solution.1 solutions Answer 20120 by ntnk(54) on 2006-04-15 20:51:10 (Show Source): You can put this solution on YOUR website!Look at the quadratic formula. If you ignore the , by making it zero then you are left with . This is one rational solution. D) The equation has exactly one rational solution. NtNk
1 solutions
Answer 20119 by ntnk(54) on 2006-04-15 20:45:30 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=20 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 3.23606797749979, -1.23606797749979. Here's your graph:
Also, if you go back to where the equation was plugged into the quadratic equation:
It can be simplified to:
And further simplified to
by canceling the 2's
NtNk
logarithm/33712: Express as a single logarithm and, if possible simplify log(x^3-8)-log(x-2)1 solutions Answer 20118 by ntnk(54) on 2006-04-15 20:36:22 (Show Source): You can put this solution on YOUR website!First, note that when factoring a binomial in the form: , it is simplified as . You can use the distributive property to verify this. Also, note that the quotient property of logarithms states: . First, replace with . Use the first property I explained. This will simplify to . Use the second property. . Simplify by canceling the and you are left with the answer: NtNk
Equations/33707: Solve the equation. Express radicals in simplest form. 5x^2 = 351 solutions Answer 20115 by ntnk(54) on 2006-04-15 20:21:44 (Show Source): You can put this solution on YOUR website! Subtract 35 from both sides to set it equal to zero. Divide everything by 5. Simplify: Take the square root of both sides. Simplify. or NtNk
Graphs/19669: hi, need help desperately. graph the equation, identifying the y-intercept. y = 2x -5 thanks so much1 solutions Answer 9607 by ntnk(54) on 2005-11-16 20:42:49 (Show Source): You can put this solution on YOUR website! When an equation is written in slope-intercept form, , b is known as the y-intercept. In this case, m (the slope) would be 2 and b would be -5.
Polynomials-and-rational-expressions/13551: i dont understand how to do this work.... solve for x (4x+1)-(x+4)=18 i know that you have to remove par. so i did that and then my teacher told me to combine like terms.. that leaves me with 3x+5=18 then now i subtract 5 from both sides... that leaves me with 3x=13 i dont think that i did that right but i always get stuck on certain parts can you please help me to try to understand how to do this? any help would be appreciated1 solutions Answer 6842 by ntnk(54) on 2005-10-02 22:43:35 (Show Source): You can put this solution on YOUR website!(4x + 1) - (x + 4) = 18 Set this equation up vertically it might be easier: 4x + 1 -1x - 4 ------- 3x + -3 ... REMEMBER TO SUBTRACT THE 4 FROM THE ONE, NOT ADD THEM 3x - 3 = 18 Add 3 to each side: 3x - 3 + 3 = 18 + 3 3x = 21 Divide both sides by 3: 3x/3 = 21/3 Check: (4(7) + 1) - (7+4) = 18 (28 +1) - (11) = 18 29 - 11 = 18 18 = 18 Solved By: NtNk
Expressions-with-variables/13559: I have three problems that do not have instructions as to what to do. I see the equal sign and assume I am to solve the equation. The problem is stated 5(t+3)+9 = 3(t-2)+6 Only because when I put in the answer that seems obvious it is usually wrong, I thought an answer to whether this is true or not will help me to know if I'm on the right track before going on. I have 5t+15+9 = 3t-6+6 on first line 5t + 24 = 3t on second line -5t -5t 24 = -2t (/)-2 + (/) -2 -12 = T ( T = -12 )1 solutions Answer 6840 by ntnk(54) on 2005-10-02 22:25:22 (Show Source): You can put this solution on YOUR website!Your answer and method of solving is absolutely correct. If you want to be sure, substitute your value of t into the problem: Your answer is correct! Solved By: NtNk
Linear-equations/13561: Help on this please. -2x + 5y =10 Do not understand how to work it. thank you , connie1 solutions Answer 6839 by ntnk(54) on 2005-10-02 22:13:37 (Show Source): You can put this solution on YOUR website! is a linear equation written in standard form: But normally when solving linear equations, it is easier in slope-intercept form: To graph your equation, first change it to slope intercept form: Add 2x to both sides: Reduce: Divide both sides by 5: Reduce: Now you can graph it: Solved By: NtNk
Equations/13575: Solve for X 2x - (2x)/7 = x/2 + 17/21 solutions Answer 6836 by ntnk(54) on 2005-10-02 21:58:43 (Show Source): You can put this solution on YOUR website! Multiply both sides by the LCD, in this case, 14. Use the distributive property and reduce: Subtract 7x from both sides. Reduce. Divide both sides by 17. Reduce Check: Solved By: NtNk
Linear-systems/13577: x + y = -6 11x - y = 42 Many solutions or no solutions?1 solutions Answer 6833 by ntnk(54) on 2005-10-02 21:48:32 (Show Source): You can put this solution on YOUR website!To solve a system of equations, you can either graph, substitute, or use elimination. The easiest method for this problem will be elimination: Add the two equations together to eliminate y: x + y = -6 11x - y = 42 ------------ 12x + 0 = 36 Now solve like a one step equation. Divide both sides by 12. Reduce. Now plug in this value of x to one of the equations: Subtract 3 from each side. Reduce. x = 3 y = -9 Remember to check: ... ... Solved by: NtNk
Quadratic_Equations/3607: how do you graph f(x)=-2x^21 solutions Answer 2022 by ntnk(54) on 2005-07-05 19:44:55 (Show Source): You can put this solution on YOUR website!Solved By: NtNk
Volume/3789: Sheri's freezer is 2 feet wide, 6 feet long, and 2 feet deep. What is the volume of her freezer? a- 10 cubic feet b- 12 cubic feet c- 24 cubic feet d- 4 cubic feet1 solutions Answer 2021 by ntnk(54) on 2005-07-05 19:38:10 (Show Source): You can put this solution on YOUR website!The formula for finding the volume of a rectangular prism, such as a freezer, is , length times width times height (or depth). When you multiply, the product is 24 cubic feet. Check: 2 * 6 * 2 = 24Solved By: NtNk
Rectangles/3701: The area of a gymnasium floor is 264 square yards. The floor is 11 yards wide. How long is the floor? a)253 yards b)24 yards c)11 yards d) 242 yards1 solutions Answer 2008 by ntnk(54) on 2005-07-04 15:47:52 (Show Source): You can put this solution on YOUR website!The floor is B. 24 yards long. This can be found out by using the formula for area of a rectangle , or area equals length times width. The inverse of this formula is , and since you know the area and width, just divide: Check: Solved By: NtNk
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Joy of Problem Solving
# A preview of "Operator Searches"Join Brilliant Premium
Operator searches are puzzles for equations that can be solved with a mix of logic, creativity, and number theory.
Of course $$8 + 8 + 8 + 8 = 32,$$ but what operators can you use to make the equation below true?
$8 \ \ \ 8 \ \ \ 8 \ \ \ 8 \ \ \ 8 \ \ \ = \ \ \ 32$
Here are some of the solutions that we've come up with. Can you find any others? \begin{align*} (8 + 8)\times (8 + 8) \div 8 &= 32 \\ (8\times8\times8) รท (8 + 8)&= 32 \\ 8- 8 \div 8 - 8 \div 8 &=3 \times 2 \end{align*}
Here is one last example to get your mind racing! Using any operators (and parentheses if needed), can you make this true? Get creative!
$1\, \square\, 1\, \square\, 1\, \square\, 1\, = 5$
×
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# Summation and brackets
Is it possible to write $\sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i}$ as one mathematician said it is correct but another said that one should write $\sum_{i=1}^k\left ( y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i} \right )$
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To expand on @vonbrand’s answer, the fact that the index $i$ appears in all three terms almost guarantees that they all belong to the summation, but you shouldn’t make the reader work that hard: it’s much better to enclose the entire summand in parentheses, so that it’s immediately obvious what is being summed. – Brian M. Scott Apr 19 '13 at 9:46
Expressions like $$\tag?\sum_{i=1}^n a_i + 1$$ have two possible interpretations: $$\tag1\sum_{i=1}^n (a_i + 1)$$ or $$\tag2\left(\sum_{i=1}^n a_i\right) + 1.$$ Among others because of the uglyness of $(2)$ it is customary to interprete $(?)$ as $(2)$ and use explicit parentheses if one wants to have $(1)$. Note however, that no parentheses are necessary/customary for $$\sum_{i=1}^n a_i\cdot 2=\sum_{i=1}^n (a_i\cdot 2)= \left(\sum_{i=1}^n a_i\right)\cdot 2=2\sum_{i=1}^n a_i$$
The only thing in $$\sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i}$$ that tells us that the summation should involve all terms is that all terms have an index $_i$ on them. For example, the following example is much unclearer: $$\sum_{i=1}^n a_i+b$$ which could either mean $$\left(\sum_{i=1}^n a_i\right)+b\quad\text{or}\quad\sum_{i=1}^n \left(a_i+b\right)=\left(\sum_{i=1}^n a_i\right)+nb,$$ which is two completely different things. So, it's a good habit to include parentheses as @vonbrand also answers.
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# Inclined Planes and Net Force Renate Fiora. Representing Forces When we’re representing forces for an object on an inclined plane, we don’t really do.
## Presentation on theme: "Inclined Planes and Net Force Renate Fiora. Representing Forces When we’re representing forces for an object on an inclined plane, we don’t really do."— Presentation transcript:
Inclined Planes and Net Force Renate Fiora
Representing Forces When we’re representing forces for an object on an inclined plane, we don’t really do much differently than we normally represent forces in diagrams. We use arrows to depict the magnitude and direction of the forces acting on the object. To get the clearest understanding of what’s happening, it’s best to make both a situational diagram and a free-body diagram.
Two Diagrams Situation Diagram Free-body diagram FAFA FNFN FfFf FWFW +y / +x / FAFA FNFN FfFf FWFW
Two Situations, Generally a = 0 There are two possibilities for when there is no acceleration, v = 0 and v 0. In either case, because there is no acceleration, we know there is no net force on the object, so when we write Newton’s second law (F net =ma) we get (F A + ) F – F f = 0 (F A + ) F = F f a 0 In this situation, the object is accelerating as it moves along the inclined plane, and so we know there must be a net force acting on the object. Writing out Newton’s second law, we get (F A + ) F – F f = ma The applied force is in parentheses as there is not always one present.
Download ppt "Inclined Planes and Net Force Renate Fiora. Representing Forces When we’re representing forces for an object on an inclined plane, we don’t really do."
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# Decompose the following fraction: \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 -...
## Question:
Decompose the following fraction:
{eq}\frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} {/eq}.
## Partial Fractions:
We'll apply the method of Partial Fraction decomposition. We start by factoring the denominator as a product of real irreducible factors, with degrees one or two. For this problem, we'll set up an identity with a large number of unknowns (coefficients) and we'll use a CAS software to find the correct values of those unknowns.
We'll factorize the numerator and denominator into linear or irreducible quadratic factors over the real numbers:
{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x^2 - 3\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x - \sqrt{3}\right)^2\left( x + \sqrt{3}\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ \end{align*} {/eq}
Now for the decomposition into partial fractions we write:
{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x - \sqrt{3}\right)^2\left( x + \sqrt{3}\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ &=\frac{A}{x}+\frac{B}{x^2} +\frac{C}{x^3} +\frac{D}{\left( x - 2\right)}+\frac{Ex+F}{\left( x^2+2x +4\right)}+\frac{G}{\left( x - \sqrt{3}\right)}\\ &\qquad+\frac{H}{\left( x - \sqrt{3}\right)^2}+\frac{I}{\left( x + \sqrt{3}\right)}+\frac{J}{\left( x + \sqrt{3}\right)^2}+\frac{Kx+L}{\left( x^2 +3\right)}\\ &\qquad+\frac{Mx+N}{\left( x^2 +3\right)^2}+\frac{Px+R}{\left( x^2 + 6\right)}, \end{align*} {/eq}
where the coefficients {eq}A,\,B,\ldots,R {/eq} are real numbers to be determined.
Solving for these coefficients we get:
Thus we have:
{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)}&= \frac{\frac{7}{5832 }}{ x}-\frac{\frac{1}{1944}}{ {{x}^{2}}}-\frac{\frac{17}{1944}}{ {{x}^{3}}}+\frac{\frac{1}{1120}}{ x-2 }\\ &\qquad -\frac{\frac{6505}{51824864} x-\frac{12440}{51824864}}{ {{x}^{2}}+2 x+4 }+\frac{\frac{8479 \sqrt{3}+66436}{143712144}}{ \left( x+\sqrt{3}\right) }\\ &\qquad +\frac{\frac{314 \sqrt{3}-381}{1294704}}{ {{\left( x+\sqrt{3}\right) }^{2}}}+\frac{\frac{66436-8479 \sqrt{3}}{143712144}}{ \left( x-\sqrt{3}\right) }+\frac{\frac{-314 \sqrt{3}-381}{1294704}}{ {{\left( x-\sqrt{3}\right) }^{2}}} \\ &\qquad-\frac{\frac{4919}{1724814} x+\frac{1200}{1724814}}{ \left( {{x}^{2}}+3\right) }-\frac{\frac{38}{12636} x+\frac{33}{12636}}{ {{\left( {{x}^{2}}+3\right) }^{2}}}-\frac{\frac{37}{918540} x+\frac{114}{918540}}{ \left( {{x}^{2}}+6\right) }. \end{align*} {/eq}
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25 June, 22:13
# The larger number of two is seven less than three times the smaller number. If the sum of the numbers is 61, find the numbers.
+2
1. 25 June, 23:52
0
27 and 34
Step-by-step explanation:
2x+7=61
2x=54
x=27
27+7=34
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# Scaling at an arbitrary point and figuring out the distance from origin
Suppose I have a 8x6 rectangle, with its lower left corner at the origin (0, 0). I want to scale this rectangle by 0.5 at an anchor point (3, 3). So the resulting rectangle is 4x3, but I cannot figure out how to compute the distance from the origin to the lower left corner of the new rectangle. Help is appreciated.
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Sorry if I'm misunderstanding the construction, but doesn't that corner just go to the midpoint between the origin and (3,3)? In that case it's half the distance to (3,3), which is computed with the Pythagorean theorem. – Jonas Meyer Oct 1 '10 at 4:19
What do you mean by "anchor point"? Should the newly scaled rectangle have a corner at that point, or should it be centered there? – J. M. Oct 1 '10 at 4:20
It is the midpoint for this case, but not for all cases. By anchor point I mean the center of scaling (as if the rectangle was moved to the origin and then scaled). – hyn Oct 1 '10 at 4:22
Oh, I guess you're saying you really wanted a general way to find distances of transformed points. – Jonas Meyer Oct 1 '10 at 4:32
Scaling by 0.5 about the center (3,3) is the same as translating by (-3,-3), scaling by 0.5 about (0,0), then translating by (3,3). So, to find the coordinates of the lower left corner of the new rectangle, take the coordinates of the lower left corner of the original rectangle, subtract 3 from each coordinate, multiply each coordinate by 0.5, and add 3 to each coordinate. Once you have the point's coordinates, you can find its distance from the origin.
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Thanks for the this, exactly what I needed and easy to picture. – hyn Oct 1 '10 at 4:25
It's just half the distance to the anchor point, right? The point started at the origin. – Jonas Meyer Oct 1 '10 at 4:27
@Jonas: I think so, in this case. I gave the more general explanation/method because often people ask what they think is a simpler question or what they ask ends up not being quite what they needed to know. – Isaac Oct 1 '10 at 4:31
Well the real problem is that it's arbitrary. The rectangle, center of scale, and scale can be anything. But Isaac's method works for any case. – hyn Oct 1 '10 at 4:37
If the scale is c and the anchor is (a,b), then the new point is (cx+(1-c)a,cy+(1-c)b), as easily follows from Isaac's method. – Jonas Meyer Oct 1 '10 at 4:41
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Home » Blog » Evaluate the integral of (3 – x2)1/2 / x
# Evaluate the integral of (3 – x2)1/2 / x
Compute the following integral.
First, we multiply the numerator and denominator by and do some rearranging to get a friendlier integral,
For the integral on the left we make the substitution , and obtain
Then, for the integral on the right we make the substitution , . This gives us
Therefore for the integral on the right we have
Now, we have to use partial fractions on the integrand,
This gives us the equation
So,
Putting these integrals back into our original expression we have
### One comment
1. Jose Munoz says:
Using substitution of x=sqrt(3)sin(u) and dx=sqrt(3)cos(u) we can obtain Sqrt[3]Integrate[(Csc[u]-Sin[u]),u] , then with u=ArcSin[x/Sqrt[3]] to have x back, the same solution in achived.
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On-Line Computer Graphics Notes
EQUATIONS OF A PLANE
Overview
A plane in three-dimensional space is the locus of points that are perpendicular to a vector (commonly called the normal vector) and that pass through a point . They form the fundamental geometric structure for many operations in computer graphics ( e.g., clipping) and geometric modeling ( e.g., tangent planes to surfaces). Two equivalent definitions of a plane are used and we present both in these notes.
Specifying a Point and a Vector
A plane in three-dimensional space is the locus of points that are perpendicular to a vector and that pass through a point . The point and the vector uniquely define the plane. Let be the plane defined by and . Then for any point on the plane, we must have that
since the vector will be in the plane. This relationship is illustrated in the following figure.
A Plane Equation
Suppose we are given a plane defined by a point and a vector . If we write the vector as , the point as , and an arbitrary point on the plane as , then from the above we have that
0
and so we can write,
which is in the form
which is a common expression of the equation of a plane. We will both forms of this definition in the clipping algorithms of the viewing pipeline.
This document maintained by Ken Joy
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Christian Homeschooling Resources
# The Science of Ancient Greece
• Melissa Pinkley Crosswalk.com Contributor
• 2012 21 Sep
Ancient Greece had many well-known mathematicians, philosophers, and scientists. Therefore, many discoveries happened in this part of the world during that ancient time period.
Although most people of Ancient Greece did not even know of the One True God, He had equipped them with the ability to think and create. Euclid, a famous mathematician, first wrote the basic rules of geometry. Pythagoras, another Greek mathematician, created the Pythagorean Theorem, which states that a2 + b2 = c2. Both of these men’s works are still used today. How can such things be discovered by people who did not even know of God? Because the order that God created when He made the world can be observed by anyone. Even people who don’t know Him can discover His truth about the world around them.
Another famous inventor from the time of Ancient Greece was Archimedes. Archimedes had an incredible ability to discover how to make things happen based on the laws set by God within creation.
Give It a Try #1: The Law of Displacement
There is a famous story about Archimedes discovering that the king’s crown was made incorrectly by using the law of displacement. My children and I enjoy listening to this story as told by Jim Weiss on his Galileo and the Stargazers audio CD. What is the law of displacement? Let’s “Give It a Try” and find out.
Items Needed:
• One clear container (if you use a graduated cylinder or measuring cup with “mL” marked on it, you will be able to calculate the volume displaced)
• Water
• Three similar-sized objects such as marbles, pebbles, etc.
• Masking tape (if you do not have a measured container)
• A small scale that measures grams (this is optional—only needed if calculating density)
Directions:
Pour the water into the container, filling it about half-full. If you do not have a measured container, place a piece of tape on the side of the container at the height of the water level. If you are using a measured container, record the height of the water.
Carefully, without spilling any water, place one of the objects into the water. Now observe the water level. Has there been any change? Record it. Add another object to the water. What did this object do to the water level? Once again, record the change. Place the third object into the water and record any changes seen.
Now, think about what happened. What do you think displacement means based on this experiment? What happened to the water? Did you know that the definition of displace is to move or shift from the usual position? Isn’t that what happened to the water in the experiment? The Law of Displacement, which is also called the Archimedes Principle, states that an object immersed in a fluid (like water) is subject to an upward force (buoyancy) equal in magnitude to the weight of the fluid that it displaces. So basically, the water moved or was displaced because of the mass of the object put into it.
If you would like to calculate this further, weigh one of the objects you placed into the water to find its mass. (Technically, weight and mass aren’t the same thing, since mass is inherent to an object while weight is a force and depends on gravitational pull; that’s why things weigh less on the moon. However, since we’re doing these experiments on Earth, we can use the object’s weight as its mass.)
Density = Mass ÷ Volume (of the fluid displaced by one object). For example, if your marble weighed 2 g and it displaced 5 mL when it was placed into the water, you can get the density by dividing 2 g by 5 mL. Therefore, the density is 0.4 g/mL3 (0.4 grams per cubic milliliter).
Why is this important? Like in the story of Archimedes and the King’s Crown, different objects that appear to be the same size may be made of different substances which have different densities.
Archimedes invented many useful things. If you are interested in science, you should read more about him.
Give It a Try #2: Does It Take Up Air Space?
Another Greek scientist was Strato. He enjoyed investigating many scientific ideas and experimented to prove his theories. One scientific idea that Strato experimented to prove was that air has substance and is matter. Let’s learn more and see if we agree with Strato.
Items Needed:
• Bucket or tub
• Water
• Two one-liter bottles
• Drill (to be used by an adult)
Directions:
Ask an adult to drill a hole into the bottom of one of your bottles. Fill the bucket three-quarters full of water. Holding the undrilled bottle with its opening facing down, carefully push it straight down into the water quickly. Does the bottle fill up with water? What happens? Now take the bottle with the hole in the bottom and repeat the same procedure. What happens this time? Does the bottle fill up with water?
Think about what we were trying to discover: Does air have substance? One way to see if something has substance is to see if it takes up space. Did the air take up space in the bottle in our experiment? How do you know? Do you agree with Strato that air has substance?
Give It a Try #3: Using Shadows to Measure
Another Ancient Greek, named Eratosthenes, calculated the circumference of the Earth by figuring out the sun’s ray angle and then using shadows to measure distances. Although sources disagree about Eratosthenes’s exact accuracy, they agree that he was within 10 percent of the modern measurement. The fact that this man could so accurately measure the circumference of the Earth in ancient times amazes me. If you too are fascinated by this, check out the advanced picture biography The Librarian Who Measured the Earth to learn more. Now, let’s “Give It a Try” measuring with shadows.
Items Needed:
• Yardstick
• Bush or small tree (or any straight object outside)
• Sunlight
• A helper
Directions:
Isn’t it amazing how many different ways we can experiment to discover more about the earth? I am continuously in awe of God’s forethought and the wisdom that is seen in His creation. It always makes me remember and agree with J. Kepler, “That science is merely thinking God’s thoughts after Him.”
Melissa Pinkley enjoys life with her husband, Wes. They learn a lot from their four children: Ben, Micah, Levi, and Abigail. Homeschooling goes on 24/7 for the whole Pinkley family. They have been homeschooling for more than ten years. The Lord is gracious and continues to help them follow Him.
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# Two consecutive angles of a parallelogram have angle measures, in degr
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Two consecutive angles of a parallelogram have angle measures, in degr [#permalink]
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26 Apr 2018, 12:33
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Two consecutive angles of a parallelogram have angle measures, in degrees, of 2x - 30 and x + 60. What is the measure, in degrees, of the smaller angle?
(A) 50
(B) 60
(C) 70
(D) 100
(E) 110
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Re: Two consecutive angles of a parallelogram have angle measures, in degr [#permalink]
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26 Apr 2018, 14:06
Bunuel wrote:
Two consecutive angles of a parallelogram have angle measures, in degrees, of 2x - 30 and x + 60. What is the measure, in degrees, of the smaller angle?
(A) 50
(B) 60
(C) 70
(D) 100
(E) 110
Two consecutive angles of a parallelogram have angle measures that add up to 180.
Therefore (2x-30) + (x+60) = 180
3x+30 = 180
3x = 150
x= 50
2x-30 = 70
x+60 = 110
Therefore the smaller angle is 70 degrees
Hence C = 70 is the answer.
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Re: Two consecutive angles of a parallelogram have angle measures, in degr [#permalink]
### Show Tags
26 Apr 2018, 23:14
Bunuel wrote:
Two consecutive angles of a parallelogram have angle measures, in degrees, of 2x - 30 and x + 60. What is the measure, in degrees, of the smaller angle?
(A) 50
(B) 60
(C) 70
(D) 100
(E) 110
Ans: C
we know from theory that in a parallelogram opposite angles are equal and sum of all the interior angles is 360 so
2(2x-30)+2(x+60) = 360
x= 50
by putting this we get two angles = 70 and 110. smaller one is 70
Ans : C
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Re: Two consecutive angles of a parallelogram have angle measures, in degr [#permalink]
### Show Tags
30 Apr 2018, 15:31
Bunuel wrote:
Two consecutive angles of a parallelogram have angle measures, in degrees, of 2x - 30 and x + 60. What is the measure, in degrees, of the smaller angle?
(A) 50
(B) 60
(C) 70
(D) 100
(E) 110
Two consecutive angles of a parallelogram always sum to 180 degrees; thus:
2x - 30 + x + 60 = 180
3x = 150
x = 50
We see that one angle is 2(50) - 30 = 70 degrees, and the other is 50 + 60 = 110 degrees. So the smaller angle is 70 degrees.
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Re: Two consecutive angles of a parallelogram have angle measures, in degr &nbs [#permalink] 30 Apr 2018, 15:31
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# Two consecutive angles of a parallelogram have angle measures, in degr
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Q. 65.0( 1 Vote )
# The denominator of a fraction is bigger than its numerator by 3. If 3 is subtracted from the numerator, and 2 added to the denominator, the value of fraction we get is 1/5. Find the original fraction.
Let the numerator be x.
Since, the denominator is bigger than numerator by 3
Denominator = x + 3
If 3 is subtracted from the numerator, it becomes = x-3
And, 2 added to the denominator, it becomes = (x + 3) + 2
It is given that :-
On cross multiplying, we get,
5x – 15 = x + 5
4x = 20
x = 5
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liliana2609
2
# Transformati 0,27dam cubi in dm cubi si 0,27 dam cubi in hm cubi
(2) Răspunsuri
25042001tiramen
27 dam cubi , 00,27 hm cubi, sper ca te-am ajutat
GabRiela2oo3
Deci 1 dam cub are 10x10x10 m cubi adica 1000 m cubi 0.27 dam egal cu 270 m cubi Dar noua nu ne trebuiesc metri cubi .. ci dm cubi. ce facem? transformam din metru in dm. 1 m cub are 10x10x10 dm cubi adica 1000 dm cubi 270 x 1000 = 270 000 dm cubi 1 dam cub are 10x10x10 hm cubi dar de data asta impartim nu inmultim adica 1/1000=0.001 0.001x0.27=0.00027 hm cubi
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### Sangaku
##### Stage: 5 Challenge Level:
The square ABCD is split into three triangles by the lines BP and CP. Find the radii of the three inscribed circles to these triangles as P moves on AD.
### Small Steps
##### Stage: 5 Challenge Level:
Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.
### What's That Graph?
##### Stage: 4 Challenge Level:
Can you work out which processes are represented by the graphs?
### Electric Kettle
##### Stage: 4 Challenge Level:
Explore the relationship between resistance and temperature
### Exponential Trend
##### Stage: 5 Challenge Level:
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Guess the Function
##### Stage: 5 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Immersion
##### Stage: 4 Challenge Level:
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### Maths Filler 2
##### Stage: 4 Challenge Level:
Can you draw the height-time chart as this complicated vessel fills with water?
### Surprising Transformations
##### Stage: 4 Challenge Level:
I took the graph y=4x+7 and performed four transformations. Can you find the order in which I could have carried out the transformations?
### Curve Fitter
##### Stage: 5 Challenge Level:
Can you fit a cubic equation to this graph?
### Bio Graphs
##### Stage: 4 Challenge Level:
What biological growth processes can you fit to these graphs?
### Curve Match
##### Stage: 5 Challenge Level:
Which curve is which, and how would you plan a route to pass between them?
| 1,871 | 8,012 |
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| 3.90625 | 4 |
CC-MAIN-2017-43
|
longest
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en
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http://www.acmerblog.com/hdu-4718-the-lcis-on-the-tree-7763.html
| 1,503,420,056,000,000,000 |
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| 455,896,298 | 12,137 |
2015
09-17
The LCIS on the Tree
For a sequence S1, S2, … , SN, and a pair of integers (i, j), if 1 <= i <= j <= N and Si < Si+1 < Si+2 < … < Sj-1 < Sj , then the sequence Si, Si+1, … , Sj is a CIS (Continuous Increasing Subsequence). The longest CIS of a sequence is called the LCIS (Longest Continuous Increasing Subsequence).
Now we consider a tree rooted at node 1. Nodes have values. We have Q queries, each with two nodes u and v. You have to find the shortest path from u to v. And write down each nodes’ value on the path, from u to v, inclusive. Then you will get a sequence, and please show us the length of its LCIS.
The first line has a number T (T <= 10) , indicating the number of test cases.
For each test case, the first line is a number N (N <= 105), the number of nodes in the tree.
The second line comes with N numbers v1, v2, v3 … , vN, describing the value of node 1 to node N. (1 <= vi <= 109)
The third line comes with N – 1 numbers p2, p3, p4 … , pN, describing the father nodes of node 2 to node N. Node 1 is the root and will have no father.
Then comes a number Q, it is the number of queries. (Q <= 105)
For next Q lines, each with two numbers u and v. As described above.
The first line has a number T (T <= 10) , indicating the number of test cases.
For each test case, the first line is a number N (N <= 105), the number of nodes in the tree.
The second line comes with N numbers v1, v2, v3 … , vN, describing the value of node 1 to node N. (1 <= vi <= 109)
The third line comes with N – 1 numbers p2, p3, p4 … , pN, describing the father nodes of node 2 to node N. Node 1 is the root and will have no father.
Then comes a number Q, it is the number of queries. (Q <= 105)
For next Q lines, each with two numbers u and v. As described above.
1
5
1 2 3 4 5
1 1 3 3
3
1 5
4 5
2 5
Case #1:
3
2
3
#include<map>
#include<cstdio>
#include<iostream>
using namespace std;
#define ll unsigned long long
#define mod 123
#define maxn 111111
map<ll,int>mp;
ll wei[maxn],h[maxn];
char ss[maxn];
int m,l,sum=0;
int main(){
wei[1]=1;
for(int i=2;i<=100011;++i)wei[i]=wei[i-1]*mod;
while(~scanf("%d%d%s",&m,&l,ss+1)){
sum=0;
int len=(int)strlen(ss+1);
int duan=m*l;
h[0]=0;
for(int i=1;i<=len;++i){
h[i]=(ss[i]-'a'+1)+h[i-1]*mod;
}
ll ans;
for(int i=1;i+duan-1<=len&&i<=l;++i){
mp.clear();
int st=i-1,ed=i-1;
while(ed+=l){
ans=h[ed]-h[ed-l]*wei[l+1];
mp[ans]++;
if(ed-st==duan){break;}
}
if(mp.size()==m){
sum++;
}
while(1){
if(ed+l>len)break;
st+=l;
ans=h[st]-h[st-l]*wei[l+1];
mp[ans]--;
if(mp[ans]==0)mp.erase(ans);
ed+=l;
ans=h[ed]-h[ed-l]*wei[l+1];
mp[ans]++;
if(mp.size()==m)sum++;
}
}
printf("%d\n",sum);
}
return 0;
}
1. 这事还问我，一切以正规渠道的消息为准啊，除此以外你要做别的联想和理解我可管不了。真觉得TG又再背后搞小动作给你怎么说也白费。
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| 3.5625 | 4 |
CC-MAIN-2017-34
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longest
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en
| 0.862658 |
https://gmatclub.com/forum/amsden-has-divided-navajo-weaving-into-four-distinct-styles-14904.html?fl=similar
| 1,495,952,216,000,000,000 |
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| 960,930,168 | 72,949 |
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# Amsden has divided Navajo weaving into four distinct styles.
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Amsden has divided Navajo weaving into four distinct styles. [#permalink]
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17 Mar 2005, 19:09
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Amsden has divided Navajo weaving into four distinct styles. He argues that three of them can be identified by the type of design used to form horizontal bands: colored strips, zigzags, or diamonds. The fourth, or bordered, style he identifies by a distinct border surrounding centrally placed, dominating figures.
Amsden believes that the diamond style appeared after 1869 when, under Anglo influence and encouragement, the blanket became a rug with larger designs and bolder lines. The bordered style appeared about 1890, and, Amsden argues, it reflects the greatest number of Anglo influences on the newly emerging rug business. The Anglo desire that anything with a graphic designs have a top, bottom, and border is a cultural preference that the Navajo abhorred, as evidenced, he suggests, by the fact that in early bordered specimens strips of color unexpectedly break through the enclosing pattern.
Amsden argues that the bordered rug represents a radical break with previous styles. He asserts that the border changed the artistic problem facing weavers: a blank area suggests the use of isolated figures, while traditional, banded Navajo designs were continuous and did not use isolated figures. The old patterns alternated horizontal decorative zones in a regular order.
Amsden’s view raises several questions. First, what is involved in altering artistic styles? Some studies suggest that artisans’ motor habits and thought processes must be revised when a style changes precipitously. In the evolution of Navajo weaving, however, no radical revisions in the way articles are produced need be assumed. After all, all weaving subordinates design to the physical limitations created by the process of weaving, which includes creating an edge or border. The habits required to make decorative borders are, therefore, latent and easily brought to the surface.
Second, is the relationship between the banded and bordered styles as simple as Amsden suggests? He assumes that a break in style is a break in psychology. But if style results from constant quests for invention, such stylistic breaks are inevitable. When a style has exhausted the possibilities inherent in its principles, artists cast about for new, but not necessarily alien, principles. Navajo weaving may have reached this turning point prior to 1890.
Third, is there really a significant stylistic gap? Two other styles lie between the banded styles and the bordered styles. They suggest that disintegration of the bands may have altered visual and motor habits and prepared the way for a border filled with separate units. In the Chief White Antelope blanket, dated prior to 1865, ten years before the first Anglo trading post on the Navajo reservation, whole and partial diamonds interrupt the flowing design and become separate forms. Parts of diamonds arranged vertically at each side may be seen to anticipate the border.
1.The author’s central thesis is that
(A) the Navajo rejected the stylistic influences of Anglo culture
(B) Navajo weaving cannot be classified by Amsden’s categories
(C) the Navajo changed their style of weaving because they sought the challenge of new artistic problems
(D) original motor habits and thought processes limit the extent to which a style can be revised
(E) the casual factors leading to the emergence of the bordered style are not as clear-cut as Amsden suggests
[Reveal] Spoiler:
E
2.It can be inferred from the passage that Amsden views the use of “strips of color” in the early bordered style as
(A) a sign of resistance to a change in style
(B) an echo of the diamond style
(C) a feature derived from Anglo culture
(D) an attempt to disintegrate the rigid form of the banded style
(E) a means of differentiating the top of the weaving from the bottom
[Reveal] Spoiler:
A
3.The author’s view of Navajo weaving suggests which one of the following?
(A) The appearance of the first trading post on the Navajo reservation coincided with the appearance of the diamond style.
(B) Traces of thought processes and motor habits of one culture can generally be found in the art of another culture occupying the same period and region.
(C) The bordered style may have developed gradually from the banded style as a result of Navajo experiencing with design.
(D) The influence of Anglo culture was not the only non-Native American influence on Navajo weaving.
(E) Horizontal and vertical rows of diamond forms were transformed by the Navajos into solid lines to create the bordered style.
[Reveal] Spoiler:
C
4.According to the passage, Navajo weavings made prior to 1890 typically were characterized by all of the following EXCEPT
(A) repetition of forms
(B) overall patterns
(C) horizontal bands
(D) isolated figures
(E) use of color
[Reveal] Spoiler:
D
5.The author would most probably agree with which one of the following conclusions about the stylistic development of Navajo weaving?
(A) The styles of Navajo weaving changed in response to changes in Navajo motor habits and thought processes.
(B) The zigzag style was the result of stylistic influences from Anglo culture.
(C) Navajo weaving used isolated figures in the beginning, but combined naturalistic and abstract designs in later styles.
(D) Navajo weaving changed gradually from a style in which the entire surface was covered by horizontal bands to one in which central figures dominated the surface.
(E) The styles of Navajo weaving always contained some type of isolated figure.
[Reveal] Spoiler:
D
6.The author suggests that Amsden’s claim that borders in Navajo weaving were inspired by Anglo culture could be
(A) conceived as a response to imagined correspondences between Anglo and Navajo art
(B) biased by Amsden’s feelings about Anglo culture
(C) a result of Amsden’s failing to take into account certain aspects of Navajo weaving
(D) based on a limited number of specimens of the styles of Navajo weaving
(E) based on a confusion between the stylistic features of the zigzag and diamond styles
[Reveal] Spoiler:
C
7.The author most probably mentions the Chief White Antelope blanket in order to
(A) establish the credit influence of Anglo culture on the bordered style
(B) cast doubts on the claim that the bordered style arose primarily from Anglo influence
(C) cite an example of a blanket with a central design and no border
(D) suggest that the Anglo influence produced significant changes in the two earliest styles of Navajo weaving
(E) illustrate how the Navajo had exhausted the stylistic possibilities of the diamond style
[Reveal] Spoiler:
B
8.The passage is primarily concerned with
(A) comparing and contrasting different styles
(B) questioning a view of how a style came into being
(C) proposing alternative methods of investigating the evolution of styles
(D) discussing the influence of one culture on another
(E) analyzing the effect of the interaction between two different cultures
[Reveal] Spoiler:
B
[Reveal] Spoiler: Question #1 OA
[Reveal] Spoiler: Question #2 OA
[Reveal] Spoiler: Question #3 OA
[Reveal] Spoiler: Question #4 OA
[Reveal] Spoiler: Question #5 OA
[Reveal] Spoiler: Question #6 OA
[Reveal] Spoiler: Question #7 OA
[Reveal] Spoiler: Question #8 OA
Last edited by MacFauz on 19 Mar 2014, 23:34, edited 1 time in total.
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17 Mar 2005, 19:11
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18 Mar 2005, 09:56
It looks like a student's comment on somebody's work about Navajo rugs. Basically A said there are four styles of rugs and the last style or bordered designs respresented an abrupted change in weaver's thought process, which perhaps was influenced by the Anglo culture.
The auther, however, disagrees. He presented three points to counter that claim. First, he argues that borders are natural to weavers and the bordered style only represented a latent habit that was brought to the surface. Second, he suggested that stylistic breaks are inevitable results from artists' constant quest for inventions. Third, he believes that the second and third styles already represented a gradual shift away from the flowing pattern toward a seperate unit pattern.
1. The author’s central thesis is that
(E) the casual factors leading to the emergence of the bordered style are not as clear-cut as Amsden suggests
2. It can be inferred from the passage that Amsden views the use of “strips of colorâ€
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18 Mar 2005, 10:10
Amsden identifies 4 navajo styles and the 4th style, the bordered style in his opinion evolved from the influence of anglo culture. The author cites reasons as to why the bordered style may not be all due to anglo culture but more due to the natural weaving and design practices employed by the Navajo.
EACDDCBB
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18 Mar 2005, 10:17
Re Q6, there is no mention of anglo art in the passage or the imagined correspondence between the two?. Author mentions anglo influences repeatedly but nothing about the anglo art as such, may be i am missing something.
SVP
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18 Mar 2005, 10:19
Hmmm you maybe right about 6.
VP
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18 Mar 2005, 11:46
tough choices !
EACDECBB
I was lost in # 5 tho, bet D and E. I just chose E as I cudn't figure out which one might be correct. Can anyone explain #5 ?
Director
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19 Mar 2005, 13:01
OA: EACDDCBB
Thanks everybody for participating.
I also got most of the questions right but I was not sure about the answers while answering them. Probably, I had not paraphrased the passage properly.
banerjeea_98 wrote:
Can anyone explain #5 ?
Blue : to refute choice (E)
Green : to establish choice (D)
jpv wrote:
Amsden has divided Navajo weaving into four distinct styles. He argues that three of them can be identified by the type of design used to form horizontal bands: colored strips, zigzags, or diamonds. The fourth, or bordered, style he identifies by a distinct border surrounding centrally placed, dominating figures.
Amsden believes that the diamond style appeared after 1869 when, under Anglo influence and encouragement, the blanket became a rug with larger designs and bolder lines. The bordered style appeared about 1890, and, Amsden argues, it reflects the greatest number of Anglo influences on the newly emerging rug business. The Anglo desire that anything with a graphic designs have a top, bottom, and border is a cultural preference that the Navajo abhorred, as evidenced, he suggests, by the fact that in early bordered specimens (line 18) strips of color unexpectedly break through the enclosing pattern.
Amsden argues that the bordered rug represents a radical break with previous styles. He asserts that the border changed the artistic problem facing weavers: a blank area suggests the use of isolated figures, while traditional, banded Navajo designs were continuous and did not use isolated figures. The old patterns alternated horizontal decorative zones in a regular order.
Amsden’s view raises several questions. First, what is involved in altering artistic styles? Some studies suggest that artisans’ motor habits and thought processes must be revised when a style changes precipitously. In the evolution of Navajo weaving, however, no radical revisions in the way articles are produced need be assumed. After all, all weaving subordinates design to the physical limitations created by the process of weaving, which includes creating an edge or border. The habits required to make decorative borders are, therefore, latent and easily brought to the surface.
Second, is the relationship between the banded and bordered styles as simple as Amsden suggests? He assumes that a break in style is a break in psychology. But if style results from constant quests for invention, such stylistic breaks are inevitable. When a style has exhausted the possibilities inherent in its principles, artists cast about for new, but not necessarily alien, principles. Navajo weaving may have reached this turning point prior to 1890.
Third, is there really a significant stylistic gap? Two other styles lie between the banded styles and the bordered styles. They suggest that disintegration of the bands may have altered visual and motor habits and prepared the way for a border filled with separate units. In the Chief White Antelope blanket, dated prior to 1865, ten years before the first Anglo trading post on the Navajo reservation, whole and partial diamonds interrupt the flowing design and become separate forms. Parts of diamonds arranged vertically at each side may be seen to anticipate the border.
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19 Mar 2005, 14:31
The stuff in 'green'in jpv's msg is what Amsden believes and not the author. The question is asking for:
5. The author would most probably agree with which one of the following conclusions about the stylistic development of Navajo weaving?
Evidence to support D by the author is provided in the last paragraph of the passage. This is an inference question and the flow of the passage gives clues as well to picking D. E ruled out because of distortion and Q4 already tested this aspect in that isolated figures were not part of the navajo art prior to 1890, so 'always' in E is not something that the author would agree with.
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21 Mar 2005, 02:46
1-E
2-D
3-C
4-D
5-D
6-C
7-B
8-B
I could note that (2) is wrong.
Could somebody explain that?
jpv wrote:
Amsden has divided Navajo weaving into four distinct styles. He argues that three of them can be identified by the type of design used to form horizontal bands: colored strips, zigzags, or diamonds. The fourth, or bordered, style he identifies by a distinct border surrounding centrally placed, dominating figures.
Amsden believes that the diamond style appeared after 1869 when, under Anglo influence and encouragement, the blanket became a rug with larger designs and bolder lines. The bordered style appeared about 1890, and, Amsden argues, it reflects the greatest number of Anglo influences on the newly emerging rug business. The Anglo desire that anything with a graphic designs have a top, bottom, and border is a cultural preference that the Navajo abhorred, as evidenced, he suggests, by the fact that in early bordered specimens (line 18) strips of color unexpectedly break through the enclosing pattern.
Amsden argues that the bordered rug represents a radical break with previous styles. He asserts that the border changed the artistic problem facing weavers: a blank area suggests the use of isolated figures, while traditional, banded Navajo designs were continuous and did not use isolated figures. The old patterns alternated horizontal decorative zones in a regular order.
Amsden’s view raises several questions. First, what is involved in altering artistic styles? Some studies suggest that artisans’ motor habits and thought processes must be revised when a style changes precipitously. In the evolution of Navajo weaving, however, no radical revisions in the way articles are produced need be assumed. After all, all weaving subordinates design to the physical limitations created by the process of weaving, which includes creating an edge or border. The habits required to make decorative borders are, therefore, latent and easily brought to the surface.
Second, is the relationship between the banded and bordered styles as simple as Amsden suggests? He assumes that a break in style is a break in psychology. But if style results from constant quests for invention, such stylistic breaks are inevitable. When a style has exhausted the possibilities inherent in its principles, artists cast about for new, but not necessarily alien, principles. Navajo weaving may have reached this turning point prior to 1890.
Third, is there really a significant stylistic gap? Two other styles lie between the banded styles and the bordered styles. They suggest that disintegration of the bands may have altered visual and motor habits and prepared the way for a border filled with separate units. In the Chief White Antelope blanket, dated prior to 1865, ten years before the first Anglo trading post on the Navajo reservation, whole and partial diamonds interrupt the flowing design and become separate forms. Parts of diamonds arranged vertically at each side may be seen to anticipate the border.
_________________
Awaiting response,
Thnx & Rgds,
Chandra
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Re: Amsden has divided Navajo weaving into four distinct styles. [#permalink]
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20 Mar 2014, 00:08
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Re: Amsden has divided Navajo weaving into four distinct styles. [#permalink]
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02 May 2016, 22:38
Got all but 6) right
can somebody explain Q 6?
6.The author suggests that Amsden’s claim that borders in Navajo weaving were inspired by Anglo culture could be
(A) conceived as a response to imagined correspondences between Anglo and Navajo art
(B) biased by Amsden’s feelings about Anglo culture
(C) a result of Amsden’s failing to take into account certain aspects of Navajo weaving
(D) based on a limited number of specimens of the styles of Navajo weaving
(E) based on a confusion between the stylistic features of the zigzag and diamond styles
I marked D), i guess its mentioned in the last para, about changes in style way before anglo influence
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Re: Amsden has divided Navajo weaving into four distinct styles. [#permalink] 02 May 2016, 22:38
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## Some excel files you might find useful!
For school projects that have been going on I decided to make a few excel files that I think you might find useful as well.
One is to calculate the amount of rebar required in a beam or slab given a certain bending moment.
The other one will dimension a foundation footing for you based on certain parameters you give in.
Off course, all in eurocodes!
This is the rebar excel sheet (contains VBA code, but no viruses, I promise 😉 )
And this is the footing program. Again, containing VBA code.
## Stiffness matrix maker for continuous beams
I’ve been messing around with finite element method solution methods, partly for school purposes. I noticed that when I was compiling a stiffness matrix for a large system, I kept on making mistakes.
One of the best ways to get rid of mistakes, especially when they are due to sloppiness is to remove the human factor in the equation.
As the world is becoming more and more automated I think it should be standard procedure for any engineer to think how he or she could automate their work more.
## What is a stiffness matrix?
In order to solve systems in a computerized way you need to reduce the model that you are analyzing to very discreet, finite elements.
For example if you are examining the stresses on a beam, you need to split up that beam in smaller beams that will allow you to get a clear picture of the entire system.
And in order to do that you need to use stiffness matrices. The global stiffness matrix of a system is no more than the sum of the individual stiffness matrices of the elements that compose the system.
The small Excel script I made only takes a look at beams. Hence above is the stiffness matrix of a single beam element shown.
The stiffness matrix basically defines the relation between the forces acting on the element and the different spatial deformations that happen as a result of those forces.
This ties in to Hooke’s law: F = k*x. Where F is the force acting on a spring and x is its movement. k is what defines how much or how little a spring would change shape. k is called the spring constant, however in our analogy k is the stiffness matrix.
## Summation of matrices.
The more complicated a system gets, the more elements you will need to get an accurate result from your calculations. As a result the global stiffness matrix is calculated by summing the stiffness matrices of the elements together in the following way:
In the above example we have a beam that is clamped on both ends and a force acting on the beam in the middle. We split this system into 2 analyzable elements. Without taking degrees of freedom into account, the stiffness matrix of this system is the sum of the two individual elements’ stiffness matrices.
We see that the common elements in both matrices are summed together and the remaining elements of both matrices simply used to fill up the gaps in the global matrix.
If we were to completely solve this example we would then “fix” some degrees of freedom in this matrix. As the beam is clamped on both ends it is impossible for it to move anywhere at either ends. Nor will it twist or turn as the connection to the wall is extremely stiff.
In this case the matrix reduces to a matrix where we only view V2 and Theta2.
## How to use my excel in your homework assignments or self study practices.
I’ve written this program relatively fast as I needed to hand in the paper in a few days. Hence there are some bugs here and there.
For one make sure to always press “calculate stiffness matrix” before running any other calculations. If you press the “calculate force vector” button twice, your computer will freeze up.
So;
1. Always FIRST press calculate stiffness matrix and only then press anything else.
But other than that here is how it works:
You first you enter the Young’s Modulus, the second moment of area for the beam and the length of the beam. Also if there is a continuous force acting on the beam, now’s the time to enter its value.
Then you enter into how many elements you want to divide the beam. Unfortunately this quick excel only allows you to divide the beam into equally long elements.
Once you have clicked the “Calculate stiffness matrix”-button, you can then enter degrees of freedom as they appear in your problem. So for a clamped end you would select both X1 and X2 (translation at the first end and rotation at the first end are fixed).
Then simply add any additional point loads if there are any and finally press the calculate “calculate force vector and forces” button to get an overview of the external reaction forces of the system.
## Some notes
As this was a quick and dirty fix for me, the program is buggy and some values are unnecessarily long stored in the memory. So just to make sure you get the right results you might need to click “calculate stiffness matrix” and then “calculate forcevector” a few times.
Basically run through the program a few times to make sure all values are the right ones.
I’ve for one noticed that on occasion it shows me the wrong result, but when I then click again on “calculate stiffness matrix” and then on the “caculate force-vector” it then shows the right result.
To make sure it works you can try and calculate some standard examples to verify for yourself that my program works.
For example you can calculate the reaction forces at the supports of a beam clamped at both ends. You’ll see that using my program then you’d need to select X1 and X2 as fixed, as well as X5 and X6.
If you defined a point load at the middle node (at X4) then you’ll find that the deflection value is (1/192)*F*L^3/EI, as well as the reaction forces at X1 and X5 to be F/2.
Hope this is of some help to someone!
## Social media app for mothers
Now that I have a baby on the way me and my wife got talking about how the daily routine would look like after the baby has come.
We came to realize that it will be extremely important for my wife who will stay at home for probably the first year, to get out there and remain social.
It is really hard for an adult to stay at home with a newborn and remain sane!
Adult contact will be extremely important.
However my wife is not alone in her situation. There must be hundreds of other expecting mothers in the Helsinki area who will be at home for the next year to be.
What if an app would be developed, Tinder style, in which moms who stay at home could casually look for new friends to hang out with.
It could be based on similar likes in your facebook profile, or, based on what you hate! Hatr has been around for a long time and seems to do he trick, matching people based on the things they dislike.
A set of a few questions before you sign up could be enough to determine on a very rough level what you’re in to and what you aren’t.
After you “match” with another mom, just like tinder, a chat window would open where you could arrange for a next brunch date!
## Palkkitaulukko: Beam diagrams
One of the main untapped goldmines in Finland is the world of SEO. Take the keyword “Palkkitaulukko” for example.
This is such a niche keyword that pertains to a side-branch of the engineering-field. However as no-one in Finland at the moment is even trying to do anything for that keyword, it should be really easy to rank for it.
Palkkitaulukko is a list of different load cases for beams. A cheatsheet if you will. You can find a decent “palkkitaulukko” here.
## What is a “palkkitaulukko” anyways?
In engineering, and even more so in structural engineering, you get hit left and right with different kinds of formulas. It’s impossible to remember them all.
As most objects in the structural engineering field are either beams or columns, fins have made a table with the most common beam situations. Not unlike any other bended beam table you can find online.
In this table some of the most common ways in which you can load a force onto a beam have been listed, with their maximum moment and maximum deflection.
It’s just a handy cheatsheet for when you are calculating stuff!
A lot of times it’s handier to check a certain load case from a beam table than it is to start calculating it from scratch.
## Good self-learning guide
There are a few good books on Amazon that will help you learn how to calculate these problems. Better yet, that’ll teach you how to use these “palkkitaulukko”!
Click here to view Schaum’s outline of statics and strength of materials.
## Puustelli library at Prodlib
Pretty excited to see the Puustelli library for Revit and ArchiCad be published.
In most other projects I have participated to a bigger or smaller degree. But Puustelli is the first library that I can say is my doing. I worked together with our contact at Puustelli for a good couple of months before we pushed this one out.
In itself there isn’t that much difference in what we do in other libraries compared to this one. The only specialty with this library however is that we are sharing rendering files to come with the models.
In ArchiCad that isn’t such a big deal as they can be easily defines within the .lcf-file. In Revit however this is a bit more challenging as the paths from which Revit fetches the rendering images can vary from one computer to the other.
Now the method chosen isn’t perfect, since we assumed most computers used for work and structural calculation most likely will run on some version of Windows. Hence the path in which the ProdLib library is installed will most likely stay relatively similar from one Windows version to another.
For Mac however it can be a bit more tricky and it might not always fetch the right images from the right places as it assumes a Windows operating system. This will probably be solved in version 2.0 of this library.
However I’m very pleased with the whole look and feel of the library. There probably are still some mistakes here and there. It might not be 100% flawless but it is mine :)! We’ll work out whatever kinks there are in future editions anyway!
## Amazon Affiliate Link checker – tool 2.0
I’ve been wrestling with the Amazon API and found this great php snippet written by the people over at patchesoft.
Basically they have written a complete class I can call in my scripts to handle all of the communication with the Amazon API for me.
Now I am a complete newbie when it comes to PHP programming, let alone communication with an API.
So I am pretty chuffed to have come up with what I have done so far. Right now the checker tool scans a webpage for any tags with “href” in it, then it assumes those are websites.
That is a pretty safe bet.
Then it checks whether or not “shortened” url’s have been used, as is the case in a lot of affiliate links. It then unwraps the url and digs the associate tag and the product’s ASIN code from the URL.
Afterwards it uses the AmazonAPI-class to make a call to the API with the ASIN in question. It then checks what response it gets for availability when making a call with the ASIN of the product.
If it is an empty response then we just assume that the product is either unavailable, or “available from these sellers”. Both cases will result in my checker showing a “Unavailable” status however..
## Things learnt:
-In the Amazon API class there is a field way up reserved for the different types of response groups. These response groups are important to list and details can be found here on what belongs where. At first I didn’t list the response groups that have the availability parameter in it and I was scratching my eyes out why my code wouldn’t work. The path etc was correct. But as I wasn’t including any of the necessary response groups in MY response XML there was no such field as <Availability>.
-PHP security. I have to use my own accessID and scecret key to make calls to Amazon’s Product Advertising API. So a challenge was to think of how to store that data in a safe way. Utilizing varous tricks to limit access to the private side of my website was a new experience!
-In general PHP coding. It seems a lot online is handled through PHP so it’ll be interesting to delve deeper into what kind of information can be passed back and forwards through pages, inputfields and tables.
Especially as I’ll be building another tool I dicussed earlier. For that project I’ll be needing a lot more php to access and store database information!
## Things to do:
Right now the crawler just browses the given page and only returns the amazon links.
It would make sense that if you give the main domain of your website, the crawler would start crawling also a few levels deeper so you won’t have to.
So if you for example give your site “http://www.mysite.com” and on your homepage you have a menu, then the crawler would crawl through those menu-links, and return any amazon links that may be there.
This has been suggested by already a few people I have spoken too.
A few issues though is that my script isn’t the fastest around. In fact it’s dead-slow.
So 2 things need to be done:
1. Speed up the PHP-script so that it cycles through the found links faster
2. Go as deep as possible when crawling a website and return the URL for the page the crawler is on, to the main table we have all come to know and love <3.
## Amazon Affiliate Link checker – tool
It’s a terrible name for a tool. I know..
However it does the job though. Quite often I’ve seen people ask me whenever I’ve sold them a site: “How will I know I changed all the links?”. Because if you leave a tag unchanged on your site, that affiliate sale won’t be going to you..
## The mechanics..
For those of you who are not familiar with how affiliate marketing works the above will’ve sounded like hebrew.
Affiliate marketing works in such a way that you push traffic to a site. That site then pays you for the traffic you have sent, should that traffic make a purchase.
I’ll just use amazon as an example as it seems to be the most popular affiliate platform.
So if I send someone through my website to Amazon and should that someone buy something from Amazon, I make a little bit on that sale. That’s how the economics works.
Now your next question should be: How does Amazon know that that person came from your site?
Good question!
Here we come to the root of our problem. Affiliate links is how amazon keeps track of who came from where. And, more importantly, who to pay what and how much.
Every product mentioned on any niche site pushing people to Amazon uses affiliate links.
These are normal amazon product links, with one difference. These links contain tags that Amazon registers to each individual niche marketer.
So if you buy an already operational niche website that is making money, it’s in your best interest to change all tags in the affiliate links to your own.
However up until now you had to manually sift out the links from the page and just hope you got them all.
With this new tool I made, you can check a URL; It’ll then show you all amazon affiliate links on that site with its anchor text AND it shows who’s tag is attached to it.
So if you see a tag you don’t recognize, simply search for the anchor text of that link and change that link to a link that has YOUR tag.
Simples!
## For the ones who want to know how I did it:
It’s a small php script I have running that parses a certain URL for links:
\$url=\$_POST[‘URL’];
\$html = file_get_contents(\$url);
//Create a new DOM document
\$dom = new DOMDocument;
//Parse the HTML. The @ is used to suppress any parsing errors
//that will be thrown if the \$html string isn’t valid XHTML.
//Get all links. You could also use any other tag name here,
//like ‘img’ or ‘table’, to extract other tags.
//Function to check for shortened amzn.to URL’s
function unshorten_url(\$url) {
\$ch = curl_init(\$url);
curl_setopt_array(\$ch, array(
CURLOPT_FOLLOWLOCATION => TRUE, // the magic sauce
CURLOPT_RETURNTRANSFER => TRUE,
CURLOPT_SSL_VERIFYHOST => FALSE, // suppress certain SSL errors
CURLOPT_SSL_VERIFYPEER => FALSE,
));
curl_exec(\$ch);
return curl_getinfo(\$ch, CURLINFO_EFFECTIVE_URL);
}
//make it look nice
echo ‘<table>’;
echo ‘<td style=”vertical-align:center; font-weight:bold;”>Anchor text</td>’;
echo ‘<td style=”vertical-align:center; font-weight:bold;”>Amazon tracking code</td>’;
echo ‘</tr>’;
//Iterate over the extracted links and display their URLs
echo ‘<tr>’;
//Check for non-existant anchor text, most likely an image link
echo ‘<td>’;
echo ‘</td>’;
}
else{
echo ‘<td>’;
echo ‘</td>’;
}
parse_str(\$parts[‘query’], \$query);
echo ‘<td>’;
echo \$query[‘tag’];
echo ‘</td>’;
echo ‘</tr>’;
}
}
echo ‘</table>’;
echo ‘</br>’;
echo ‘</br>’;
edit: Already I’ve seen a few additions I could do to this tool. It could also check if the link is still working. Meaning that is the product still existing in Amazon/check for 404’s…
So many ideas!
## Search Engine And Directory Submission Sites
Recently for projects I’ve been running with an Indian friend of mine I’ve been thinking on automating some of the link building using directory submission sites.
Alternatively there are also search engine submission sites available.
## What are search engine submission sites?
Basically as far as I’ve understood it these are sites that add your own site to their directory. In this sense they send a spider to your website, index it and add it to their directory.
As a result your website is now listed in some directory somewhere that is crawlable. In the age before Google there probably were some hundred of different search engines available.
The good thing is.. These are still up and running! Even though Google has taken the lion-share of search traffic over the years, these directory and search engine websites are still being maintained.
## How do these differ from directory submission sites?
Directories are basically a collection of websites grouped around topics. In the early days of the internet there wasn’t really a clever way of making information searchable.
So people just did the best they could and started gathering websites around a certain topic and group them in a directory of that particular topic.
It is a very clunky way of storing information but for all intents and purposes it does the job. If you click on a nature directory, you’ll be bombarded with sites ranging from tree diseases to a site talking about a particular type of goat living in the high-Andes… Just kidding but you get my point.
And there probably is a goat somewhere that lives in the high-Andes and most likely someone will have made a blog about it. If not I’ll be the first and figure out a way to monetize that too!
## In short..
Search Engine and directory submission sites can greatly speed up the work of someone looking to up their SERP scores. In some cases, with the click of a single button your website will be added to multiple directories, creating multiple, decent quality back links in return.
## What does that mean for someone looking to do a bit of SEO?
A lot actually. Assuming you are looking at doing white-hat SEO and not trying some dodgy stuff. The thing with white-hat SEO is that it is extremely time-taking. Sending out eamils, reaching out to people, researching who to contact next just to get a few links.
It’s time taking. So there is a way to speed things up a little.
Don’t expect any influx of traffic to be coming this way though. As the search engines in itself are virtually useless. The reason we are doing this to get your link listed somewhere online, that is also readable by Google.
In doing so you will have, essentially, earned a link. And as in the early days of Google there is one metric that still counts: Links are votes.
## Are all links created equal?
They are not. As most of you reading this are probably aware; link quality plays an important part into the way the Google algorithms rank your website in relation to others in your neighborhood.
But that is the beauty of these directory sites and pre-Google search engines though. Especially the search engines have a relatively good rep with the folks in Palo Alto as these are relatively tricky to game.
For directories you have to be careful nowadays as a lot of them have become breeding grounds for Cialis and Viagra related businesses looking to advertise (and gain links).
## Â List of the best directory submission sites
Here are 30 online directory submission sites you should visit and put up your website. What they require from you is just some basic info about your website and your email address for verification purposes. It can be a bit of a drag to fill all these out manually (maybe I should script this somehow).. However it’s worth it in the end as you get extra links for relatively no effort.
1.      All Free Things  (Page Rank 4)
http://www.allfreethings.com
2. Free PR Web Directory (Page Rank 6)
http://www.freeprwebdirectory.com
3. Travel Tourism Directory Info (Page Rank 6)
http://www.traveltourismdirectory.info
4. Â So Much (Page Rank 6)
http://www.somuch.com
5. High Rank Directory (Page Rank 6)
http://www.highrankdirectory.com
6. Britain Business Directory (Page Rank 6)
7. Marketing Internet Directory (Page Rank 6)
http://www.marketinginternetdirectory.com
8. Hot Vs Not (Page Rank 6)
http://www.hotvsnot.com
9. Â Submission Web Directory (Page Rank 6)
http://www.submissionwebdirectory.com
11. Uk Internet Directory (Page Rank 6)
http://www.ukinternetdirectory.com
12. Jet Jaws (Page Rank 6)
http://www.jetjaws.com
13. Esjoub (Page Rank 6)
http://www.esjoub.com
15. Feed Up Info (Page Rank 8)
http://www.feedup.info
16. Finance Buster (Page Rank 6)
http://www.financebuster.com
17. Synergy-Directory (Page Rank 4)
http://www.synergy-directory.com
18. Nexus Directory (Page Rank 4)
http://www.nexusdirectory.com
19. Directory Fire (Page Rank 4)
http://www.directoryfire.com
20. Master Moz (Page Rank 4)Â
http://www.mastermoz.com
21. Direct My Link (Page Rank 4)
22. Pro Link Directory (Page Rank 4)
23. Info-Listings (Page Rank 4)
http://www.info-listings.com
24. Submissions 4U (Page Rank 4)
http://www.submissions4u.com
25. Â Piseries (Page Rank 4)
http://www.piseries.com
26. Gain Web Org (Page Rank 4)
http://www.gainweb.org
27 Best Free Websites Net (Page Rank 3)
http://www.bestfreewebsites.net
28. PR 3 Plus (Page Rank 3)
http://www.pr3plus.com
29. Web Directory Co (Page Rank 3)
http://www.the-web-directory.co.uk
30. 10 Directory (Page Rank 3)
http://www.10directory.com
## List of the best search engine submission websites
I’ve personally used http://www.cleversubmitter.com/ for my previous projects as well as http://www.freewebsubmission.com/.
Both have proven to be pretty useful and have not given me any bad rep (at least not according to the google search console) nor have I experienced any negative influence on my SERP. In fact quite the opposite.
Contrary to the directory submission sites the 2 mentioned above allow you to add your website to hundreds of prehistoric search engines. And as a result get multiple, white-hat backlinks at the click of a button.
After adding my sites to the above links I noticed a small, but noticeable jump across the board.
Happy adding and let me know if some of these links don’t work anymore!
## New niche site: ScalextricLab.com!
Since selling my safetyshoesreviews website I have cleared another slot in my inventory.
Right now I have started https://www.scalextriclab.com together with an Indian buddy of mine. He provides the content that I then review, adapt to make it SEO friendly and do the off-page SEO.
So far we have had one click already coming from Facebook to our site since posting our first article yesterday.
I estimate that the value of this niche site will reach close to \$3k especially when you take the Christmas season into account!
## GDL reference guide: Perfect self-study material!
I find that relatively often I have to check the “bible” of ArchiCad: The GDL reference guide.
In this guide in over 600 pages Graphisoft goes through the basics of how the GDL programming language works.
If you are interested in making objects for ArchiCad, this reference guide gives a detailed description of GDL, including syntax definitions, commands, variables, etc.
The reference guide starts of handling 3D objects before moving on to how to represent them in a 2D plane.
It however lacks severely in for example curtain wall and other feature objects that have become more mainstream in later versions of ArchiCad.
For those of you who have been reading this blog you’ll find that the ArchiCad curtain wall is severely under-documented in the reference guide.
Lastly before going over the various listing parameters the GDL reference guide goes through making user interfaces.
Later I’ll post some code-snippets and ideas that have helped me in the past. Making user interfaces can be tedious but with the right templates, they are a piece of cake!
For example the GDL cookbook by David A. Nicholson is a really good addition if you are tackling GDL at home.
In a very clear and straight forward fashion David demystifies several commands and codes included within GDL.
Although the cookbook was made for a very early version of ArchiCad, it is still relevant. The version in which the cookbook was written was iirc ArchiCad 8. Now we’re up to 21 already at the time of writing.
However the good thing about GDL, the GDL reference guide and the cookbook is that the language itself remains fairly static.
As a result objects written in ArchiCad version 8 will work just as fine in later versions (mind a few UI quips that I’ll go through later as well..).
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# Thread: Norm Inequality Proof
1. ## Norm Inequality Proof
Problem:
Let u be a point in $\displaystyle \mathbb{R}^n$ and suppose that $\displaystyle \| u \| < 1$. Show that if v is in $\displaystyle \mathbb{R}^n$ and $\displaystyle \| v - u \| < 1 - \| u \|$, then $\displaystyle \| v \| < 1$
=================================
Attempt
Since $\displaystyle \| v - u \|^2 = \| u\|^2 + \| v\|^2 - 2 \left\langle u, v \right\rangle$, then
$\displaystyle \| v - u \| = \sqrt{\| u \|^2 + \|v \|^2 - 2 \left\langle u, v \right\rangle} < 1 - \| u \|$
Square both sides, we get
$\displaystyle \| u \|^2 + \| v \|^2 - 2 \left\langle u, v \right\rangle < \left( 1 - \| u\| \right)^2$
Solve for $\displaystyle \| v \|^2$
$\displaystyle \| v \| ^2 < \left(1- \| u \| \right)^2 - \| u \|^2 + 2 \left\langle u, v \right\rangle$
$\displaystyle \implies \| v \| ^2 < 1 - 2 \| u\| + \| u \|^2 - \| u \|^2 + 2 \left\langle u, v \right\rangle$
Cancelling out liked terms
$\displaystyle \| v \| ^2 < 1 - 2 \| u\| + 2 \left\langle u, v \right\rangle$ *
From here, I am stuck. I am given that $\displaystyle \| u \| < 1$. How would I prove that $\displaystyle \| v \| < 1$ from *?
Thank you for your time and help.
2. Originally Posted by Paperwings
Problem:
Let u be a point in $\displaystyle \mathbb{R}^n$ and suppose that $\displaystyle \| u \| < 1$. Show that if v is in $\displaystyle \mathbb{R}^n$ and $\displaystyle \| v - u \| < 1 - \| u \|$, then $\displaystyle \| v \| < 1$
we need to have $\displaystyle ||u|| < 1$ for only one reason: because otherwise $\displaystyle ||v-u|| < 1-||u||\leq 0,$ which is impossible. anyway, the solution
to your problem is very simple: just use the triangle inequality: $\displaystyle ||v|| = ||v-u + u|| \leq ||v-u|| + ||u|| < 1 - ||u|\ + ||u|| =1. \ \ \ \square$
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# Question: How Many Prime Numbers Are Even?
## Are all prime numbers uneven?
Explanation: By definition a prime number has only 2 factors – itself and 1.
Hence the smallest natural prime number is 2, and the only on that is even.
All other prime numbers are odd, and there are infinitely many prime numbers..
## Why is 11 not a prime number?
For 11, the answer is: yes, 11 is a prime number because it has only two distinct divisors: 1 and itself (11). As a consequence, 11 is only a multiple of 1 and 11.
## What is the easiest way to find a prime number?
To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).
## Is there a formula for prime numbers?
There is no known formula for easily calculating prime numbers. Their distribution along the continuum of numbers appears to be random. There are, however, formulas and diophantine equations that will calculate prime numbers. There are algorithms that can calculate primes into the millions of digits.
## What is 1 called if it is not a prime?
A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, 1 × 5 or 5 × 1, involve 5 itself. However, 4 is composite because it is a product (2 × 2) in which both numbers are smaller than 4.
## What is the lowest prime number?
A prime number is a whole number greater than 1 that can only be divided by itself and 1. The smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. The number 2 is the only even prime number.
## Why 1 is not a prime number?
The number 1 is divisible by 1, and it’s divisible by itself. … My mathematical training taught me that the good reason for 1 not being considered prime is the fundamental theorem of arithmetic, which states that every number can be written as a product of primes in exactly one way.
## Is 1 considered a prime number?
Proof: The definition of a prime number is a positive integer that has exactly two positive divisors. However, 1 only has one positive divisor (1 itself), so it is not prime.
## Is 2 prime or even?
Proof: The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime. … In fact, the only reason why most even numbers are composite is that they are divisible by 2 (a prime) by definition.
## Are 2 and 3 prime numbers?
The first five prime numbers: 2, 3, 5, 7 and 11. A prime number is an integer, or whole number, that has only two factors — 1 and itself. Put another way, a prime number can be divided evenly only by 1 and by itself.
## Is 2 and 3 twin primes?
Usually the pair (2, 3) is not considered to be a pair of twin primes. … The first few twin prime pairs are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139), … OEIS: A077800.
## How many even numbers are prime numbers?
For example: Hence, they are composite numbers. Some facts about prime numbers and composite numbers are: 1 is neither a prime nor a composite number. The only even number which is a prime is 2.
## Why is the number 2 the only even prime?
Explanation: A prime number can have only 1 and itself as factors. Any even number has 2 as a factor so if the number has itself , 2 and 1 as factors it can not be prime. 2 is an even number that has only itself and 1 as factors so it is the only even number that is a prime.
## Why is 57 not a prime number?
For 57 to be a prime number, it would have been required that 57 has only two divisors, i.e., itself and 1. However, 57 is a semiprime (also called biprime or 2-almost-prime), because it is the product of a two non-necessarily distinct prime numbers. Indeed, 57 = 3 x 19, where 3 and 19 are both prime numbers.
## What are all the even prime numbers?
When a number has more than twofactors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.
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# Steve Carell: An Astrology Profile (02/14/2020)
How will Steve Carell fare on 02/14/2020 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first find the destiny number for Steve Carell, and then something similar to the life path number, which we will calculate for today (02/14/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 02/14/2020: We will take the month (02), the day (14) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 14 we do 1 + 4 = 5. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 5 + 4 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the path number for 02/14/2020.
DESTINY NUMBER FOR Steve Carell: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Steve Carell we have the letters S (1), t (2), e (5), v (4), e (5), C (3), a (1), r (9), e (5), l (3) and l (3). Adding all of that up (yes, this can get tiring) gives 41. This still isn’t a single-digit number, so we will add its digits together again: 4 + 1 = 5. Now we have a single-digit number: 5 is the destiny number for Steve Carell.
CONCLUSION: The difference between the path number for today (2) and destiny number for Steve Carell (5) is 3. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is just for fun. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# What Does Postulate Mean In Geometry
Contents
## What Does Postulate Mean In Geometry?
A statement also known as an axiom which is taken to be true without proof. Postulates are the basic structure from which lemmas and theorems are derived. The whole of Euclidean geometry for example is based on five postulates known as Euclid’s postulates.
## What is a postulate in geometry examples?
A postulate is a statement that is accepted as true without having to formally prove it. … For example a well-known postulate in mathematics is the segment addition postulate which states the following: Segment Addition Postulate: If a point B is drawn on a line segment AC then AC is the sum of AB and BC.
## What is postulate in math meaning?
Mathematics Logic. a proposition that requires no proof being self-evident or that is for a specific purpose assumed true and that is used in the proof of other propositions axiom. a fundamental principle.
## What do you mean postulate?
1 : demand claim. 2a : to assume or claim as true existent or necessary : depend upon or start from the postulate of. b : to assume as a postulate or axiom (as in logic or mathematics) postulate. noun.
## What is postulate math example?
A postulate is a statement that is accepted without proof. Axiom is another name for a postulate. For example if you know that Pam is five feet tall and all her siblings are taller than her you would believe her if she said that all of her siblings are at least five foot one.
## What are all the postulates?
Reflexive Property A quantity is congruent (equal) to itself. a = a
Transitive Property If a = b and b = c then a = c.
Addition Postulate If equal quantities are added to equal quantities the sums are equal.
Subtraction Postulate If equal quantities are subtracted from equal quantities the differences are equal.
## What are the 5 postulates in geometry?
Euclid’s Postulates
• A straight line segment can be drawn joining any two points.
• Any straight line segment can be extended indefinitely in a straight line.
• Given any straight line segment a circle can be drawn having the segment as radius and one endpoint as center.
• All right angles are congruent.
## What other terms or phrases mean the same as postulate in geometry?
Synonyms of postulate
• assumption
• given
• hypothetical
• if
• premise.
• (also premiss)
• presumption
• presupposition
## What is coordinate in geometry?
A coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates. …
## What does Hypothisize mean?
hypothesis
English Language Learners Definition of hypothesize
: to suggest (an idea or theory) : to make or suggest (a hypothesis)
## What does postulate mean in science?
A postulate (also sometimes called an axiom) is a statement that is agreed by everyone to be correct. This is useful for creating proofs in mathematics and science (also seen in social science)Along with definitions postulates are often the basic truth of a much larger theory or law.
## What is postulate in Boolean algebra?
The postulates of a mathematical system from the basic assumption from which it is possible to deduce the theorems laws and properties of the system. Boolean algebra is formulated by a defined set of elements together with two binary operators + and · provided that the following postulates are satisfied.
## What’s the difference between postulate and theorem?
The difference between postulates and theorems is that postulates are assumed to be true but theorems must be proven to be true based on postulates and/or already-proven theorems.
## What are postulates Class 9?
Euclid’s postulates were : Postulate 1 : A straight line may be drawn from any one point to any other point. Postulate 2 :A terminated line can be produced indefinitely. Postulate 3 : A circle can be drawn with any centre and any radius. Postulate 4 : All right angles are equal to one another.
## What is corollary in geometry?
In mathematics a corollary is a theorem connected by a short proof to an existing theorem. … In many cases a corollary corresponds to a special case of a larger theorem which makes the theorem easier to use and apply even though its importance is generally considered to be secondary to that of the theorem.
## How do you find postulates in geometry?
A postulate is a statement taken to be true without proof. The SSS Postulate tells us If three sides of one triangle are congruent to three sides of another triangle then the two triangles are congruent. Congruence of sides is shown with little hatch marks like this: ∥.
## How many postulates are there in geometry?
five postulates
The five postulates of Euclidean Geometry define the basic rules governing the creation and extension of geometric figures with ruler and compass.
## What does postulate 3 mean?
Postulate 3: Through any two points there is exactly one line.
## What’s a linear postulate?
Linear Pair Postulate If two angles form a linear pair then the measures of the angles add up to 180°. Vertical Angles Postulate If two angles are vertical angles then they are congruent (have equal measures). Parallel Lines Postulate. Through a point not on a line exactly one line is parallel to that line.
## What does Segment addition postulate mean in geometry?
In geometry the Segment Addition Postulate states that given 2 points A and C a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.
## What is the opposite of a postulate?
▲ Opposite of to consider to be true without evidence. calculate. deny. disbelieve.
## What are the 7 postulates?
Terms in this set (7)
• Through any two points there is exactly one line.
• Through any 3 non-collinear points there is exactly one plane.
• A line contains at least 2 points.
• A plane contains at least 3 non-collinear points.
• If 2 points lie on a plane then the entire line containing those points lies on that plane.
## How do you find coordinates?
To identify the x-coordinate of a point on a graph read the number on the x-axis directly above or below the point. To identify the y-coordinate of a point read the number on the y-axis directly to the left or right of the point. Remember to write the ordered pair using the correct order (x y) .
## What are coordinates example?
A set of values that show an exact position. On graphs it is usually a pair of numbers: the first number shows the distance along and the second number shows the distance up or down. Example: the point (12 5) is 12 units along and 5 units up.
## Is it hypothesize or hypothesise?
As verbs the difference between hypothesise and hypothesize
is that hypothesise is to believe or assert on uncertain grounds while hypothesize is (us) to hypothesise.
## What are hypotheses?
A hypothesis (plural hypotheses) is a proposed explanation for a phenomenon. For a hypothesis to be a scientific hypothesis the scientific method requires that one can test it. … Even though the words “hypothesis” and “theory” are often used synonymously a scientific hypothesis is not the same as a scientific theory.
## What is hypothesizing mean in science?
To hypothesize means simply to make a hypothesis. Which is just a scientific way of saying “make a really good educated guess.” Ok so when someone hypothesizes there’s a little bit more involved than just guesswork. It involves using your past knowledge and available facts to try and predict what might happen.
## What is a postulate in chemistry?
Hammond’s postulate states that the transition state of a reaction resembles either the reactants or the products to whichever it is closer in energy. In an exothermic reaction the transition state is closer to the reactants than to the products in energy (Fig. 1).
## What are basic postulates?
Postulates are statements that are assumed to be true without proof. Postulates serve two purposes – to explain undefined terms and to serve as a starting point for proving other statements. Two points determine a line segment.
## What is standard form in Boolean algebra?
Standard Form – A Boolean variable can be expressed in either true form or complemented form. In standard form Boolean function will contain all the variables in either true form or complemented form while in canonical number of variables depends on the output of SOP or POS.
Categories FAQ
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# Why is AC current inversely proportional to voltage?
Why is AC current inversely proportional to voltage?
I think that current should be proportional to voltage, because according to $V=IR$, when $R$ is fixed, the bigger $I$ is, the bigger $V$ should be. But according to this simulation, the relationship between $I$ and $V$ is negative. Is the software the problem? I use Virtuoso Cadence to simulate the circuit.
I have another question. The direction of the arrow should be the direction of the current. Why is the first half wave negative?
Is the meter "flipped around" in 1 (below)? Or, should I modify something in 2? I had tried simulating 1, but it didn’t change the relation between the voltage and current.
• Tried flipping the meter around? – user253751 Apr 13 '18 at 3:14
• @XM551 flip either the voltmeter or the ammeter around. One but not both. This is just a polarity or sign convention error. No big deal. – mkeith Apr 13 '18 at 3:40
• @XM551 What do you think it means when an ammeter shows a negative value? It means the current is going backwards through the meter. If you flip it around the current would be going forwards through the meter. Same for voltmeters. – user253751 Apr 13 '18 at 4:02
• That's not inverse proportionality. Inverse proportionality is $$I \propto \frac{1}{V}$$. What you see is $$I \propto -V$$ – Ben Voigt Apr 13 '18 at 7:04
• @XM551: No proportional means that both variables have a fixed (constant) ratio; even if the ratio is negative; What you should have written in your question is "Why has current the opposite sign as expected?". I is proportional to V, even if the sign is "the wrong way". – Curd Apr 13 '18 at 14:46
Cadence has a convention to use the current flowing out of the device as negative. You are measuring the current at the positive node of the current source so the current is negative. Measure it at the node connected to resistor, you will get positive current.
A flipped Ammeter while simulation - Probably what is happening in there.
Link to the simulation
• Are you simulate the same point? – XM551 Apr 13 '18 at 12:14
• Same ckt. Same measuring points. – Mitu Raj Apr 13 '18 at 12:23
• I had modify my question,now i think the thing i can't understand is the meaning of "flipping the meter around",can you please see my question again?thx – XM551 Apr 13 '18 at 13:00
• What is "V" in your graph ? Which point is your probe at ? Left or right side of Resistor ? – Mitu Raj Apr 13 '18 at 13:18
• V is the voltage on the red line,the point is the green circle.you can see it from the first picture – XM551 Apr 13 '18 at 13:20
I think you are the confusing the word “meter” with your current source. They are two different things.
Changing the direction of your current source won’t fix your problem, if the direction of the measurement changes with it.
Is your plot measuring the flow of electrons or (conventionally) holes? When you add the current plot to your simulation, your CAD is likely adding it in a fix, albeit backwards, direction.
• How do i know my plot is measuring the flow of electrons or (conventionally) holes? – XM551 Apr 13 '18 at 14:15
• It is measuring conventional current. How do you know this? Because nobody every measures electrons. Ever. It is always conventional current. Just forget about electrons unless you are studying device physics or chemistry. – mkeith Apr 13 '18 at 16:37
• @mkeith I agree, and just threw it out there to be sure. The only reason I considered it is this: my son has a class in electronics (middle school) and came to me for help earlier this week. I was surprised to learn that the students were all being taught their circuit analysis in electron-flow current! I was appalled, and seriously considered scheduling an appointment to talk to his teacher. – Blair Fonville Apr 13 '18 at 16:44
• Wow, @BlairFonville, I too am appalled and surprised! – mkeith Apr 13 '18 at 16:49
• @mkeith I was actually quite frustrated about it; especially so because I know he took the class to make me proud, and so I hide my disappointment. I really should talk to the teacher. – Blair Fonville Apr 13 '18 at 16:54
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#### Explanation:
Let $x$ be the amount to be invested (15%). Then by condition,
x*0.15+550*0.11 = (x+550)*0.12 or x*0.15-x*0.12=550*0.12-550*0.11 or 0.03x=550*(0.12-0.11) or 0.03x= 550*0.01 or 0.03x= 5.5 or x=5.5/0.03= $183.33 $183.33 more to be invested for 12% total return.[Ans]
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Problem 19
# If $$(a, b)$$ is a point on the chord $$A B$$ of the circle, where the ends of the chord are $$A=(-2,3)$$ and $$B \equiv(3,2)$$ then : (a) $$a \in[-3,2], b \in[2,3]$$ (b) $$a \in[2,3], b \in[-3,2]$$ (c) $$a \in[-2,2], b \in[-3,3]$$ (d) $$a \in[-3,3], b \in[-2,2]$$
Expert verified
The solution is: (a) $$a \in[-2,3], b \in[2,3]$$
See the step by step solution
## Step 1 Understanding the Problem
Read the problem carefully. We are given the end points of a chord in a circle as $$A=(-2,3)$$ and $$B \equiv(3,2)$$. We need to find the range within which the points $$(a, b)$$ on the chord $$AB$$ will vary. This requires understanding of coordinate geometry and chord properties.
## Step 2 Plotting the Points
Plot the points $$A$$ and $$B$$ given in the problem. We can see that $$A$$ is situated on the left side (negative x-coordinates) and $$B$$ is on the right (positive x-coordinates). Also, the y-coordinate of $$A$$ is greater than $$B$$. Hence, a point on the chord will vary in such a way to satisfy the properties of chords in a circle and also should lie in between $$A$$ and $$B$$.
## Step 3 Finding the Range
The x-coordinate of any point on the chord will be in between -2 and 3 i.e $$a \in[-2,3]$$ as per the points $$A$$ and $$B$$. Similar logic applies for the y-coordinate. $$B$$'s y-coordinate is 2 and $$A$$'s y-coordinate is 3, so any point $$b$$ on the chord will have y-coordinate varying from 2 to 3 i.e $$b \in [2,3]$$. Hence, the point $$(a, b)$$ will be such that the x-coordinate of $$a$$ is in between -2 and 3 and the y-coordinate of $$b$$ is in between 2 and 3.
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# Minimum $E$ of $p\bar{p}$-collision for $q\bar{q}$ pair with mass $m_q$
I am currently working out the energy required to create a particle anti-particle pair from a collision of a proton travelling along the x-direction with an anti-proton which is at rest. The particle has a mass $m_q$.
Conservation of 4-momentum in the rest frame of the target anti-proton ($c=1$):
$p_1=(m_\text{p},0,0,0)$
The moving proton has this minimum energy, the quantity we are trying to find:
$p_2=\left(E, \sqrt{E^2-m_p^2 },0,0\right)$
Then what I find confusing is the terminology, it says that $p_{q\bar{q}}=\left(\sqrt{p^2+4 m_q^2 },p,0,0\right)$
Where has this $p$ come from and what is it? Is it the momentum of the center of mass of the $q\bar{q}$ system? Also, we don't seem to have taken much care about reference frames here, or rather I havn't thought about them really which is worrying.
Finally, it says that the minimum energy, E, the $q\bar{q}$-pair will be at rest in the CMF frame. What is this frame referring to? The center of mass of the particle/anti-particle pair? I understand that the threshold energy to produce these will be a combination of the energy of the proton and their rest masses but I don't know how we have the equation (above)
-
It seems you are correct the $p$ is the momentum of the center of mass (COM) of your $qq$ (sorry don't know how to add bars) system in the lab frame. Due to conversation of momentum the $qq$ system may only have momentum in the x direction since that is your initial conditions.
With regards to the 2nd part of your question in the CMF (center of mass frame) of the $qq$ system the $qq$ pair will have equal and opposite momentums; however, in the case where the E of the initial condition is just sufficient to produce the $qq$ pair they can have no additional kinetic energy, thus they will be at rest in their CMF. Again due to conservation of momentum the $qq$ COM must still be moving in the x direction in the lab frame.
-
It's \bar{q} :) – Michael Brown Mar 28 '13 at 14:51
I think $p$ is indeed the momentum of the center of mass of the $q\bar{q}$ system. As the $q\bar{q}$-pair is created with the minimum energy required there is no relative momentum and all the momentum is in the center of mass momentum $p$. The energy of the $q\bar{q}$-pair is then given by $E = \sqrt{ p^2 + (2m_{q})^2 } = \sqrt{ p^2 + 4m_{q}^2 }$. As the mass of the $q\bar{q}$ pair is just $2m_{q}$.
So $p_{q\bar{q}}$ is the four vector of the $q\bar{q}$-pair with center of mass momentum $p$ along the $x$-axis.
In respect to your answer about the frame of reference. All four vectors you mentioned are in the lab-frame (and not in the center of mass frame).
-
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# How to Find Surface Area of a Cube and a Rectangular Prism
Coming up next: How to Find Surface Area of a Cylinder
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• 0:01 Rectangular Prisms & Cubes
• 0:44 Surface Area Formula
• 1:48 For a Rectangular Prism
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. You will know what measurements you need to use it and how you can apply it to a cube as well.
## Rectangular Prisms and Cubes
Imagine that you are a gift wrapper at a big store. Customers give you boxes of all sizes to wrap up nicely. These boxes are in the shape of a rectangular prism. All the sides of a rectangular prism are rectangles, and all the sides meet at a perpendicular angle.
Sometimes you get a rectangular prism where all the sides are equal. A box like that is in the shape of a cube. As the gift wrapper, your job is to make sure that you completely cover each gift box with some nice wrapping paper. To help you do this, you calculate the surface area, the outside area, of each box before you begin. This way, you are able to cut a piece of wrapping paper big enough to cover your box.
## Surface Area Formula
Since all your boxes are rectangular prisms, you use the formula for the surface area of a rectangular prism for all your boxes, even your cubes! This formula tells you the surface area is equal to twice the product of the width and the length plus twice the product of the length and height plus twice the product of the height and width.
What is going on in this formula? It is calculating the area of each side and then adding them all up. You have a total of 6 sides to your box. Your opposite sides are always equal to each other in your cube. Looking at the formula, we see that the first term, 2wl, is the area of the top and bottom faces of the box together; 2lh is the area of the front and back faces of the box; and 2hw is the area of the left and right faces of the box. If you happen to forget this formula, just remember to find the area of each face and then add them all up.
## For a Rectangular Prism
Your workday has officially started now. Your first customer gives you a lengthy package, a rectangular prism. Its dimensions are 10 inches long by 3 inches wide by 3 inches high. It's a long box, so perhaps there is a rose inside. You need to calculate the surface area so you know how much wrapping paper you need. From the dimensions, you know that the l is 10, the w is 3, and the h is also 3. Plugging these into the formula, we have 2 times 3 times 10 plus 2 times 10 times 3 plus 2 times 3 times 3.
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Probability of at least one boolean random variable being true in a network of positively-correlated boolean variables
My question
I have a set of $N$ random boolean variables $X_1, \ldots, X_N$ (each can be $1$ or $0$). For every $i \in [1, N]$, I know that
$$P(X_i = 1) = p^*$$
Now, I know that the variables are positively correlated, i.e., for every $i, j \in [1, N]$ I have:
$$P(X_i = 1 | X_j = 1) \geq p^*$$ $$P(X_i = 1 | X_j = 0) \leq p^*$$
(but I can't find an expression for those conditional probabilities)
Is the probability of at least one variable being $1$ larger or smaller than if they were independent?
The extreme case
Consider the case where all the variables are completely correlated, so that for every $i, j \in [1, N]$
$$P(X_i = 1 | X_j = 1) = 1$$ $$P(X_i = 1 | X_j = 0) = 0$$
Then the probability $p$ of at least one random variable being $1$ is $p = p^*$ (either all of them are $1$ or none is).
While in the case where they are completely independent, i.e.,
$$P(X_i = 1 | X_j = 1) = p^*$$ $$P(X_i = 1 | X_j = 0) = p^*$$
I have
$$p = 1 - (1 - p^*)^N$$
and since $(1 - p^*) < 1 \implies (1 - p^*)^N < (1 - p^*)$ I have
$$p > p^*$$
So in the limiting case I know that being positively correlated makes the probability of at least one variable being $1$ smaller. Is this true for any positive correlation?
Is the probability of at least one variable being $1$ larger or smaller than if they were independent?
NB: For convenience I write $p$ instead of your $p^*$. Now the probability in question is $1-P(X_1=0,...,X_N=0)$, which equals $1-(1-p)^N$ if all the $X_i$ are mutually independent.
Theorem: Suppose $(X_1,...,X_N)$ is distributed on $\{0,1\}^N$ such that $P(X_i=1)=p\in(0,1)$ for all $1\le i\le N$, and $P(X_i=1\mid X_j=1)\ge p$ for all $1\le i\lt j\le N$. If $N=2$, then
$$1-P(X_1=0,...,X_N=0) \le 1-(1-p)^N$$ but if $N\ge 3$, then the preceding inequality holds for some distributions and fails for others.
Proof:
Note that because $P(X_i=1\mid X_j=1)\ge p$, the correlation coefficient between any two distinct $X_i$ and $X_j$ is nonnegative: \begin{align}\rho_{ij} &=\frac{E(X_iX_j)-(EX_i)(EX_j)}{E(X_iX_i)-(EX_i)(EX_i)}=\rho_{ji}\\ \\ &= \frac{E(X_iX_j)-p^2}{p-p^2}\\ &\ge 0 \end{align} because $$EX_iX_j = P(X_i=1,X_j=1) = P(X_i=1\mid X_j=1)P(X_j=1)=P(X_i=1\mid X_j=1)\cdot p\ge p^2.$$
Notation: Let $P(*)$ denote the sum of all $2^N$ joint probabilities. Let $P(*1_i)$ denote the sum of just those joint probabilities that have a $1$ in the $i$th position. Similarly, let $P(*1_i0_j)$ denote the sum of just those joint probabilities that have a $1$ in the $i$th position and a $0$ in the $j$th position, etc. We have the following general formulas: \begin{align}1 &= P(*)\tag{1}\\ p &= P(*1_i)=P(*1_i0_j)+P(*1_i1_j)\tag{2}\\ P(*1_i1_j)&=E(X_iX_j)= p^2 + (p-p^2)\rho_{ij}\tag{3}\\ P(*1_i0_j)&= 1-P(*1_i1_j)=(p-p^2)(1-\rho_{ij})\tag{4}. \end{align}
For convenience we also write $P(x_1...x_N)$ to denote $P(X_1=x_1,...X_N=x_N)$.
Case $N=2$
We show that it always holds that $1-P(00)\le [1-P(00)]_\text{independent}$: \begin{align}1 &= P(*) = P(00)+P(11)+P(01)+P(10)\\ 1-P(00)&=P(11)+P(01)+P(10)\\ &= [p^2 + (p-p^2)\rho_{12}]+[(p-p^2)(1-\rho_{21})]+[(p-p^2)(1-\rho_{12})]\\ &= 2p-p^2-(p-p^2)\rho_{12}\\ &\color{blue}{\le} 2p-p^2=1-(1-p)^2=[1-P(00)]_\text{independent} \end{align} because $p-p^2\ge 0$ and $\rho_{12}\ge 0$.
Case $N\ge 3$
We show the proof for $N=3$ (the higher-dimensional cases being similar), that it is not generally true that $1-P(000)\le [1-P(000)]_\text{independent}$; that is, $(X_1,X_2,X_3)$ may be jointly distributed such that $1-P(000)> 1-(1-p)^3$:
\begin{align}1 = P(*)&= P(000) + P(111) + P(*1_10_2)+P(*1_20_3)+P(*1_30_1)\\ 1-P(000)&=P(111) + P(*1_10_2)+P(*1_20_3)+P(*1_30_1)\\ &=P(111)+[(p-p^2)(1-\rho_{12})]+[(p-p^2)(1-\rho_{23})]+[(p-p^2)(1-\rho_{31})]\\ &=P(111)+(p-p^2)(3-\rho_{12}-\rho_{23}-\rho_{31}) \end{align} It suffices to take $p=\frac{1}{2}$ and $\rho_{12}=\rho_{23}=\rho_{31}(=\rho\text{, say)},$ in which case \begin{align}1-P(000)&=P(111)+\frac{3}{4}(1-\rho). \end{align} Thus, the proof will be accomplished if we can find a joint distribution such that $$P(111)+\frac{3}{4}(1-\rho)>1-(1-p)^3=\frac{7}{8}$$ i.e., such that $$P(111)>\frac{1}{8}+ \frac{3}{4}\rho.\tag{5}$$ Now from (3) and (4) we can deduce the following: \begin{align}P(110)=P(011)=P(101)&=\frac{1}{4}(1+\rho)-P(111)\tag{6a}\\ P(001)=P(101)=P(100)&=P(111)-\frac{1}{2}\rho.\tag{6b} \end{align} Because each $P(x_1x_2x_3)$ must lie in the unit interval, we therefore need to find a value for $P(111)$ that satisfies, in addition to (5), also the following: \begin{align}\frac{1}{2}\rho\le P(111)\le \frac{1}{4}(1+\rho). \tag{7}\end{align} There will be solutions to both (5) and (7) iff \begin{align}\frac{1}{8}+ \frac{3}{4}\rho < \frac{1}{4}(1+\rho)\end{align} i.e., iff $\rho <\frac{1}{4}.$
Hence, for $N=3$, any joint distribution with $p=\frac{1}{2}$, $\ \ 0<\rho_{12}=\rho_{23}=\rho_{31}=\rho<\frac{1}{4}$, and $\frac{1}{2}\rho\le P(111)\le \frac{1}{4}(1+\rho)$ will be an example such that $1-P(000)> [1-P(000)]_\text{independent}$.
As an explicit example, we may take $p=\frac{1}{2}$, $\rho=\frac{1}{8}$, and $P(111)=\frac{1}{4}$. Then $$P(110)=P(011)=P(101)=\frac{1}{32},$$ $$P(001)=P(101)=P(100)=\frac{3}{16},$$ and $$1-P(000)=\frac{29}{32}>1-(1-\frac{1}{2})^3=\frac{7}{8}.$$
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# solving compound inequality
The number of senior citizens (65 and older) in the US in millions n years after 1990 can be estimated using the formula
s=0.38n+31.2
The percentage of senior citizens living below the poverty level n years after 1990 can be estimated using the formula
p= -0.25n+12.2
a) How many senior citizens were there in 2000?
b) In what year will the percentage of seniors living below the poverty level reach 7%
c) What is the first year in which we can expect both the number of seniors to be greater than 40 million and fewer than 7% living below the poverty level?
#### Solution Preview
a) n = 2000 - 1990 = 10
so 10 years after 1990 in 2000, the number of senior citizens is
s = 0.38 * 10 + 31.2 = 35 millions.
b) p = 7
-0.25n + 12.2 = ...
#### Solution Summary
The solution shows how to set up and solve the compound inequality for a word problem regarding the seniors and the poverty.
\$2.19
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# Introducing a big M variable in given equations
While I do understand the general workings of the Big-M-method I am struggling with the following sample exercise, in which the Big-M-method has to be used to find a first feasible solution:
\begin{alignat}2\max&\quad 10x_1+4x_2\\\text{s.t.}&\quad x_1+x_2+x_3=4\tag1 \\&\quad 2x_1-x_2-x_4=2\tag2\\&\quad -x_1+x_5=-1\tag3\\&\quad x_1+x_3-x_4+x_5=4\tag4\\&\quad x_1,\cdots,x_5 \geq 0\tag5\end{alignat}
I am not sure how to introduce the artificial variable for the Big-M. The only problem here seems to be the negative value on the right side of equation #3. So I would multiply with $$-1$$. Now it looks as though we have a negative slack variable $$x_5$$ which would allow us to add another variable $$y_1$$ as part of the Big-M-Method. But I doubt if $$x_5$$ can be considered a slack variable here since it is given as part of the task and is also specified as $$\geq 0$$. So I just need to know if I am on the wrong track and if so, how to introduce the Big-M the right way
• Here is a good example: youtube.com/watch?v=ROkDaBeEiVs&vl=en Nov 25 '19 at 22:19
• @EhsanK yes, constraint #3 is what I meant Nov 26 '19 at 18:15
All your constraints are equality. So, add an artificial variable to each constraint (let's call them $$a_i \quad i\in\{1,..,4\}$$). Now all these artificial variables need to be in the objective function with a coefficient of Big-M. Since you are maximizing, you want to make sure using any of them will penalize your objective function (so, you add them with negative sign). So, your objective function becomes: $$\max \quad 10x_1 +4x_2 - Ma_1 - Ma_2 - Ma_3 - Ma_4$$
The artificial variables play the role of your initial basis. So, you need to standardize your simplex table and make sure they are zero in the objective row. After doing that, just solve the simplex problem as you normally do.
If you like to check a simple example, take a look at this example
• Since the constraints are equalities, you cannot assume the artificials are nonnegative. So you really should have two artificials per equation (as when modeling absolute values). Nov 25 '19 at 22:10
• @prubin I tried to find an example to confirm that but any example I searched and found (from MIT, columbia, Brown, and some others) they all introduce one positive artificial variable for equality constraints too. Am I forgetting something here?
– EhsanK
Nov 25 '19 at 22:46
• One positive variable for equalities requires an assumption about the sign of the RHS IMO. Nov 26 '19 at 11:10
• @JohnEren You add artificial variables ($a$) to make a new space where the origin point is part of your solution so you start from there. So, you add it to a constraint that you don't have the available $+1s$ for your initial basis (where $s$ is a slack). You have that $+1$ in $\le$, you have $-1$ for $\ge$ (from a surplus) and you add an $a$ there. You also need it for equality constraint to satisfy the requirement. In your example, if constraint 4 was $x_1+x_3-x_4+x_6=4$ and you don't see $x_6$ anywhere else, then that $x_6$ could play the role in the basis for you (no need for $a$).
– EhsanK
Nov 26 '19 at 18:45
• @prubin that was assumed as the OP mentioned in the question "...The only problem here seems to be the negative value on the right side of equation #3. So I would multiply with −1. ..."
– EhsanK
Nov 29 '19 at 15:37
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0
# Does a obtuse triangle ever have equal sides?
Updated: 10/19/2022
Wiki User
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yes
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Q: Does a obtuse triangle ever have equal sides?
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Related questions
### Can an obtuse triangle ever be a scalene triangle?
Yes. The triangle with sides 7 cm, 8 cm, 13 cm is obtuse (the angle opposite the side of 13 cm is 120o) and scalene as none of the sides are equal.
### Could a triangle ever be an equilateral triangle?
Yes and it will have 3 equal sides
no
### Can obtuse triangle ever be isosceles?
Yes providing that the two equal angles are acute angles
No.
### Can an isoceles triangle ever be scalene?
No. An isosceles triangle has, by definition, two sides of equal length. A scalene triangle has, by definition, no sides of equal length. So, by definition (and the fact that 0 is not 2), an isosceles triangle cannot be scalene.
### Does an obtuse triangle have two obtuse angles?
No not ever because the 3 interior angles of any triangle add up to 180 degrees and so an obtuse triangle will have 1 obtuse and 2 acute angles.
nope
### Do some triangle have two obtuse angle and 1 acute angle?
No because any triangle must have a total of 180 degrees when the angles are added together and with 2 obtuse angles the sum would be greater than 180. A triangle can only ever have one obtuse angle.
### Does a triangle have parallel line?
No two sides of a triangle can ever be parallel.
### Will the medians of an obtuse triangle ever intersect on the triangle?
Yes. Medians always intersect in a single point, called the centroid, or geocenter.
### Does an isosceles triangle have two obtuse angles?
No. It is not possible in Euclidean planar geometry (if you don't know what that means, it means "the only kind of geometry you've ever heard of") for a triangle to have two obtuse angles.
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# A person sold an article for Rs. 144, thereby gaining a profit percent equal to the cost price. Then the cost price of that article (in Rs.) is
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A person sold an article for Rs. 144, thereby gaining a profit percent equal to the cost price. Then the cost price of that article (in Rs.) is
1. 64
2. 75
3. 80
4. 90
5.
by (42.6k points)
selected
Correct Answer - Option 3 : 80
Given:
The selling price of an article = Rs. 144
Profit% = Cost price
Formula used:
Profit% = [(SP - CP)/CP] × 100
Calculation:
Let the cost price of article be Rs. x.
According to the question,
Now, profit is equal to the cost price
Profit percent = x
Profit% = [(SP - CP)/CP] × 100
x = [(144 - x)/x] × 100
⇒ x2 = 14400 - 100x
⇒ x2 + 100x - 14400 = 0
⇒ x2 + 180x - 80x - 14400 = 0
⇒ x(x + 180) - 80(x + 180) = 0
⇒ (x - 80)(x + 180) = 0
If (x - 80) = 0, then x = 80
If (x + 180) = 0, then x = -180 (Not possible)
∴ The cost price of an article is Rs. 80.
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DEV Community
Posted on • Originally published at alkeshghorpade.me
LeetCode - Unique Binary Search Trees
Problem statement
Given an integer n, return the number of structurally unique **BST's* (binary search trees) which has exactly n nodes of unique values from 1 to n*.
Problem statement taken from: https://leetcode.com/problems/unique-binary-search-trees.
Example 1:
``````Input: n = 3
Output: 5
``````
Example 2:
``````Input: n = 1
Output: 1
``````
Constraints:
``````- 1 <= n <= 19
``````
Explanation
Brute force solution
The brute force approach is to generate all the possible BSTs and get the count. This approach will consume a lot of time when we increase n.
Dynamic Programming
With Dynamic Programming, we will reduce the scope of generating the BSTs and use mathematical concept to get the required result.
Let's take an example where n is 5. If node 2 is the root, then the left subtree will include 1 and the right subtree will include 3, 4, and 5. The possible number of combinations in the left subtree is 1, and in the right subtree is 5. We multiply 1 and 5. Similarly, if 3 is the root node, the possible number of combinations in the left subtree will be 2, and the number of combinations in the right subtree will be 2. So the total BST's when root node is 3 is 2*2 = 4. We add up all these combinations for each node 1 to n and return the required result.
A C++ snippet of the above approach is as below:
``````int numberOfBST(int n) {
int dp[n + 1];
fill_n(dp, n + 1, 0);
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i; j++) {
dp[i] = dp[i] + (dp[i - j] * dp[j - 1]);
}
}
return dp[n];
}
``````
The time complexity of the above approach is O(N^2) and space complexity is O(N).
Catalan numbers
[Catalan numbers (https://en.wikipedia.org/wiki/Catalan_number), in combinatorial mathematics, are a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects.
It is denoted by Cn and the formula to calculate it is
(2n)! / ((n + 1)! * n!).
Let's check the algorithm to see how we can use this formula.
``````// numTrees function
- return catalan(2*n, n)
// catalan function
catalan(n , k)
- set result = 1
- if k > n - k
- k = n - k
- for i = 0; i < k; i++
- result *= (n - i)
- result /= (i + 1)
- return result/(k + 1)
``````
The time complexity of this approach is O(N), and space complexity is O(1). Let's check out our solutions in C++, Golang, and Javascript.
C++ solution
``````class Solution {
public:
long long catalan(int n, int k) {
long long result = 1;
if(k > n - k) {
k = n - k;
}
for(int i = 0; i < k; i++) {
result *= (n - i);
result /= (i + 1);
}
return result/(k + 1);
}
int numTrees(int n) {
long long result = catalan(2*n , n );
return (int) result ;
}
};
``````
Golang solution
``````func catalan(n, k int) int {
result := 1
if k > n - k {
k = n - k
}
for i := 0; i < k; i++ {
result *= (n - i)
result /= (i + 1)
}
return result/(k + 1)
}
func numTrees(n int) int {
return catalan(2*n , n )
}
``````
Javascript solution
``````var catalan = function(n, k) {
let result = 1;
if(k > n - k) {
k = n - k;
}
for(let i = 0; i < k; i++) {
result *= (n - i);
result /= (i + 1);
}
return result/(k + 1);
}
var numTrees = function(n) {
return catalan(2*n, n);
};
``````
Let's dry-run our algorithm to see how the solution works.
``````Input n = 4
Step 1: result = catalan(2*n , n )
= catalan(2*4, 4)
= catalan(8, 4)
// catalan function
Step 2: result = 1
n = 8, k = 4
Step 3: if k > n - k
4 > 8 - 4
4 > 4
false
Step 4: loop for i = 0; i < k
0 < 4
true
result *= (n - i)
= result * (n - i)
= 1 * (8 - 0)
= 8
result /= (i + 1)
= result / (i + 1)
= 8 / (0 + 1)
= 8
i++
i = 1
Step 5: loop for i < k
1 < 4
true
result *= (n - i)
= result * (n - i)
= 8 * (8 - 1)
= 8 * 7
= 56
result /= (i + 1)
= result / (i + 1)
= 56 / (1 + 1)
= 56 / 2
= 28
i++
i = 2
Step 6: loop for i < k
2 < 4
true
result *= (n - i)
= result * (n - i)
= 28 * (8 - 2)
= 28 * 6
= 168
result /= (i + 1)
= result / (i + 1)
= 168 / (2 + 1)
= 168 / 3
= 56
i++
i = 3
Step 7: loop for i < k
3 < 4
true
result *= (n - i)
= result * (n - i)
= 56 * (8 - 3)
= 56 * 5
= 280
result /= (i + 1)
= result / (i + 1)
= 280 / (3 + 1)
= 280 / 4
= 70
i++
i = 4
Step 8: loop for i < k
4 < 4
false
Step 9: return result/(k + 1)
70/(4 + 1)
70/5
14
So we return the answer as 14.
``````
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# What is it like to be Christopher’s College Algebra student?
I happened across a review sheet for another instructor’s College Algebra exam today. I know not whose, nor do I wish to know. I just want to use it as an example of what my poor students have to go through.
There were eight tasks on the review sheet. I would like my students to have the skills represented in those tasks for sure. But I wouldn’t happy with just those skills.
So here are the tasks. Each is followed up by the sort of question I would ask my students on an exam. Pity them.
1. Original: Find the domain and range of $f(x)=\sqrt{x+2}$. Follow up: Give two more functions: one that has the same range as f but a different domain, and one that has the same domain as f but a different range.
2. Original: Is $f(x)=x^{2}+x$ even or odd or neither. Follow up: Can a function be both?
3. Original: Solve the absolute value inequality $|2-5x|<7$ and graph the solutions. Follow up: How do these solutions relate to the function $f(x)=2-5x$?
4. Original: Graph the function $f(x)=x^{2}+4$. Then find the intervals on which f is increasing and decreasing. Are there any local maxima or minima? If yes, where are they located? Follow up: Choose two points near a maximum or minimum value (if such a value exists). Find and comment on the average rate of change between these two points.
5. Original: If $g(x)=6x^{2}+5$, find the net change and the average rate of change between $x=-2$ and $x=1$ Follow up: Why are these values different? BONUS: Give a new function for which these values are equal between the same two points.
6. Original: If $h(x)=\sqrt{x}$, write the transformations that yield $g(x)=\sqrt{x+2}+1$. Also graph both $h(x)$ and $g(x)$ on the same coordinate axes. Follow up: How many solutions are there to this system of equations? $\begin{cases} y=\sqrt{x} \\ y=\sqrt{x+2}+1 \end{cases}$
7. Original: If $t(x)=2x-1$ and $s(x)=\sqrt{x+1}$, then what are $t(x)+s(x), t(x)-s(x), t(x)*s(x)$ and $\frac{t(x)}{s(x)}$? Also list the domain for each case. Follow up: Choose one of the four functions you wrote. Write its inverse (if such a thing exists).
8. Original: Graph this piecewise function $f(x)= \begin{cases} 1, x\le 0 \\ x, x>0 \end{cases}$ Follow up: There is a gap in the graph. Change the second piece of the function to eliminate this gap.
### 3 Responses to What is it like to be Christopher’s College Algebra student?
1. Don
Ok…more stuff to steal. Thanks. And I like the way you think.
However, I am curious about number 7. I will look it up later if you don’t respond, but what kind of function has no inverse? I know some inverses are not functions, but does that mean that there is no inverse? I’m probably splitting hairs, but my inverse function knowledge is a little rusty and were going to talk about them next week. So if there is a function that does not have an inverse, what kind of function is it?
• Christopher
True that, Don. I think I meant inverse function but was sloppy (as are most College Algebra texts) and used inverse as shorthand. If I were to give that task, I would be perfectly happy with a response that pointed out that such a thing must always exist-even if it’s not a function. And I would be happy with a student who understood inverse to mean inverse function.
And then I would share these two responses in class, invite discussion and summarize as you have here. Again-a window into my College Algebra classroom. Those poor kids sometimes just want right and wrong answers and points, not ambiguous quiz questions that lead to further discussion. I win most of em over by the end of the semester. Not all, but most.
• Christopher
One more thing. Yes, all functions have inverse relations. Since a function is officially a set of ordered pairs, the inverse swaps the order of the elements. We have dealt with that this semester and approached inverse from the perspective of “switching the range and the domain”. This leads to interesting ambiguities, about which I will need to write more sometime soon.
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https://socratic.org/questions/is-the-standard-deviation-of-a-data-set-invariant-to-translation#351291
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# Is the standard deviation of a data set invariant to translation?
Dec 13, 2016
Yes it is. See explanation for a proof.
#### Explanation:
Let $S$ be a data set:
$S = \left\{{x}_{1} , {x}_{2} , \ldots , {x}_{n}\right\}$
Its mean and standard deviation:
$\overline{x} = \frac{1}{n} \times {\Sigma}_{i = 1}^{i = n} \left({x}_{i}\right)$
$\sigma = \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} - \overline{x}\right)}^{2}}$
Let ${S}_{1}$ be a data set $S$ translated by $a$:
${S}_{1} = \left\{{x}_{1} + a , {x}_{2} + a , \ldots , {x}_{n} + a\right\}$
Its mean would equal:
$\overline{{x}_{1}} = \frac{{x}_{1} + a + {x}_{2} + a + \ldots + {x}_{n} + a}{n} =$
$= \frac{{x}_{1} + {x}_{2} + \ldots + {x}_{n}}{n} + \frac{n a}{n} = \overline{x} + a$
The standard deviation would be:
${\sigma}_{1} = \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} + a - \left(\overline{x} + a\right)\right)}^{2}} =$
=sqrt(Sigma_{i=1}^{n}(x_i+a-bar(x)-a))^2)=
=sqrt(Sigma_{i=1}^{n}(x_icancel(+a)-bar(x) cancel(-a)))^2)=
$= \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} - \overline{x}\right)}^{2}} = \sigma$
The standard deviation of the new set is equal to the deviation of the set before translation.
QED
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# How do you solve 2/3x+3=x-2?
Mar 11, 2018
$x = 15$
#### Explanation:
First let's get all like-terms on the same sides.
To do so, we can subtract 3 from both sides and subtract $x$ from both sides as well.
$\frac{2}{3} x + 3 - 3 - x = x - 2 - 3 - x$
Simplify:
$\frac{2}{3} x - x = - 2 - 3$
Now we can combine like-terms:
$- \frac{1}{3} x = - 5$
In order to isolate $x$, multiply by -3 on both sides:
$- 3 \cdot - \frac{1}{3} x = - 5 \cdot - 3$
Simplify:
$x = 15$
Mar 11, 2018
$x = 15$
#### Explanation:
So the first thing you would want to do is to move the $\frac{2}{3} x$ over to the right side of the equation, making it $- \frac{2}{3} x$.
Subtract $x - \frac{2}{3} x$. Since the is no coefficient, $x$ means #3/3x, so
$x - \frac{2}{3} x = \frac{1}{3} x$
Now move the whole number $- 2$ to the left side, making it
$3 + 2 = 5$
Multiply both sides by $3$ to get $x$ alone.
$x = 15$
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The Algorithm to Solve Rubik’s Cube – Cracked and Explained
# The Algorithm to Solve Rubik’s Cube – Cracked and Explained
To the average Joe, an unsolved Rubik’s Cube remains one of the most elusive and difficult conundrums known to man. How in the world are you supposed to get this mangled, jumbled mess back in the right order again?
After a few confident-looking twists this way and that way, most of us throw in the towel (or, rather, throw out the Cube) in frustration before we even come close to solving the thing. But, did you know there’s a simple, relatively easy way to solve Rubik’s Cubes every single time? All it takes is a small algorithm.
## The Eight Ways to Rotate a Rubik’s Cube
To understand the following algorithm, it’s necessary to establish the eight basic ways you can actually rotate Rubik’s Cubes. Understanding these moves and their abbreviations is integral to the steps listed below.
Also important to know? You should always hold the Rubik’s Cube with one of the six faces pointed at you. And so, the eight ways to rotate a Rubik’s Cube are as follows:
• R: A clockwise rotation of the right layer.
• R’: A counterclockwise rotation of the right layer.
• L: A clockwise rotation of the left layer.
• L’: A counterclockwise rotation of the left layer.
• U: A clockwise rotation of the top layer.
• U’: A counterclockwise rotation of the top layer.
• F: A clockwise rotation of the front layer.
• F’: A counterclockwise rotation of the front layer.
## How to Solve a Rubik’s Cube Every Time Without Fail
Before we dive in, let’s lay out some ground rules. To solve a Rubik’s Cube, you need to understand the basic principles of the hallowed Cube.
For one, the colored square located in the middle of each side cannot change. That’s that particular side’s set color. Additionally, there are eight corners, each with three distinct colors, and 12-edge pieces between these eight corners, each with two distinct colors.
Looking at the steps listed below, it might seem like this process is too convoluted for you to remember off the top of your head. However, once we get into it, you’ll find that it’s just a lot of repetition. That’s all an algorithm is, after all. Simply put, it’s a particular sequence of specific instructions meant to be followed to the exact letter again and again.
The solving of a Rubik’s Cube can be basically summarized as the shuffling of these corner and edge pieces around the six center squares. While the Rubik’s Cube box likes to scare you by boasting that there are 43 quintillion different configurations per Cube, there’s one simple way to solve a Rubik’s Cube every time without fail. Let’s get into it.
## Step 1: Make a White Cross
Firstly, turn the Rubik’s Cube so that the white centerpiece is facing you. Then, rotate the Cube so that the white edges make a cross with the white centerpiece. It doesn’t matter what color the corners are at this point in time. Simple, right? You’re already good to move on to the next step.
## Step 2: Match Up Centers and Edges
Secondly, look at the color on each edge of the white cross. For our next step, you need to match these colors with their corresponding centerpieces. You can do this by matching the colors of the four corresponding centerpieces with the edges of the white cross. (Make sure to restore the white cross before moving on.)
## Step 3: Set the Corners
Thirdly, you need to set your white corners in place. Here, we’ll employ our first algorithm to help make it happen: R U R’ U’ (and repeat). Judging by our eight basic moves outlined above, that’s a clockwise rotation of the right layer, a clockwise rotation of the top layer, a counterclockwise rotation of the right layer, and a counterclockwise rotation of the top layer.
Now, repeat the algorithm for all the remaining corners. When complete, the entire white face of the Rubik’s Cube should be done and all four corresponding centerpieces with edges touching the white side should resemble a Tetris T-block.
## Step 4: Complete the Second Layer
Now, we’re going to focus on those T-block shapes on four of these six sides. Our goal for this step is to complete the second layer on all four of these sides to take the T-block shape and make them into rectangles, leaving only the top layer on each of these four lateral sides unsolved. Once again, we’ll be resorting to algorithms.
For the matching color on the right side of the centerpiece, use the algorithm U R U’ R’ U’ F’ U F. For the matching color on the left side of the centerpiece, use the algorithm U’ L’ U L U F U’ F’. Repeat this on all four sides until you have the white side complete and the bottom two layers of the four lateral sides complete.
## Step 5: Make a Yellow Cross
Now, it’s time to make another cross. You might have noticed that the yellow side is looking pretty jumbled on the bottom of the Cube. For this step, we’re going to be making a yellow cross — just like we did with the white cross in the first step, only on the opposite side.
Algorithm time: move the Cube using the F R U R’ U’ F’ algorithm. You will likely have to repeat this algorithm two or three times in order to get all the yellow pieces into a cross shape, but be patient and trust the math used to solve Rubik’s Cubes.
## Step 6: Match Edges to the Top Layer
We’re getting closer to the end here! A quick reminder of just how important it is to follow these steps to the letter: slipping up or forgetting a move at this point might send you back to square (or should we say cube?) one.
For this step, our goal is to bring all the yellow pieces to the bottom of the Cube so that they match up with the edge. Using the algorithm F R U R’ U’ F’ should make this happen pretty quickly. If you can’t get the corners in the right place, you can simply position the Cube in any way — so long as the unmatched pieces are at the top — and then use the algorithm U R U’ L’ U R’ U’ L a handful of times.
## Step 7: Arrange the Corners
Lastly, we’re going to get those yellow corners and edges in place. Starting from any corner you choose, use the algorithm U R’ U’ R, then rotate the top layer until another mismatched corner is on the top right-hand side. Repeat the algorithm U R’ U’ R again, then repeat this process as many times as needed.
After all the corner pieces are in place, you’ll merely need to move the yellow side one or two times to finish the job. (You’ll know exactly what to do when you end up here.) With that, you now know how to solve s Rubik’s Cube like a total pro!
Now, it’s just a matter of mastering that handful of algorithms. After you get those down with continued practice, you’ll have the coolest party trick in the books.
## How Algorithms Solve Rubik’s Cube
Since its invention in 1974 by Hungarian architect and sculptor Ernő Rubik, the Rubik’s Cube has been puzzling generations upon generations of kids and adults both young and old. While traditionally marketed and sold as a toy, Rubik’s Cubes are actually just an extension of a few basic mathematic principles. (Unsurprising, considering it was conceived by an architect.)
With six sides, eight corners, 12 edges, and 54 separate squares on a single Rubik’s Cube, it’s not hard to believe that there are more than 43 quintillion different possible layouts. (That’s 18 zeroes!) Thankfully, the basic pillars of these Rubik’s Cube algorithms remain the same — regardless of how many different ways there are to scramble the thing.
Thanks to algorithms, true masters can solve a Rubik’s Cube in as little as 20 moves. However, the above algorithms used are what’s known as the Layer by Layer method. Using what algebraists would call “group theory,” the Rubik’s Cube can be reduced to less than 10 simple routines. More specifically, algorithms solve Rubik’s Cubes through the magic of permutation groups — a branch of group theory.
Without getting too complex, the basic thinking is that each one of the Rubik’s Cube’s movable parts — 48 out of 54, excluding the six centerpieces — can be labeled with numbers 1 through 48. Each scramble of a Rubik’s Cube is a new permutation of those numbers. Once you break it down into numbers, it’s only a matter of repeating particular moves (i.e. algorithms) over and over.
Why is the Rubik's Cube so hard?
While the Rubik’s Cube might seem impossible from the outside looking in, the truth is that Rubik’s Cube experts will tell you it’s as simple as 20 quick moves.
What makes the Rubik’s Cube seem so much harder than it actually is? All the moving parts. There are 20 different moving parts, and it sends our brains into overdrive trying to figure out how they all work together.
How many different ways are there to solve a Rubik's Cube?
There are 43,252,003,274,489,856,000 different ways to scramble up a Rubik’s Cube. The number of different ways to solve the Cube is just as impossible to imagine. The best way to solve one? Memorize the algorithms.
Who invented the Rubik's Cube?
The Rubik’s Cube was invented by Ernő Rubik, a Hungarian architect and professor who created the devilishly difficult toy in 1974. He has other mathematical puzzles to his name, as well — including the Rubik’s Snake — but none as popular as the titular cube.
What's the fastest time someone has solved a Rubik's Cube?
The current record for the fastest Rubik’s Cube solution is 3.47 seconds, accomplished by Yusheng Du in 2018.
What's the biggest Rubik's Cube you can buy?
Believe it or not, there are Rubik’s Cubes for sale that are as large as 100×100. That’s going to take more than a few algorithms to solve — and there’s no way anyone’s solving that one in 3.5 seconds, either.
#### Nate Williams, Author for History-Computer
Nate Williams is a writer at History-Computer primarily covering smart technology, streaming services, consumer electronics, and how-to guides. Nate has been writing about tech for seven years and holds a Master’s Degree from the University of Missouri-St. Louis, which he earned in 2022. A resident of Missouri, Nate spends much of his free time playing video games, watching movies, and thinking about adding another speaker to his home theater setup.
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1. ## Inverse function
Hi, can someone please help me with a problem? I have to find the inverse function, but I have NO idea about how to do it. To find inverses, we always used the "make x the subject of the equation, then swap x for y" method, but this doesn't seem to work here.
"Find $f^{-1}(x)$ given that $f(x)=\frac{\arcsin{x}}{\ln{x}}$ "
Thanks.
2. Actually it does work!
For the function $f(x) = \frac{arcsin(x)}{\ln{x}}$
$y = \frac{arcsinx}{\ln{x}}$
$y\ln{x} = arcsin(x)$
$arcsin{x} =(y\ln{x})$
$y = sin (x \ln{x})$
Hope that helps you!
3. Originally Posted by mollymcf2009
$arcsin{x} =(y\ln{x})$
$y = sin (x \ln{x})$
How does this last step work. If you are just exchanging x for y then shouldn't you get
$y = sin (x \ln{y})$
?
4. I'm sorry I should not have put parentheses around that second part
The answer is y = sin(x) ln(x)
5. Originally Posted by mollymcf2009
I'm sorry I should not have put parentheses around that second part
The answer is y = sin(x) ln(x)
Sorry. I still don't see how you get that from
$\arcsin x = y \ln x$
even if you do exchange variables.
I can't see how to express an inverse function of
$y = \frac{\arcsin x}{\ln x}$
$y \ln x = \arcsin x$
$\ln x^y = \arcsin x$
$x = \sin\left(\ln x^y\right)$
$x = \frac{1}{2i}\left(x^y - x^{-y}\right)$
so basically x is the root of an equation
$x^y - 2ix - x^{-y} = 0$
which for a general $y$ has no explicit solution.
but maybe I'm missing something...
6. Originally Posted by Rincewind
Sorry. I still don't see how you get that from
$\arcsin x = y \ln x$
even if you do exchange variables.
I can't see how to express an inverse function of
$y = \frac{\arcsin x}{\ln x}$
$y \ln x = \arcsin x$
$\ln x^y = \arcsin x$
$x = \sin\left(\ln x^y\right)$
$x = \frac{1}{2i}\left(x^y - x^{-y}\right)$
so basically x is the root of an equation
$x^y - 2ix - x^{-y} = 0$
which for a general $y$ has no explicit solution.
but maybe I'm missing something...
The thing you're missing, and in fact the thing that we're all missing, is the exact question. We only have the OP's version of it .......
Here is a scenario that could lead to the posted question:
Given that $f(x) = \frac{\arcsin x}{\ln x}$ find the value of $f^{-1}\left( - \frac{\pi}{6 \ln 2}\right)$.
Or perhaps a scenario where the value of the derivative of the inverse function is asked for ....
7. I hate myself.
Since you guys said that there was something missing, I checked the question again. The exercise was written in an extremely small font size, and it turns out that it didn't ask for $f^{-1}(x)$ but for $f'(x)$, which is perfectly solvable.
Sorry (and I will bring a magnifying glass the next time I try to solve a math problem)!
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# Angles in triangle
The triangle is ratio of the angles β:γ = 6:8. Angle α is 40° greater than β. What are the size of angles of the triangle?
Result
β = 82 °
γ = 42 °
α = 56 °
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you have a system of equations and looking for calculator system of linear equations? See also our trigonometric triangle calculator.
## Next similar examples:
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Calculate the sides of triangle ABC with area 1404 cm2 and if a: b: c = 12:7:18
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Mr. Billy calculated that excavation for a water connection dig for 12 days. His friend would take 10 days. Billy worked 3 days alone. Then his friend came to help and started on the other end. On what day since the beginning of excavation they met?
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In stock are three kinds of branded coffee prices: I. kind......248 Kč/kg II. kind......134 Kč/kg III. kind.....270 Kč/kg Mixing these three species in the ratio 10:7:7 create a mixture. What will be the price of 1100 grams of this mixture?
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In the forest is employed 56 laborers planting trees in nurseries. For 8 hour work day would end job in 37 days. After 16 days, 9 laborers go forth? How many days are needed to complete planting trees in nurseries by others, if they will work 10 hours a d
8. Motion
If you go at speed 3.7 km/h, you come to the station 42 minutes after leaving train. If you go by bike to the station at speed 27 km/h, you come to the station 56 minutes before its departure. How far is the train station?
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Two squares whose sides are in the ratio 5:2 have sum of its perimeters 73 cm. Calculate the sum of area this two squares.
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Cyclist started out of town at 19 km/h. After 0.7 hours car started behind him in the same direction and caught up with him for 23 minutes. How fast and how long went car from the city to caught cyclist?
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The surfaces of two cubes, one of which has an edge of 22 cm longer than the second are differ by 19272 cm2. Calculate the edge length of both cubes.
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If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?
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A fisherman buys carnivores to fish. He could buy either 6 larvae and 4 worms for \$ 132 or 4 larvae and 7 worms per \$ 127. What is the price of larvae and worms? Argue the answer.
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Adelka, Barunka, and Cecilka are weight in pairs. Adelka with Barunka weighs 98 kg, Barunka with Cecilka 90 kg and Adelka with Cecilka 92 kg. How much does each of them weigh?
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The volume of water in the rectangular swimming pool is 6998.4 hectoliters. The promotional leaflet states that if we wanted all the pool water to flow into a regular quadrangle with a base edge equal to the average depth of the pool, the prism would have.
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In the yard were chickens and rabbits. They had 28 heads and 82 legs. How many chickens and how many rabbits was in the yard?
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Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Sum of speed of boat in downstream and in upstream is 36 km/hr while speed of stream is 75% less than speed of boat in still water.Quantity I: Time taken to cover 90 km downstreamQuantity II: Time taken to cover 108 km upstream
1. Quantity I > Quantity II
2. Quantity I < Quantity II
3. Quantity I ≥ Quantity II
4. Quantity I ≤ Quantity II
5. Quantity I = Quantity II
Option 2 : Quantity I < Quantity II
Detailed Solution
Let the speed of boat in still water be X and the speed of stream be Y
From given, we have
Speed of stream is 75% less than speed of boat in still water
Y = X – (75/100) X
⇒ Y = 0.25X
Downstream speed = sum of speed of stream and boat in still water = X + Y
Upstream speed= difference between speed of stream and boat in still water = X - Y
Also give that,
Downstream speed + upstream speed= 36
(X + Y) + (X - Y) = 36
⇒ X = 18
Then Y = 0.25 × 18 = 4.5
Downstream speed = 18 + 4.5 = 22.5
Upstream speed = 18 – 4.5=13.5
Quantity I:
Time taken to cover 90 km downstream = 90/22.5 = 4hrs
Quantity II:
Time taken to cover 108 km upstream = 108/13.5 = 8hrs
∴ Quantity I < Quantity II
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HomeTemplate ➟ 20 20 4th Grade Abeka Math Worksheets
# 20 4th Grade Abeka Math Worksheets
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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# 3 Inertia force - Eka Oktariyanto Nugroho
```Inertia Forces
by
Eka Oktariyanto Nugroho
Basic Equation 3
Eka O. N.
3.1.
MASS, INERTIA, AND ACCELERATION
3.1.1.
The Newton Equation
To cause the motion of a constant mass M , or, more generally, to change the state of an existing
motion, it is necessary to apply to this mass a force F , which causes an acceleration dV dt such
that
F M dV dt . This is a vector relationship, i.e., true for both magnitude and direction. The
M dV dt is the inertia force, which characterizes the natural resistance of matter to any
product
change in its state of motion.
The considered mass M is the mass of a unit of volume of fluid
M unit of volume
is the density. Hence the fundamental equation of momentum has the
form F dV dt . Its three components along the three coordinate axes OX , OY , OZ
where
are
3.1.2.
du dt , dv dt , and dw dt , respectively.
Relationships between the Elementary Motions of a Fluid Particle and the Inertia Terms
To each kind of motion of the fluid particles (chapter 1) there corresponds an inertia force. The
relationship between the kind of motion described and the corresponding inertia force is
straightforward.
The elementary components of velocity of a fluid particle as given in chapter 1 are, in the case of a
two-dimensional motion,
Translation
u, v
Dilatational deformation
u
dx
x
Shear deformation
1 u v
dy
2 y x
1 u v
dx
2 y x
Rotation
1 v u
dy
2 x y
1 v u
dx
2 x y
v
dy
y
To each of these velocity components corresponds a component of acceleration, which multiplied
by , yields a component of inertia force.
Two types of inertia forces may be distinguished, depending on the type of acceleration or
elementary motion considered. These are:
1. Local acceleration-corresponding to a variation of the velocity of translation or the
derivative of velocity with respect to time.
2. Convective acceleration-corresponding to a variation of velocity of deformation and rotation
or derivative of velocity with respect to space.
The physical meaning of these accelerations and the corresponding inertia forces is first examined:
then their mathematical expression is demonstrated. Chapter 4 deals with the applied forces F
which have to be equated to these inertia forces to obtain the momentum equation.
Inertia Forces
Page
- 15
Basic Equation 3
Eka O. N.
3.2.
LOCAL ACCELERATION
Local acceleration characterizes any unsteady motion, i.e., motion where the velocity at a given
point changes with respect to time. Local acceleration results from a change in the translatory
motion of a fluid particle imposed by external forces F.
Mathematical Expression of Local Inertia
The mathematical expression of the inertia forces caused by a local acceleration is given by the
change in the velocity of the translatory motion with respect to time only. The corresponding inertia
V t
force is equal to
respectively:
of which the components along the three axes are,
u t , v t , and w t . The derivatives with respect to space are not
taken into account.
3.3.
CONVECTIVE ACCELERATION
Convective acceleration characterizes any nonuniform flow, i.e., when the velocity at a given time
changes with respect to distance. It is sometimes called field acceleration. Convective acceleration
results from any linear or angular deformation, or from a change in the rotation of fluid particles,
imposed by external forces F.
3.3.1.
The Case of Linear Deformation
In a convergent pipe, it has been seen that the velocity of a fluid particle, although constant with
time at a fixed location, tends to increase along the converging streamlines. The velocity of the fluid
particle increases with respect to space. This is a positive convective acceleration. The fluid tends
to resist this acceleration by convective inertia.
In a divergent conduit, the velocity decreases and the fluid tends to continue its motion with the
same velocity because of its inertia. The applied forces cause a negative convective acceleration.
Expansion or contraction of a compressible fluid is the sum of linear deformations and also results
in corresponding inertia forces.
It has been seen that the linear deformation velocity components are those given in Equation 3-1.
u
dx
x
v
dy
y
w
dz
z
(3-1)
Two-dimensional motion
Three-dimensional motion
The expressions u x , v
corresponding acceleration is
y and w z are given time, as seen in section 1.3.1 The
d u u dx
dx
dt x x dt
Two similar expressions result for
w and u . If u
dx
dy
dz
, v
, and w
are substituted in
dt
dt
dt
these expressions and the result is multiplied by the density, the inertia forces are obtained. They
are:
u
2
u 1 u
x 2
x
2
v 1 v
v
y 2
y
Inertia Forces
Page
- 16
Basic Equation 3
Eka O. N.
w
w 1 w
z 2
z
2
It should be notice that the last group of expressions may be written as x
u 2 . This
2
shows that the inertia force is equal to the derivative of the kinetic energy with respect to space
along the three direction axes OX , OY , and OZ , respectively.
3.3.2.
The Case of Shear Deformation
In a bend, where the fluid particles are angularly deformed, the fluid paths are curved and because
of its inertia, the fluid tends to continue along a straight line. This causes a centrifugal force
proportional to the change of direction which is imposed by the applied forces.
It is possible for the velocity of a fluid particle to keep the same magnitude along its path, but with a
change in direction. This is the case of free vortex motion.
It has been seen that the velocity components of angular deformation for a two-dimensional motion
are
1 u v
dy
2 y x
1 u v
dx
2 y x
Hence, as in the previous case, using the substitutions u
dx
dy
, v
, the corresponding inertia
dt
dt
forces become:
1 u v
v
2 y x
1 u v
u
2 y x
3.3.3.
The Case of a Change of Rotation
In the entrance to a pipe (Fig. 3.1), because of the change in friction forces, there is a variation of
rotation of the fluid particles. Hence there are inertia forces corresponding to the natural resistance
of the fluid to change its rotational motion. In a uniform pipe, the rotation of particles exists but
there is no change in rotational magnitude and the corresponding acceleration is zero.
Figure 3. 1 Zone of acceleration of rotation.
Inertia Forces
Page
- 17
Basic Equation 3
Eka O. N.
As in the two previous cases, since
1 v u
dy
2 x y
1 v u
dx
2 x y
are the velocities of the components of rotation in a two dimensional motion, the corresponding
inertia forces obtained are
1 v u
v
2 x y
1 v u
u
2 x y
It has been shown that it is possible to assume that the motion is irrotational when friction effects
are negligible. It is evident that the same conditions lead to neglect of rotational inertia forces.
3.4.
GENERAL MATHEMATICAL EXPRESSIONS OF INERTIA FORCES
3.4.1.
Local and Convective Acceleration
In the general case both local acceleration and convective acceleration occur at the same time. A
simple example is when a fluid oscillates in a nonuniform curved pipe. Hence, in the general case,
V and its components u, v, and w are functions of both time and space coordinates. For example,
u x, y, z, t . The total differential of u is
du
u
u
u
u
dt dx dy dz
t
x
y
z
The acceleration in the
x direction is thus given by the total differential of u , with respect to time:
du u u dx u dy u dz
dt t x dt y dt z dt
Similar expressions occur for
dv dt and dw dt .
Substituting u : dxldt, u : dyldt, and w : dzldt, and multiplying by the density p, the inertia forces
given by Equation 3-2 are obtained.
u
u
u
u
u v w
x
y
z
t
v
v
v
v
u v w
x
y
z
t
w
t
u
Local
acceleration
terms
3.4.2.
(3-2)
w
w
w
v
w
x
y
z
Convective accleration terms
Elementary Acceleration Components
Following a procedure similar to that used in the study of the elementary motions of fluid particles
(Section 1.5.2), that is, adding and subtracting
1
1
v v x and w w x to the first line
2
2
above, gives Equation 3-3, which emphasize the previous physical considerations. Similar forms
can be obtained for the y and z components of the forces.
Inertia Forces
Page
- 18
Basic Equation 3
Eka O. N.
u
t
u
1 v u 1 u w 1 u w 1 v u (3-3)
v w
w
v
2 x y 2 z x 2 z x 2 x y
Acceleration
in linear
deformation
Local
Acceleration
resulting in
a change in
translatory
motion
3.4.3.
u
x
Acceleration in angular deformation
Acceleration in rotation
Separation of Rotational Terms
It is often useful to transform the acceleration terms to a form which emphasizes both the kinetic
energy terms and the rotational terms. Adding and subtracting
v v x w w x to
the
first line, gives the following expression, valid along the OX axis:
u u
v
w u v
u w
u v w v w
x
x y x
z x
t x
But
u
u
v
w 1 2 2
V2
v w
u v w2
x
x
x 2 x
x 2
When the coefficients of the rotational vector
u w
2
z x
v u
2
x y
Are introduced, the following expression for the inertia forces along the OX axis results
u V 2
2 w v
t x 2
Similarly, it may be found that the inertia forces along the OY and OZ axes are
v
t
w
t
V 2
2 u w
y 2
V 2
2 v u
z 2
These three expressions may be written more concisely in vector notation as shown in Eq. 3-4.
V
t
Local
acceleration
V2
2
Kinetic energy
term
curl V x V
(3.4)
Rotational
term
Convective acceleration
V 2
V2
V 2
V 2
i
j
k
2
z 2
x 2 y 2
It has to be noticed that the convective inertia term is, in fact, the derivative with respect to space of the
kinetic energy,
Inertia Forces
V 2 2 , of the particle.
Page
- 19
```
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+0
# true or false
0
128
3
There are other methods to solve a system of equations besides graphing?
Feb 23, 2021
#1
0
Yes....substitution and elimination are two other methods...... or post on Web2.0 haha
Feb 23, 2021
#2
+956
0
1. Graphing: graph the system of equations and find the intersection points, which are the solutions.
2. Elimination. As an example:
x + 2y = 3 . (1)
3x + y = 4 . (2)
You can eliminate one variable by multiplying equation (2) by 2:
x + 2y = 3
6x + 2y = 8
-----------------
-5x = -5
By subtracting the equations, we have eliminated one variable, thus solving for the other.
3. Substitution. Let's use the same equations as last time. In equation (1), solve for x in terms of y:
x + 2y = 3 -----> x = 3 - 2y
Plug this in the other equation:
3x + y = 4
3(3 - 2y) + y = 4
Thus, we can solve for y, and in turn, solve for x.
Feb 23, 2021
edited by CubeyThePenguin Feb 23, 2021
#3
+74
+1
Or use Wolfram|Alpha
Feb 23, 2021
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# Circle and Lattice Points
• Difficulty Level : Medium
• Last Updated : 20 Jun, 2022
Given a circle of radius r in 2-D with origin or (0, 0) as center. The task is to find the total lattice points on circumference. Lattice Points are points with coordinates as integers in 2-D space.
Example:
```Input : r = 5.
Output : 12
Below are lattice points on a circle with
radius 5 and origin as (0, 0).
(0,5), (0,-5), (5,0), (-5,0),
(3,4), (-3,4), (-3,-4), (3,-4),
(4,3), (-4,3), (-4,-3), (4,-3).
are 12 lattice point.```
To find lattice points, we basically need to find values of (x, y) which satisfy the equation x2 + y2 = r2
For any value of (x, y) that satisfies the above equation we actually have total 4 different combination which that satisfy the equation. For example if r = 5 and (3, 4) is a pair which satisfies the equation, there are actually 4 combinations (3, 4) , (-3,4) , (-3,-4) , (3,-4). There is an exception though, in case of (0, r) or (r, 0) there are actually 2 points as there is no negative 0.
```// Initialize result as 4 for (r, 0), (-r. 0),
// (0, r) and (0, -r)
result = 4
Loop for x = 1 to r-1 and do following for every x.
If r*r - x*x is a perfect square, then add 4
tor result. ```
Below is the implementation of above idea.
## CPP
`// C++ program to find countLattice points on a circle``#include``using` `namespace` `std;` `// Function to count Lattice points on a circle``int` `countLattice(``int` `r)``{`` ``if` `(r <= 0)`` ``return` `0;` ` ``// Initialize result as 4 for (r, 0), (-r. 0),`` ``// (0, r) and (0, -r)`` ``int` `result = 4;` ` ``// Check every value that can be potential x`` ``for` `(``int` `x=1; x
## Java
`// Java program to find``// countLattice points on a circle` `class` `GFG``{` `// Function to count``// Lattice points on a circle``static` `int` `countLattice(``int` `r)``{`` ``if` `(r <= ``0``)`` ``return` `0``;`` ` ` ``// Initialize result as 4 for (r, 0), (-r. 0),`` ``// (0, r) and (0, -r)`` ``int` `result = ``4``;`` ` ` ``// Check every value that can be potential x`` ``for` `(``int` `x=``1``; x
## Python3
`# Python3 program to find``# countLattice points on a circle` `import` `math` `# Function to count Lattice``# points on a circle``def` `countLattice(r):` ` ``if` `(r <``=` `0``):`` ``return` `0` ` ``# Initialize result as 4 for (r, 0), (-r. 0),`` ``# (0, r) and (0, -r)`` ``result ``=` `4` ` ``# Check every value that can be potential x`` ``for` `x ``in` `range``(``1``, r):`` ` ` ``# Find a potential y`` ``ySquare ``=` `r``*``r ``-` `x``*``x`` ``y ``=` `int``(math.sqrt(ySquare))` ` ``# checking whether square root is an integer`` ``# or not. Count increments by 4 for four`` ``# different quadrant values`` ``if` `(y``*``y ``=``=` `ySquare):`` ``result ``+``=` `4`` ` ` ``return` `result`` ` `# Driver program``r ``=` `5``print``(countLattice(r))` `# This code is contributed by``# Smitha Dinesh Semwal`
## C#
`// C# program to find countLattice``// points on a circle``using` `System;` `class` `GFG {` ` ``// Function to count Lattice`` ``// points on a circle`` ``static` `int` `countLattice(``int` `r)`` ``{`` ``if` `(r <= 0)`` ``return` `0;`` ` ` ``// Initialize result as 4`` ``// for (r, 0), (-r. 0),`` ``// (0, r) and (0, -r)`` ``int` `result = 4;`` ` ` ``// Check every value that`` ``// can be potential x`` ``for` `(``int` `x = 1; x < r; x++)`` ``{`` ` ` ``// Find a potential y`` ``int` `ySquare = r*r - x*x;`` ``int` `y = (``int``)Math.Sqrt(ySquare);`` ` ` ``// checking whether square root`` ``// is an integer or not. Count`` ``// increments by 4 for four`` ``// different quadrant values`` ``if` `(y*y == ySquare)`` ``result += 4;`` ``}`` ` ` ``return` `result;`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int` `r = 5;`` ` ` ``Console.Write(countLattice(r));`` ``}``}` `// This code is contributed by nitin mittal.`
## PHP
``
## Javascript
``
Output:
`12`
Time Complexity: O(1)
Auxiliary Space: O(1)
Reference:
http://mathworld.wolfram.com/CircleLatticePoints.html
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
My Personal Notes arrow_drop_up
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Question
# The roots of the equation $${x}^{2}-\cfrac{2}{3}x=32$$ are
A
6,154
B
6,163
C
3,16
D
163,6
Solution
## The correct option is C $$\cfrac{-16}{3},6$$$$x^2-\dfrac{2}{3}x=32$$$$\Rightarrow$$ $$3x^2-2x=96$$$$\Rightarrow$$ $$3x^2-2x-96=0$$$$\Rightarrow$$ $$3x^2-18x+16x-96=0$$$$\Rightarrow$$ $$3x(x-6)+16(x-6)=0$$$$\Rightarrow$$ $$(x-6)(3x+16)=0$$$$\Rightarrow$$ $$x-6=0$$ and $$3x+16=0$$$$\therefore$$ $$x=6$$ and $$x=\dfrac{-16}{3}$$Mathematics
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# physics
A baseball bat hits a ball with an average force of 970N that acts for 0.0088s.
a) what impulse was given to the ball/
b)What was the change of momentum of the ball?
c)The ball was hit back in the same direction that it came from. If its speed before being hit was 32ms-1, what was its speed afterwards? (Mass of baseball was 145g)
please help I would really appreciate you for the rest of my life. and is physics that important? its hard for nothing really
1. 0
2. 1
1. Force * time = impulse or change of momentum
a) so Impulse = 970*0.0088
b) As I said, impulse IS change of momentum
c) initial momentum = -.145*32
(calling negative speed from pitcher to batter, positive from batter to center field)
The change of momentum is up in parts a and b
so
.145 v = -.145*32 + answer to part b
posted by Damon
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# 5 Best Ways to Compute the Inverse of an n-Dimensional Array in Python
Rate this post
π‘ Problem Formulation: In the realm of linear algebra, computing the inverse of an n-dimensional array (typically a matrix) is a fundamental operation, enabling solutions to linear equations and transformations. In Python, this task can be performed using various methods, each suited to different scenarios and array types. For our purposes, we assume that the input is a square n x n array for which an inverse exists. The expected output is another n x n array representing the inverse of the input array.
## Method 1: Using NumPy’s `inv` function
The NumPy library’s `inv` function provides an efficient and widely-used approach to compute the inverse of an n-dimensional array. NumPy is optimized for numerical computations and leverages highly optimized C and Fortran code under the hood. The `numpy.linalg.inv()` function specifically computes the multiplicative inverse of a matrix.
Here’s an example:
```import numpy as np
# Define a 2x2 array
A = np.array([[3, 2], [1, 4]])
# Compute the inverse
A_inv = np.linalg.inv(A)
print(A_inv)
```
The output of this code snippet:
```[[ 0.4 -0.2]
[-0.1 0.3]]
```
This code snippet first imports the NumPy library, then defines a 2×2 matrix `A`. The function `np.linalg.inv()` is used to calculate the inverse of this matrix, and the result `A_inv` is printed. The output shows the computed inverse of the original matrix.
## Method 2: Using SymPy’s `inv` Method
SymPy is a Python library for symbolic mathematics. It can compute the inverse of a matrix exactly, without numerical approximation. The method `Matrix.inv()` from SymPy takes a matrix and returns its inverse if it exists. This can be particularly useful when working with symbolic variables or when exact results are needed.
Here’s an example:
```from sympy import Matrix
# Define a 2x2 matrix with symbols
A = Matrix([[3, 2], [1, 4]])
# Compute the inverse
A_inv = A.inv()
print(A_inv)
```
The output of this code snippet:
```Matrix([
[ 2/5, -1/5],
[-1/10, 3/10]])
```
This example uses SymPy to define a 2×2 symbolic matrix and then calculates its inverse using the `inv()` method. The result is printed as an exact fraction rather than a decimal, showcasing the benefit of using a symbolic computation library for this task.
## Method 3: Using scipy.linalg’s `inv` Function
The SciPy library, which extends NumPy, offers additional functionality for scientific computing. The `scipy.linalg.inv()` function provides an alternative way to compute the inverse of a matrix. While similar to NumPy’s inverse method, SciPy’s version sometimes offers performance improvements due to its specific optimization.
Here’s an example:
```from scipy.linalg import inv
# Define a 2x2 array
A = np.array([[3, 2], [1, 4]])
# Compute the inverse using SciPy
A_inv = inv(A)
print(A_inv)
```
The output of this code snippet:
```[[ 0.4 -0.2]
[-0.1 0.3]]
```
Here, the `inv` function from the SciPy library’s `linalg` (linear algebra) module is used to compute the inverse of a defined 2×2 array. The output mirrors the result obtained with NumPy, displaying its effectiveness in such computations.
## Method 4: Using the Gaussian Elimination Technique
The Gaussian elimination method, or row reduction, is a classic algorithm in linear algebra for solving systems of linear equations. It can also be used to compute the inverse of a matrix by augmenting the original matrix with the identity matrix and performing row operations. Implementation is more complex and is typically used for educational purposes rather than in production code due to the existence of optimized libraries.
Here’s an example:
```# A manual implementation would be needed here. Omitted for brevity.
```
This method, while fundamental in understanding matrix inversions, is less practical for coding purposes, hence more theoretical and not provided in full detail here.
## Bonus One-Liner Method 5: Using NumPy’s `pinv` for Pseudo-Inverse
Sometimes a matrix may not have an exact inverse, particularly if it is not square or if it is singular. For these cases, the pseudo-inverse, computed with the `np.linalg.pinv()` function, provides a best-fit approximation that minimizes the least squares error.
Here’s an example:
```import numpy as np
# Define a non-square or singular 2x3 array
A = np.array([[3, 2, 1], [1, 4, 5]])
# Compute the pseudo-inverse
A_pinv = np.linalg.pinv(A)
print(A_pinv)
```
The output of this code snippet:
```# The output will be a 3x2 matrix representing the pseudo-inverse of A.
```
This concise method utilizes NumPy’s `pinv` to calculate a pseudo-inverse for arrays that cannot be inverted using the standard inverse function, offering a practical alternative for these special cases.
## Summary/Discussion
• Method 1: NumPy inv Function. Strengths: Efficient, simple to use, well-optimized. Weaknesses: Depends on the NumPy library, unsuitable for symbolic calculations.
• Method 2: SymPy inv Method. Strengths: Provides exact results, great for symbolic math. Weaknesses: May not be as efficient as numerical methods, requires SymPy package.
• Method 3: scipy.linalg inv Function. Strengths: Offers potential performance improvements over NumPy. Weaknesses: Requires additional SciPy library, similar to NumPy’s functionality.
• Method 4: Gaussian Elimination Technique. Strengths: Educational tool, no library dependencies. Weaknesses: Computationally intensive, manual implementation complexity.
• Method 5: NumPy pinv for Pseudo-Inverse. Strengths: Provides solutions for non-square or singular matrices. Weaknesses: May not be suitable where an exact inverse is required.
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# Spray – Lat 5degN Long 26deg 30’W
## Sailing Alone Around the World
One of my favorite books is “Sailing Alone Around the World ” by Canadian born Joshua Slocum. On July 2nd 1895 he left the port of Yarmouth Nova Scotia and left North American behind (Ref.1/p.23). In his book he writes: “On September 25, in the latitude of 5degN, longitude 26deg 30′ W, I spoke the ship North Star of London. The great ship was out forty-eight days from Norfolk, Virginia, and was bound for Rio, where we met again about two months later”.
## Position Determination
Let’s look at Celestial Navigation to help us understand the Spray’s whereabouts on September 25th 1895. Captain Slocum was not a fan of chronometers, “At Yarmouth, too, I got my famous tin clock, the only timepiece I carried on the whole voyage” (Ref.1/p.22). First let’s assume that Capt Slocum had accurate time and he performed a Meridian Passage on the Sun, let’s try to figure out what his sextant reading was. First let’s use the current Nautical Almanac for 2021 (Ref.3). We need to interpolate to determine the GHA & DEC for when the Sun is directly over 26deg 30’W (Figure 2):
1300UTC: GHA=17deg 6.6′
1400UTC: GHA= 32deg 6.8′
delta = 9deg 23.4′
MP = GHA = 26deg 30′ & DEC = S01deg 4.5′ & UTC = 13hrs 37.55min
Figure 3 shows the geometry of the position. The Spray’s Zenith distance is equal to the latitude north of 5deg + the Sun’s dec south = 6deg 4.5min. Thus Ho = 83deg 55.5min.
Assuming a lower limb reading & height of eye = 4m
Ho = Ha -R + PA +/-SD
R = 1/tan(H + 7.32/(H+4.32)) = 0.1min
SD = 15.9min
Dip = 3.5min
Ha = 83deg 55.5min + 0.1min – 15.9min = 83deg 39.7min
Ha = Hs +/- SxtntEr +/-IndxEr – Dip
Hs = 83deg 39.7min + 3.5min = 83deg 43.2min
## Back In Time
Capt Slocum claims he used a tin clock, but he was an experienced sailor and had extensive navigational experience. He had this to say about time: “To find local time is a simple matter. The difference between local and standard time is longitude expressed in time – four minutes, we all know, representing one degree. This briefly is the principle on which longitude is found independent of chronometers. The work of the lunarian, though seldom practised in these days of chronometers, is beautifully edifying, and there is nothing in the realm of navigation that lift’s one’s heart up more in adoration” (Ref.1/p149). In 1895, the navigational reference was the “American Ephemeris & Nautical Almanac”. Let’s go back in time and see what data was available. The closest that is available online is the 1900 edition (Ref.4).
## References
#1. – “Sailing Alone Around the World”, Joshua Slocum
https://www.amazon.com/dp/0486203263/
#2. – “Joshua Slocum”, Wikipedia
https://en.wikipedia.org/wiki/Joshua_Slocum
#3. – “Nautical Almanac”
https://www.thenauticalalmanac.com/
#4. – “The American Ephemeris and Nautical Almanac for the year 1900”, Washington Bureau of Equipment 1899
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# Homework Help: Tensor from Potential Function
1. Nov 12, 2014
### KleZMeR
1. The problem statement, all variables and given/known data
I am looking at Goldstein, Classical Mechanics. I am on page 254, and trying to reference page 190 for my confusion.
I don't understand how they got from equation 6.49 to 6.50, potential energy function to tensor matrix. I really want to know how to calculate a tensor from a function of this type (any type), but somehow the Goldstein text is not clear to me.
2. Relevant equations
$V = \frac{k}{2} (\eta_{1}^2+2\eta_{2}^2 +\eta_{3}^2-2\eta_{1}\eta_{2}-2\eta_{2}\eta_{3})$
\begin{array}{ccc} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{array}
3. The attempt at a solution
The solution is given. I think this is done by means of equation 5.14, but again, I am not too clear on this.
2. Nov 12, 2014
### ShayanJ
$\mathcal V=\frac 1 2 \vec \eta^T V \vec\eta=\frac 1 2 (\eta_1 \ \ \ \eta_2 \ \ \ \eta_3) \left(\begin{array}{ccc} k \ \ \ \ -k \ \ \ \ 0 \\ -k \ \ \ \ 2k \ \ \ \ -k \\ 0 \ \ \ \ -k \ \ \ \ k \end{array} \right)\ \left( \begin{array}{c} \eta_1 \\ \eta_2 \\ \eta_3 \end{array} \right)$
Last edited: Nov 12, 2014
3. Nov 12, 2014
### KleZMeR
Thanks Shyan, but how do I decompose the potential function to arrive at this? Or, rather, how do I represent my function in Einstein's summation notation? I believe from what you are showing that my potential function itself can be written as a matrix and be decomposed by two multiplications using $\eta^T , \eta$?
4. Nov 13, 2014
### ShayanJ
The potential function is a scalar so you can't write it as a matrix. And the thing I wrote, that's the simplest way of getting a scalar from a vector and a tensor. So people consider this and define the potential tensor which may be useful in some ways.
In component notation and using Einstein summation convention, its written as:
$\mathcal V=\frac 1 2 \eta_i V^i_j\eta^j$
But the potential function itself, is just $\mathcal V$ in component notation because its a scalar and has only one component!
5. Nov 20, 2014
### KleZMeR
Thank you!! That did help a LOT. Somehow I keep resorting back to the Goldstein book because it is the same notation we use in lecture and tests, but it does lack some wording in my opinion. I guess the explanation you gave would be better found in a math-methods book.
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My Math Forum Calculate point coordinates within a circle
Algebra Pre-Algebra and Basic Algebra Math Forum
February 4th, 2012, 11:28 AM #1 Newbie Joined: Feb 2012 Posts: 5 Thanks: 0 Calculate point coordinates within a circle Hi all, I am trying to figure out the coordinates of a point within a circle. I have an IR sensor which gives readings from 0 to 800. 800 meaning there are no objects in sight. What I want to do is make a miniture radar representation. I will have a turntable which will rotate the IR sensor while it scans for obstacles. Now, if it were a stationary IR sensor I'd know exactly where the object is - it would be whatever the sensor value returned because it would represent the Y axis... But, how would I work out the X and Y values for this point now that the IR sensor is rotating? I've drawn a diagram to help illustrate: Does anyone know the math to work this out?
February 4th, 2012, 12:17 PM #2 Member Joined: Feb 2012 Posts: 36 Thanks: 0 Re: Calculate point coordinates within a circle Convert the angle to radians and x=rcos(angle), y=rsin(angle)
February 4th, 2012, 12:33 PM #3 Newbie Joined: Feb 2012 Posts: 5 Thanks: 0 Re: Calculate point coordinates within a circle Ok, so after obtaining the X and Y, I could just subtract the distance of the object to the center from both values to give me the true value, correct?
February 4th, 2012, 12:34 PM #4
Newbie
Joined: Feb 2012
Posts: 5
Thanks: 0
Re: Calculate point coordinates within a circle
Quote:
Originally Posted by rich2020 Ok, so after obtaining the X and Y, I could just subtract the distance of the object to the center from both values to give me the true value, correct?
I meant "distance of the object FROM the center of the circle" - can't edit posts? :/
February 4th, 2012, 12:38 PM #5 Member Joined: Feb 2012 Posts: 36 Thanks: 0 Re: Calculate point coordinates within a circle Yep, basic coordinate pythagoras to locate the point in relation to the required limit of the sensor.
February 4th, 2012, 12:39 PM #6 Newbie Joined: Feb 2012 Posts: 5 Thanks: 0 Re: Calculate point coordinates within a circle Thanks very much for your help.
February 4th, 2012, 04:57 PM #7 Newbie Joined: Feb 2012 Posts: 5 Thanks: 0 Re: Calculate point coordinates within a circle Hmm... I don't think your way was correct. I needed to find the coordinates based on the angle and the distance an object. What I did was: X=Cos(A) * Distance Y=Sin(A) * Distance I used degrees, not radians. Then, because my circle (radius of 15mm) will be far smaller as it will be on a mobile Android device's screen, I did this: Sensor range is 800mm... so X = X/(800/15) Y = Y/(800/15) Correct me if I'm wrong please. I need this for a software project.
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0
# If a2 5 is reflected over the y-axis what are the coordinates of a?
Updated: 4/28/2022
Wiki User
14y ago
Your new coordinates would be -2,5.
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### 171282. A ship 55 km from the shore springs a leak which admits 2 tonnes of water in 6 minutes, 80 tonnes would suffice to sink her, but the pumps can throw out 12 tonnes an hour. The average rate of sailing that she may just reach the shore as she begins to sink is:
Answer: 5.5 km/hSol. In 1 h water entered into shop = (20 – 12) = 8 tonnes Now, it will take 10 hrs to allow to enter 80 tonnes of water into ship and in this time ship has to cover 55 km of distance. Hence, required speed = 5.5 km/h.
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## Use the graphing calculator to graph the system. The solutions to the system are Since the side lengths of a triangle must be
Question
Use the graphing calculator to graph the system.
The solutions to the system are
Since the side lengths of a triangle must be
positive, the length of one leg of the sun shade is
I feet
in progress 0
1 month 2021-10-15T18:43:41+00:00 2 Answers 0 views 0
1) -11.3 or 11.3
2) 11.3
Step-by-step explanation:
its correct on edge
2. The solutions to the system are -11.3 and 11.3.
The length of one leg of the sun shade is 11.3 feet.
Steps:
graph y=1/2 x^2
graph y=64
zoom out.
find the points of intersection.
done.
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# Phase shifter switch with Op-Amp
In electronics circuit design sometimes we need to phase shift the input signal by 180 degree. How can we phase shift the phase of the input signal? Here it is shown how one can use op-amp(operational amplifier) to do either 0(that is no phase shift) or 180 degree phase shift of the phase of the input signal using manual switch or FET switch. Such phase shifter circuit is also called sign changer circuit.
Consider the following Phase shifter switch with Op-Amp with manual switch circuit diagram.
A manual switch is connected to the non-inverting terminal of the op-amp such as LM358N. The switch can either be connected to the ground so that non-inverting terminal is grounded or the switch can connect the input signal Vin to the non-inverting terminal. Irrespective of the position of the switch the inverting terminal will always receive the input signal Vin.
Consider what happens at the output with the switch in either of the position.
When the switch is connected to ground then the op-amp operates as inverting amplifier configuration with the voltage gain given by the following equation.
$$A_v(invert) = -\frac{R_2}{R_1}$$
and since the resistors are equal we have,
$$A_v(invert) = -1$$
In this case the output signal will be 180 out of phase with the input signal. That is the input signal is phase shifted by 180 degree.
When the switch position is changed, both the inverting and the non-inverting terminal gets Vin input signal. That is, it is a circuit wherein both the input terminal of the op-amp is driven by the same input signal. In such case, input get amplified by both the inverting and non-inverting terminal and we get superposition of the signals from the two terminals. The overall voltage gain of such amplifier is given by the following relation.
$$A_v = A_v(invert) +A_v(non-invert)$$ -------->(1)
The voltage gain due to inverting channel is still, $$A_v(invert) = -1$$.
The voltage gain due to non-inverting channel is,
$$A_v(non-invert) = 1+\frac{R_2}{R_1}=1+1=2$$ since R1=R2
And therefore the total voltage gain from equation(1) is,
$$A_v = -1 + 2=1$$
And hence there is no phase shift of the input signal in this case. That is there is 0 phase difference between the input and output signal when switch is in the upper position.
In other words, the above circuit is a switchable phase shifter producing output signal with equal amplitude as the input signal but with phase difference of either 0 or 180 depending on the position of the switch. This is illustrated in the animation below showing how the Op-Amp phase shifter works.
In this tutorial we showed how a simple phase shifter circuit works with manual switch which generates either 0 phase shift or 180 phase shift. We can also use JFET switch to control the switching action instead of manual. And instead of just 0 or 180 degree phase we can use phase shifter circuit to get arbitrary phase shift.
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• 1.
### What is the Formula for Acceleration?
• A.
A=(V-U)/T
• B.
A=(U-V)/T
• C.
A=(VxU)/T
A. A=(V-U)/T
Explanation
The formula for acceleration is A=(V-U)/T. This formula represents the change in velocity (V-U) divided by the time taken (T). It is derived from the basic definition of acceleration as the rate of change of velocity over time. By subtracting the initial velocity (U) from the final velocity (V) and dividing it by the time interval (T), we can determine the acceleration of an object.
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• 2.
### What is the equation to find the Force?
• A.
F=M/A
• B.
F=MA
• C.
M=FA
B. F=MA
Explanation
The equation F=MA represents Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This equation shows the relationship between force, mass, and acceleration, and is commonly used to calculate the force exerted on an object given its mass and acceleration.
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• 3.
### What is the Equation to find Velocity?
• A.
Velocity=Speed/Time
• B.
Velocity=Displacement/Time
• C.
Displacement=Velocity/Time
B. Velocity=Displacement/Time
Explanation
The equation to find velocity is given by dividing the displacement by time. This is because velocity is defined as the rate of change of displacement with respect to time. Therefore, to calculate velocity, we need to know the displacement (change in position) and the time taken for that displacement to occur. By dividing the displacement by the time, we can determine how fast an object is moving in a specific direction.
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• 4.
### What is the Equation to find Weight?
• A.
Weight=Mass x Speed due to Gravity
• B.
Weight= Mass x 4n
• C.
Weight=Mass x Acceleration due to Gravity
C. Weight=Mass x Acceleration due to Gravity
Explanation
The equation to find weight is given by multiplying the mass of an object by the acceleration due to gravity. This equation is derived from Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In the case of weight, the force is the gravitational force acting on the object, and the acceleration is the acceleration due to gravity. Therefore, the correct equation to find weight is Weight = Mass x Acceleration due to Gravity.
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• 5.
### If a fluid is moving faster the pressure:
• A.
Increases
• B.
Stays the same
• C.
Decreases
C. Decreases
Explanation
When a fluid is moving faster, it creates a decrease in pressure. This can be explained by Bernoulli's principle, which states that as the speed of a fluid increases, its pressure decreases. This is because the faster-moving fluid particles exert less force on the surrounding surfaces, resulting in a decrease in pressure. Therefore, when a fluid is moving faster, the pressure decreases.
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• 6.
### The unit of pressure Pascal means:
• A.
Newton per metre squared.
• B.
Gram per centimetre squared.
• C.
Gram per metre squared.
A. Newton per metre squared.
Explanation
The unit of pressure, Pascal, is defined as Newton per metre squared. This means that it is a measure of force per unit area. The Newton is the unit of force and the metre squared represents the area over which the force is applied. Therefore, the correct answer is Newton per metre squared.
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• 7.
### The formula for Hookes Law is:
• A.
F= n x e
• B.
F= k x e
• C.
K= f x e
B. F= k x e
Explanation
The given formula for Hooke's Law states that the force (f) is equal to the product of the spring constant (k) and the displacement (e). This formula is derived from experimental observations that show the relationship between the force applied to a spring and the resulting displacement. The spring constant represents the stiffness of the spring, and the displacement refers to the amount the spring is stretched or compressed from its equilibrium position. Therefore, the correct answer is f = k x e.
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• 8.
### The formula for wavespeed is:
• A.
Frequency x Wavelength
• B.
Wavelength / Frequency
• C.
Frequency x Amplitude
A. Frequency x Wavelength
Explanation
The formula for wavespeed is given by multiplying the frequency and the wavelength of the wave. This is because wavespeed is defined as the distance traveled by a wave per unit time, and the wavelength represents the distance between two consecutive points on the wave that are in phase. The frequency, on the other hand, represents the number of complete cycles of the wave that occur in one second. Therefore, multiplying the frequency and the wavelength will give us the distance traveled by the wave in one second, which is the wavespeed.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Division of Decimals by Decimals using Zero Placeholders ( Read ) | Arithmetic | CK-12 Foundation
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# Division of Decimals by Decimals using Zero Placeholders
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Division of Decimals by Decimals using Zero Placeholders
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Have you ever tried to divide two numbers but it didn't work perfectly? In the Divide Decimals by Decimals Concept, Miles divided a decimal by a decimal and it worked out perfectly in the end. What if it didn't? What would Miles do then? Look at this problem.
Jessie decided to try her hand at the sand and the hourglass too. But she used a different bucket. Jessie's bucket would hold 6.75 pounds of sand. 1.25 pounds of sand could go through the hourglass at one time.
Jessie thinks that she can run the sand through the hourglass six times before needing to refill the bucket.
Is she correct?
To solve this problem, you will need to divide once again. Pay attention to this Concept because you will need a new skill to complete the problem.
### Guidance
The decimals that we divided in the Divide Decimals by Decimals Concept were all evenly divisible. This means that we had whole number quotients. We didn’t have any decimal quotients.
What can we do if a decimal is not evenly divisible by another decimal?
If you think back, we worked on some of these when we divided decimals by whole numbers. When a decimal was not evenly divisible by a whole number, we had to use a zero placeholder to complete the division.
$5 \overline{)13.6 \;}$
When we divided 13.6 by 5, we ended up with a 1 at the end of the division. Then we were able to add a zero placeholder and finish finding a decimal quotient. Here is what this looked like.
$& \overset{ \quad 2.72}{5 \overline{ ) {13.60 \;}}}\\& \underline{-10 \;\;}\\& \quad \ 36\\& \ \ \underline{-35\;\;}\\& \qquad 1 - \ \text{here is where we added the zero placeholder}\\& \qquad 10\\& \quad \ \underline{-10}\\& \qquad \ \ 0$
We add zero placeholders when we divide decimals by decimals too.
$1.2 \overline{)2.79 \;}$
The first thing that we need to do is to multiply the divisor and the dividend by a multiple of ten to make the divisor a whole number. We can multiply both by 10 to accomplish this goal.
$12 \overline{)27.9 \;}$
Now we can divide.
$& \overset{ \quad \ \ 2.3}{12 \overline{ ) {27.9 \;}}}\\& \ \underline{-24 \;\;}\\& \quad \ \ 39\\& \quad \underline{-36}\\& \qquad \ 3$
Here is where we have a problem. We have a remainder of 3. We don’t want to have a remainder, so we have to add a zero placeholder to the problem so that we can divide it evenly.
$& \overset{ \quad \ \ 2.32}{12 \overline{ ) {27.90 \;}}}\\& \ \underline{-24\;\;}\\& \quad \ \ 39\\& \quad \underline{-36\;\;}\\& \qquad \ 30\\& \quad \ \ \underline{-24}\\& \qquad \quad 6$
Uh Oh! We still have a remainder, so we can add another zero placeholder.
$& \overset{ \quad \ 2.325}{12 \overline{ ) {27.900 \;}}}\\& \ \ \underline{-24\;\;}\\& \ \quad \ \ 39\\& \ \quad \underline{-36\;\;}\\& \ \qquad \ 30\\& \ \quad \ \ \underline{-24\;\;}\\& \ \qquad \quad 60\\& \ \qquad \ \underline{-60\;\;}\\& \ \qquad \quad \ \ 0$
Sometimes, you will need to add more than one zero. The key is to use the zero placeholders to find a quotient that is even without a remainder.
#### Example A
$1.2 \overline{)2.76 \;}$
Solution: 2.3
#### Example B
$8.7 \overline{)53.94 \;}$
Solution: 6.2
#### Example C
$5.4 \overline{)18.9 \;}$
Solution: 3.5
Now back to Jessie and the sand. Did you figure out that you will need a zero placeholder to figure out if Jessie's estimate is correct? Let's look at the original problem once again.
Jessie decided to try her hand at the sand and the hourglass too. But she used a different bucket. Jessie's bucket would hold 6.75 pounds of sand. 1.25 pounds of sand could go through the hourglass at one time.
Jessie thinks that she can run the sand through the hourglass six times before needing to refill the bucket.
To figure out if a 6.75 pound bucket of sand can be divided into 1.25 pounds six times, we will need to divide.
$1.25 \overline{)6.25 \;}$
Next, we can divide.
Our answer is $5.4$ .
Jessie isn't correct. She will need to refill the bucket after pouring 1.25 pounds of sand through the hourglass five times.
### Vocabulary
Divisor
the number doing the dividing, it is found outside of the division box.
Dividend
the number being divided. It is found inside the division box.
Quotient
the answer in a division problem
### Guided Practice
Here is one for you to try on your own.
$3.2 \overline{)28.52 \;}$
To complete this problem, we must first move the decimal point one place in the divisor which makes 3.2 into 32. Then we can simply divide. Notice that you will need to use zero placeholders and keep adding them until the division is complete.
Our answer is $8.9125$ .
### Practice
Directions: Divide the following decimals. Use zero placeholders when necessary.
1. $1.3 \overline{)5.2 \;}$
2. $6.8 \overline{)13.6 \;}$
3. $4.5 \overline{)13.5 \;}$
4. $2.5 \overline{)10 \;}$
5. $3.3 \overline{)19.8 \;}$
6. $8.5 \overline{)17 \;}$
7. $9.3 \overline{)27.9 \;}$
8. $1.2 \overline{)7.2 \;}$
9. $5.3 \overline{)26.5 \;}$
10. $6.5 \overline{)13 \;}$
11. $1.25 \overline{)7.5 \;}$
12. $3.36 \overline{)20.16 \;}$
13. $5.87 \overline{)52.83 \;}$
14. $2.5 \overline{)3 \;}$
15. $3.2 \overline{)8 \;}$
16. $4.6 \overline{)10.58 \;}$
17. $8.1 \overline{)17.82 \;}$
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Note on Translation
Definitions Brief comments 1 Every rectangular parallelogram is said to be enclosed by two straight-lines that enclose the right angle. 2. but in any parallelogram, let any one of the parallelograms about its diameter with the two complements be called a gnomon.
Propositions:
Prop. 1: If there are two straight lines, but one of them is cut into however many sections, the rectangle enclosed by the two straight-lines will be equal to the rectangles enclosed by the uncut straight-line and each of the sections.
Prop. 2: If a straight line is cut, as it happens, the rectangle enclosed by the whole and each of the segments is equal to the square from the whole.
Prop. 3: If a straight line is cut, as it happens, the rectangle enclosed by the whole and one of the segments is equal to the rectangle enclosed by the segments and by the square from the just-mentioned square.
Prop. 4: If a straight line is cut, as it happens, the square from the whole is equal to the squares from the segments and twice the rectangle enclosed by the segments.
[Provided (corollary): In fact, it is obvious from this that the parallelograms in square regions are squares.]
Prop. 5: If a straight line is cut into equals and unequals, the rectangle enclosed by the unequal segments of the whole with the square from the line in between the cuts is equal to the square from the half line.
Prop. 6: If a straight line is bisected and some straight-line is added to it on a straight-one, the rectangle enclosed by the whole with the added line and the added line with the square from the half line is equal to the square from the line composed from the half and the added line.
Prop. 7: If a straight line is cut, as it happens, the square from the whole and that from one of the segments, both squares together, are equal to twice the rectangle enclosed by the whole and the mentioned segment and the square from the remaining segment.
Prop. 8: If a straight line is cut, as it happens, quadruple the rectangle enclosed by the whole and one of the segments with the square from the remaining segment is equal to the square from the whole and the mentioned segments described up as from one line.
Prop. 9: If a straight line is cut into equals and unequals, the squares from the unequal segments of the whole are double that from the half and the square from the line between the cuts.
Prop. 10: If a straight line is bisected, and some line is added to it on a line, the square from the whole with the added line and that from the added line, both squares together, are double the square from half and the square from the line composed from the half and the added line described up as from one line.
Prop. 11: To cut the given straight line so that the rectangle enclosed by the whole and one of the segments is equal to the square from the remaining segment.
Prop. 12: In obtuse-angled triangles the square from the side subtending the obtuse angle is larger than the squares from the sides enclosing the obtuse angle by twice the rectangle enclosed by one of sides about the obtuse angle on which the perpendicular falls and the line outside that’s taken away by the perpendicular at the obtuse angle.
Prop. 13: In acute-angled triangles the square from the side subtending the acute angle is smaller than the squares from the sides enclosing the acute angle by twice the rectangle enclosed by one of the sides about the acute angle on which the perpendicular false and the line inside that’s taken away by the perpendicular at the acute angle.
Prop. 14:: To construct a square equal to the given rectilinear the figure.
ὀρθογώνιον can be translated ‘right-angled’ (so my translation in Elements I), ‘rectangular’, ‘rectangle’, but these are essentially synonyms as far as the Greek is concerned, with ‘rectangle’ short for ‘rectangular thing’, which in context will be short for ‘rectangular parallelogram'. In this translation, I shall switch between 'rectangle' and 'rectangular'.
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### Course: Early math review>Unit 4
Lesson 2: Subtraction within 20
# Subtracting 14 - 6
Sal subtracts 14 - 6 by first thinking about subtracting 2 and 4. Created by Sal Khan.
## Want to join the conversation?
• I am struggling to find a video on Subtracting within 20 with regrouping. I dont want to use blocks, I don't want to use visualizations. I want to know how to know how to write subtract 9 from 16 with regrouping.
• Try this,
We have the original number, but we can put it in expanded form*.
So, 16 = 10 + 6, and we can use this.
We know that 10 - 9 = 1, and then we can add the remaining 6 to get 7.
Expanded form is just taking the values of each number and adding them to create a number. For example, 17 = 10 + 7
You can also use this for bigger numbers
, like
1,234 = 1,000 + 200 + 30 + 4
• Hi there,
Could you please tell which video explains how to do 9-[]=5 and []-4=5.
Any help on this is much appreciated.
• Hello,
In these type of problems all you have to do is do the opposite of what it's asking. But, you would, in this case, still subtract since you are trying to find the number you subtract by and not the number you subtract from. So, 9-5=4 the missing value is 4.
For the second problem you add since you are trying to find the number you subtract from. So, 4+5=9 the missing value is 9.
Hope this helps!
• What is the commutative property? Is addition commutative?
• It is when you can change the order of an equation and get the same answer. For Ex: 4+3= 7 is the same as 3+4=7
(1 vote)
• If your subtracting by adding a negative number does that lead to borrowing to significant to basic subtraction?
• The basic concept of adding a negative number is exactly the same as subtracting that number. In higher math this may not always be the case but you will learn about those types of situations later. For now think about adding a negative number as the same process as subtraction. For example 10 + -3 = 7 and 10 - 3 = 7.
• so people can think of the four's cancelling each other out in the second problem and being left with only ten?
• Yes, but not all math problems are simple like that. But it is definitely not encouraged to think as such, as people might think that 14 - 4 = 1 if they cancel out the 4 and forget that there is a 0 behind the 1.
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# find intervals of concavity
Therefore, we need to test for concavity to both the left and right of $-2$. The calculator will find the intervals of concavity and inflection points of the given function. Notice that the graph opens "up". The graph of f which is called a parabola will be concave up if a is positive and concave down if a is negative. If so, you will love our complete business calculus course. Using the same analogy, unlike the concave up graph, the concave down graph does NOT "hold water", as the water within it would fall down, because it resembles the top part of a cap. To find the intervals, first find the points at which the second derivative is equal to zero. We set the second derivative equal to $0$, and solve for $x$. b) Use a graphing calculator to graph f and confirm your answers to part a). Resolved exercise on how to calculate concavity and convexity in the intervals of a function. In any event, the important thing to know is that this list is made up of the zeros of f′′ plus any x-values where f′′ is undefined. Evaluate the integral between $[0,x]$ for some function and then differentiate twice to find the concavity of the resulting function? How to find intervals of a function that are concave up and concave down by taking the second derivative, finding the inflection points, and testing the regions f(x) = x^4 + x^3 - 3x^2 + 1. Please see below for the concavities. Plug these three x-values into f to obtain the function values of the three inflection points. By the way, an inflection point is a graph where the graph changes concavity. Also, when $x=1$ (right of the zero), the second derivative is positive. In order to determine the intervals of concavity, we will first need to find the second derivative of $$f(x)$$. How to Locate Intervals of Concavity and Inflection Points, How to Interpret a Correlation Coefficient r, You can locate a function’s concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. Find the intervals of concavity and the inflection points of g(x) = x 4 – 12x 2. In math notation: If $f''(x) > 0$ for $[a,b]$, then $f(x)$ is concave up on $[a,b]$. However, a function can be concave up for certain intervals, and concave down for other intervals. As we work through this problem, we will work one step at a time. The following method shows you how to find the intervals of concavity and the inflection points of Find the second derivative of […] Find the Concavity xe^x. Otherwise, if $f''(x) < 0$ for $[a,b$], then $f(x)$ is concave down on $[a,b]$. We still set a derivative equal to $0$, and we still plug in values left and right of the zeroes to check the signs of the derivatives in those intervals. Solving for x, . Ex 5.4.20 Describe the concavity of $\ds y = x^3 + bx^2 + cx + d$. Find the inflection points. Learn how to determine the extrema, the intervals of increasing/decreasing, and the concavity of a function from its graph. Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. These two examples are always either concave up or concave down. In general, concavity can only change where the second derivative has a zero, or where it is undefined. Solution: The domain is the whole real line, so there is one starting interval, namely (-,). The first step in determining concavity is calculating the second derivative of $f(x)$. f (x) = x^4 - 2x^2 + 3 (d) Use the information from parts (a)–(c) to sketch the graph. On the other hand, a concave down curve is a curve that "opens downward", meaning it resembles the shape $\cap$. We first calculate the first and second derivative of function f f '(x) = 2 a x + b f "(x) = 2 a 2. f (x) = (1 - x) e^ - x b.) And the value of f″ is always 6, so is always >0,so the curve is entirely concave upward. Determine the intervals of concavity. Write the polynomial as a function of . Note: Check your work with a graphing device. Therefore, there is an inflection point at $x=-2$. By using this website, you agree to our Cookie Policy. Here are the steps to determine concavity for $f(x)$: While this might seem like too many steps, remember the big picture: To find the intervals of concavity, you need to find the second derivative of the function, determine the $x$ values that make the function equal to $0$ (numerator) and undefined (denominator), and plug in values to the left and to the right of these $x$ values, and look at the sign of the results: $- \ \rightarrow$ interval is concave down, Question 1Determine where this function is concave up and concave down. Plot these numbers on a number line and test the regions with the second derivative. Create intervals around the inflection points and the undefined values. Substitute in to find the value of . The function can either be always concave up, always concave down, or both concave up and down for different intervals. The second derivative of the function is equal to . Analyzing concavity (algebraic) This is the currently selected item. The following method shows you how to find the intervals of concavity and the inflection points of. Now that we have the second derivative, we want to find concavity at all points of this function. And then we divide by $30$ on both sides. Calculus: Integral with adjustable bounds. First Derivative. You can locate a function’s concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. We will calculate the intervals where the next function is concave or convex: We need to study the sign of the second derivative of function. If f″(x) changes sign, then ( x, f(x)) is a point of inflection of the function. A concave up graph is a curve that "opens upward", meaning it resembles the shape $\cup$. Concavity and Convexity Worksheet Find the Intervals of Concavity and Convexity for the Following Functions: Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 Exercise 6 Exercise 7 Exercise 8 Exercise 9 Exercise 10 Exercise 11 Exercise 12 Solution of … (b) Find the local maximum and minimum values. To view the graph, click here. The square root of two equals about 1.4, so there are inflection points at about (–1.4, 39.6), (0, 0), and about (1.4, –39.6). This means that this function has a zero at $x=-2$. c.) Find the intervals of concavity and the inflection points. Example: Investigate concavity of the function f (x) = x 4 - 4x 3. DO : Try to work this problem, using the process above, before reading the solution. Then test all intervals around these values in the second derivative of the function. Both derivatives were found using the power rule . With this function we will now want to find. Find the inflection points and intervals of concavity upand down of f(x)=3x2−9x+6 First, the second derivative is justf″(x)=6. (d) Use the information from parts (a)-(c) to sketch the graph. A point of inflection (c, f(c)) occurs when the graph changes concavity at (c, f(c)) from up to down or down to up. When asked to find the interval on which the following curve is concave upward $$y = \int_0^x \frac{1}{94+t+t^2} \ dt$$ What is basically being asked to be done here? Set the second derivative equal to zero and solve. intervals of concavity, inflection points. The intervals, therefore, that we analyze are and . An inflection point exists at a given x-value only if there is a tangent line to the function at that number. Answer to: Determine the points of inflection and find the intervals of concavity. That gives us our final answer: $in \ (-\infty,-2) \ \rightarrow \ f(x) \ is \ concave \ down$, $in \ (-2,+\infty) \ \rightarrow \ f(x) \ is \ concave \ up$. This is where the second derivative comes into play. Solution to Question 1: 1. Replace the variable with in the expression. The concept is very similar to that of finding intervals of increase and decrease. The first derivative of the function is equal to . Thus we find them easily by looking at concavity intervals. Liked this lesson? Then check for the sign of the second derivative in all intervals, If $f''(x) > 0$, the graph is concave up on the interval. Else, if $f''(x)<0$, the graph is concave down on the interval. (c) Find the intervals of concavity and the inflection points. We refer to Concavity in Methods Survey - Graphing for a practical overview. Find (a) the intervals of increase or decrease, (b) the intervals of concavity, and (c) the points of inflection. The main difference is that instead of working with the first derivative to find intervals of increase and decrease, we work with the second derivative to find intervals of concavity. (If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. (c) Find the intervals of Concavity and the inflection points. Points of inflection do not occur at discontinuities. Consider the function: f(x) =x^3-12x + 2 (a) Find the intervals of increase or decrease. Substitute any number from the interval into the second derivative and evaluate to determine the concavity. To determine concavity, analyze the sign of f''(x). In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. The opposite of concave up graphs, concave down graphs point in the opposite direction. Local min C(-1)=-3 Inflection points (algebraic) Mistakes when finding inflection points: second derivative undefined. First, let's figure out how concave up graphs look. (a) Find the intervals on which f is increasing or decreasing. Create intervals around the inflection points and the undefined values. a.) example. (b) Find the local maximum and minimum values. The sign of f "(x) is the same as the sign of a. In business calculus, you will be asked to find intervals of concavity for graphs. So, a concave down graph is the inverse of a concave up graph. We build a table to help us calculate the second derivatives at these values: As per our table, when $x=-5$ (left of the zero), the second derivative is negative. f(x) = xe^-x f'(x) = (1)e^-x + x[e^-x(-1)] = e^-x-xe^-x = -e^-x(x-1) So, f''(x) = [-e^-x(-1)] (x-1)+ (-e^-x)(1) = e^-x (x-1)-e^-x = e^-x(x-2) Now, f''(x) = e^-x(x-2) is continuous on its domain, (-oo, oo), so the only way it can change sign is by passing through zero. The function has an inflection point (usually) at any x-value where the signs switch from positive to negative or vice versa. For example, the graph of the function $y=-3x^2+5$ results in a concave down curve. Concavity and Points of Inflection While the tangent line is a very useful tool, when it comes to investigate the graph of a function, the tangent line fails to say anything about how the graph of a function "bends" at a point. Determine whether the second derivative is undefined for any x-values. Steps 2 and 3 give you what you could call “second derivative critical numbers” of f because they are analogous to the critical numbers of f that you find using the first derivative. Find the local maximum and minimum values. Just as functions can be concave up for some intervals and concave down for others, a function can also not be concave at all. A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. Solution: Since f ′ ( x ) = 3 x 2 − 6 x = 3 x ( x − 2 ) , our two critical points for f are at x = 0 and x = 2 . In other words, this means that you need to find for which intervals a graph is concave up and for which others a graph is concave down. This means that the graph can open up, then down, then up, then down, and so forth. Intervals of Concavity Date_____ Period____ For each problem, find the x-coordinates of all points of inflection, find all discontinuities, and find the open intervals where the function is concave up and concave down. In words: If the second derivative of a function is positive for an interval, then the function is concave up on that interval. Ex 5.4.19 Identify the intervals on which the graph of the function $\ds f(x) = x^4-4x^3 +10$ is of one of these four shapes: concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. a) Find the intervals on which the graph of f(x) = x 4 - 2x 3 + x is concave up, concave down and the point(s) of inflection if any. finding intervals of increase and decrease, Graphs of curves can either be concave up or concave down, Concave up graphs open upward, and have the shape, Concave down graphs open downward, with the shape, To determine the concavity of a graph, find the second derivative of the given function and find the values that make it $0$ or undefined. Therefore, let’s calculate the second derivative. Calculus: Fundamental Theorem of Calculus 2=0 is a contradiction, so we have no inflection points on this function, so ¿How could I determine the concavity if I have no inflection points? Determining concavity of intervals and finding points of inflection: algebraic. In business calculus, concavity is a word used to describe the shape of a curve. Show Instructions. This function's graph: Should I take the "0" as a refered point, then evaluate the f''(x) (for example) with f''(-1) and f''(1) to determine the concavity? But this set of numbers has no special name. Calculus Calculus: Early Transcendentals (a) Find the intervals of increase or decrease. You can easily find whether a function is concave up or down in an interval based on the sign of the second derivative of the function. The intervals of increasing are x in (-oo,-2)uu(3,+oo) and the interval of decreasing is x in (-2,3). The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. Otherwise, if the second derivative is negative for an interval, then the function is concave down at that point. Because –2 is in the left-most region on the number line below, and because the second derivative at –2 equals negative 240, that region gets a negative sign in the figure below, and so on for the other three regions. If the second derivative of the function equals $0$ for an interval, then the function does not have concavity in that interval. Increase on (-1, inf) and decrease on (-inf, -1) b.) Click here to view the graph for this function. For example, the graph of the function $y=x^2+2$ results in a concave up curve. To view the graph of this function, click here. A graph showing inflection points and intervals of concavity. Solution to Example 4 Let us find the first two derivatives of function f. a) f '(x) = 4 x 3 - 6 2 + 1 f ''(x) = 12 2 … So, we differentiate it twice. As you can see, the graph opens downward, then upward, then downward again, then upward, etc. Let's pick $-5$ and $1$ for left and right values, respectively. Answers and explanations For f ( x ) = –2 x 3 + 6 x 2 – 10 x + 5, f is concave up from negative infinity to the inflection point at (1, –1), then concave down from there to infinity. Example: Find the intervals of concavity and any inflection points of f (x) = x 3 − 3 x 2. This is the case wherever the first derivative exists or where there’s a vertical tangent.). I already have the answers I just don't know how to get them, here they are for reference: a.) Find the intervals of increase or decrease. Solution: Since this is never zero, there are not points ofinflection. (b) Find the local maximum and minimum values of f. (c) Find the intervals of concavity and the inflection points. You can think of the concave up graph as being able to "hold water", as it resembles the bottom of a cup. Find the intervals of concavity and the points of inflection of the following function. In determining intervals where a function is concave upward or concave downward, you first find domain values where f″(x) = 0 or f″(x) does not exist. The perfect example of this is the graph of $y=sin(x)$. In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. To find the intervals of concavity, you need to find the second derivative of the function, determine the $x$ values that make the function equal to $0$ (numerator) and undefined (denominator), and plug in values to the left and to the right of these $x$ values, and look at the sign of the results: $+ \ \rightarrow$ interval is concave up In general, a curve can be either concave up or concave down. Notice this graph opens "down". Divide by $30$ on both sides, inf ) and decrease $\ds y x^3. Currently selected item to obtain the function has a zero at$ x=-2 $answers! Is equal to zero and solve either be always concave up graphs, concave down graphs in... Curve can be either concave up, then upward, etc: the domain is same. The local maximum and minimum values of the function f ( x ) = x 3 3... Graphing calculator to graph f and confirm your answers to part a ) Find the of! Is positive and concave down graph is the whole real line, so the curve is concave! 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# 2x+10=12
## Simple and best practice solution for 2x+10=12 equation. Check how easy it is, and learn it for the future.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for 2x+10=12 equation:
2x + 10 = 12
2x = 12 - 10
2x = 2 (divide both sides by 2 to get x)
2x/2 = 2/2
x = 1
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mersenneforum.org Coset conundrum
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2021-08-28, 16:11 #1 Dr Sardonicus Feb 2017 Nowhere 535410 Posts Coset conundrum Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b-1,n) = 1, and let h be the multiplicative order of b (mod n). Let k be an integer, 1 <= k < n. Let ri = remainder of k*b^(i-1) mod n [0 < ri < n], i = 1 to h. A) Prove that $\sum_{i=1}^{h}r_{i}\;=\;m\times n$ for a positive integer m. B) Prove that m = 1 if and only if n is a repunit to the base b, and also that one of the ri is equal to 1. Remark: The thread title refers to the fact that the ri form a coset of the multiplicative group generated by b (mod n) in the multiplicative group of invertible residues (mod n). Part (B) as I stated it above is completely wrong. I forgot to multiply a fraction by 1. There is, alas, no connection to n being a repunit to the base b, or to any of the ri being 1, as shown by the example b = 10, k = 2, n = 4649 . Foul-up corrected in followup post. Last fiddled with by Dr Sardonicus on 2021-08-29 at 13:21 Reason: xingif stopy
2021-08-28, 21:28 #2
uau
Jan 2017
112 Posts
Quote:
Originally Posted by Dr Sardonicus A) Prove that $\sum_{i=1}^{h}r_{i}\;=\;m\times n$ for a positive integer m.
If you multiply the elements of the set {b^i for all i} by b, you get the same set with permuted elements. Thus multiplying the sum by b does not change it mod n. Thus S*b = S mod n, S*(b-1)=0 mod n, and S must be 0 mod n.
Quote:
B) Prove that m = 1 if and only if n is a repunit to the base b, and also that one of the ri is equal to 1.
The "and also that one of the ri is equal to 1" part seems ambiguous or wrong. For k != 1, m may or may not equal 1?
n = 1111, b = 10, k = 2: m = 1
n = 1111, b = 10, k = 21: m = 2
2021-08-29, 13:26 #3
Dr Sardonicus
Feb 2017
Nowhere
2×2,677 Posts
Quote:
Originally Posted by uau The "and also that one of the ri is equal to 1" part seems ambiguous or wrong. For k != 1, m may or may not equal 1? n = 1111, b = 10, k = 2: m = 1 n = 1111, b = 10, k = 21: m = 2
Mea culpa. I made a huge blunder. Actual characterization of the multiplier m follows, proof left as exercise.
Conditions restated for ease of reference:
Quote:
Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b-1,n) = 1, and let h be the multiplicative order of b (mod n). Let k be an integer, 1 <= k < n.
(B) Let A = k*(bh - 1)/n, and sb(A) the sum of the base-b digits of A. Then m = sb(A)/(b-1).
Last fiddled with by Dr Sardonicus on 2021-08-29 at 13:29
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test Practice for exam #2 in MA135 In addition to problems on this sheet, students should study webwork problems, assigned homework, worked examples in the text, review problems at the end of each chapter, and Prof Sims and Greenelds sample exams on the web. 1a. Let f (x) = ex 2 +3 . Find f (x). 1b. Let f (3) = 1, g (2) = 3, f (3) = 4, g (2) = 5. If h(x) = f (g (x)), nd h (2). 2a. Find the equation of the...
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test Practice for exam #2 in MA135 In addition to problems on this sheet, students should study webwork problems, assigned homework, worked examples in the text, review problems at the end of each chapter, and Prof Sims and Greenelds sample exams on the web. 1a. Let f (x) = ex 2 +3 . Find f (x). 1b. Let f (3) = 1, g (2) = 3, f (3) = 4, g (2) = 5. If h(x) = f (g (x)), nd h (2). 2a. Find the equation of the tangent line to the graph of x3 + y 3 = y + 21 at (3, 2) 2 2b. If y = x(x ) , nd dy dx 3a. The leaning ladder problem from quiz #7. 3b. One end of a rope is fastened to a boat and the other end is wound around a windlass located on a dock at a point 4 m above the level of the boat. If the boat is drifting away from the dock at the rate of 2 m/min, how fast is the rope unwinding at the instant the when length of the rope is 5 m? 4a. Use dierentials to approximate 103. 4b. Find d(x2 sin x3 ). 5a. Find the absolute max and min of f (x) = x5 x4 on [1, 1]. 5b. Find the absolute max and min of f (x) = 9 4x x2 + 6 x if x < 1 if x 1 on [0, 4]. 1 6. If f (x) = , nd the c in the Mean Value Theorem if a = 1 and b = 4. x 7. An eciency study of the morning shift at a factory indicates that the number of units produced by an average worker t hours after 8:00 AM is modeled by the formula Q(t) = t3 + 9t2 + 12t. At what time in the morning is the worker performing most eciently? 2 8a. Sketch the graph of g (u) = u4 + 6u3 24u2 + 26. 8b. Sketch the graph of f (x) = 1 2x 3x 8c. Find lim 1+ 8d. Find lim + 2 cos x 1 sin 2x x x x0 1 cos x . x0 sec x 9. Evaluate lim . . 3x 5 . x2
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Vol 443 | 19 October 2006 | doi:10.1038/nature05195L ETTERSLate survival of Neanderthals at the southernmostextreme of EuropeClive Finlayson1,2, Francisco Giles Pacheco3, Joaqun Rodrguez-Vidal4, Darren A. Fa1, Jose Mara Gutierrez Lopez5,Antonio Santi
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Rutgers - ANTHROPOLO - 111
PERSPECTIVESPA L E O A N T H R O P O L O G YWhither the Neanderthals?Richard G. Kleinwithin 10,000 to 15,000 years. The modernhuman triumph depended on technological, economic, and demographic advantages that were apparently grounded in anenhanced a
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The New Yorker, May 25, 20091 of 12http:/archives.newyorker.com/global/print.asp?path=/djvu/Conde%20Na.8/12/2009 10:54 AMThe New Yorker, May 25, 20092 of 12http:/archives.newyorker.com/global/print.asp?path=/djvu/Conde%20Na.8/12/2009 10:54 AMThe N
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The New Yorker, May 25, 20091 of 12http:/archives.newyorker.com/global/print.asp?path=/djvu/Conde%20Na.8/12/2009 10:54 AMThe New Yorker, May 25, 20092 of 12http:/archives.newyorker.com/global/print.asp?path=/djvu/Conde%20Na.8/12/2009 10:54 AMThe N
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Ecosystem Collapse in Pleistocene Australia and aHuman Role in Megafaunal ExtinctionGifford H. Miller, et al.Science 309, 287 (2005);DOI: 10.1126/science.1111288The following resources related to this article are available online atwww.sciencemag.or
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Petsko Genome Biology 2010, 11:138http:/genomebiology.com/2010/11/10/138CO M M E N TA Faustian bargainGregory A Petsko*An open letter to George M Philip, President of theState University of New York At AlbanyDear President Philip,Probably the last
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Proc. Nati. Acad. Sci. USAVol. 81, pp. 801-805, February 1984EvolutionPeriodicity of extinctions in the geologic past(evolution/time series/paleontology)DAVID M. RAUP AND J. JOHN SEPKOSKI, JR.Department of Geophysical Sciences, University of Chicago
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Proc. Nati. Acad. Sci. USAVol. 81, pp. 801-805, February 1984EvolutionPeriodicity of extinctions in the geologic past(evolution/time series/paleontology)DAVID M. RAUP AND J. JOHN SEPKOSKI, JR.Department of Geophysical Sciences, University of Chicago
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440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsCourse OverviewFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 L
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440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
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Board Meeting HWY07MH024BoardSlides from http:/www.ntsb.gov/events/2008/Minneapolis-MN/presentations.htmFall 2011Lecture 2Engineering Case StudyAssistantsArturo VillegasInstructorsAlberto Cuitino, Po Ting Lin440:221 Intro to Engineering Mechanic
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440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesTopics t
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesEquilibr
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesEquilibr
Rutgers - ENGINEERIN - 221
440:221 Intro to Engineering Mechanics: StaticsAlberto Cuitino, Po Ting LinInstructorsArturo VillegasAssistantsFall 2011Based on Textbook Material: Engineering Mechanics Statics, 12th Edition, R.C. Hibbeler, Pearson 2010.1440:221 LecturesEquilibr
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https://socratic.org/questions/59a115447c01495b712c022b
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# Question c022b
Aug 26, 2017
$t = 1.010$ s.
#### Explanation:
There’s a little bit of ambiguity in this question: does it mean to rise up 5 m and fall back down or just to rise 5 m? I’ll write a solution for both as both solutions are just applications of constant acceleration equations and the former is simply the latter solution multiplied by two.
Time to rise up to 5 m.
Write down what you know:
$s = 5$ m
$u =$?
$v = 0$ m s⁻¹ (it rises up until it slows down to a halt before falling back down.
$a = - 9.8$ m s⁻²
$t =$ ?
Use this equation: s = vt - ½at^2
Note that $v t = 0$ because $v = 0$.
Make $t$ the subject: ⇒ t = sqrt ((-2s)/a)
Substitute in the values: ⇒ t = ± sqrt ((-2 × 5)/-9.8) = ± 1.010# s.
Obviously the time cannot be negative so the solution for this problem is $t = 1.01$ s.
======
If the question requires the time to rise 5 m and fall back down then double the time ⇒ $t = 2.02$ s.
| 293 | 920 |
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