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The required distance is, d = |10–5|/√(-3)^2+(10)^2 = 5/√109. Example 1: Find the distance between P (3, -4) and Q(-5, -1). You can use the distance formula calculator to calculate any line segment. Rewrite y = 3x + 2 as ax + by + c = 0. y - y = 3x - y + 2. Thus, the line joining these two points i.e. d = a2 +b2 +c2. Example 5: If the distance between the points (a, 2) and (3, 4) be 8, then find the value of a. To find the distance between two points ( x 1, y 1) and ( x 2, y 2 ), all that you need to do is use the coordinates of these ordered pairs and apply the formula pictured below. find the distance from the point to the line, so my task was to find the distance between point A(3,0,4) to plane (x+1)/3 = y/4 = (z-10)/6 So heres how i tried to do this 1) Found that direction vector is u = ( 3, 4, 6) and the normal vector is the same n = (3,4,6) took the equation n * v = n * P Or normal vector * any point on a plane is the same as n * the point. Let d be the distance between both the lines. The length of the straight line from point A to point B above, can be found by using the Distance Formula which is: AB = … Comparing given equation with the general equation of line Ax + By + C = 0, we get, We know, the formula to find the distance between a point and a line is. The distance between two points of the xy-plane can be found using the distance formula. Example 2: Find the distance between the parallel lines -3x + 10y + 5 = 0 and -3x + 10y + 10 = 0. Example #1. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. So, PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ=\sqrt{(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2}PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2. Using y = 3x + 2, subtract y from both sides. Share with friends Previous RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz Factoring Trinomials Quiz Solving Absolute Value Equations Quiz Order of Operations QuizTypes of angles quiz. The steps to take to find the formula are outlined below.1) Write the equation ax + by + c = 0 in slope-intercept form.2) Use (x1, y1) to find the equation that is perpendicular to ax + by + c = 0 3) Set the two equations equal to each other to find expressions for the points of intersection (x2, y2)4) Use the distance formula, (x1, y1), and the expressions found in step 3 for (x2, y2) to derive the formula. Use the point (5, 1) to find b by letting x = 5 and y = 1.1 = (-1/3) à 5 + b, 1 = -5/3 + bb = 1 + 5/3 = 8/3y = (-1/3)x + 8/3Now, set the two equations equal to themselves, 3x + (1/3)x = 8/3 - 2(10/3)x = (8 - 6)/3(10/3)x = 2/3x = (2/3) à 3/10 = 2/10 = 1/5y = 3x + 2 = 3 à 1/5 + 2 = 3/5 + 2 = 13/5The point of intersection between y = 3x + 2 and y = (-1/3)x + 8/3 is. An ordered pair (x, y) represents co-ordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the centre and y-coordinate (or ordinate) is the distance of the point from the centre. the distance between the line and point is Proof. To find the closest points along the lines you recognize that the line connecting the closest points has direction vector n = e 1 × e 2 = (− 20, − 11, − 26) If the points along the two lines are projected onto the cross line the distance is found with one fell swoop Example 3: Find the distance of line 2x + 3y – 13 = 0 from the point (1, –2). The distance formula from a point to line is as given below:. The distance between two points is the length of the line segment joining them (but this CANNOT be the length of the curve joining them). The distance from the point to the line, in the Cartesian system, is given by calculating … Basic-mathematics.com. The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. For example, if $$A$$ and $$B$$ are two points and if $$\overline{AB}=10$$ cm, it means that the distance between $$A$$ and $$B$$ is $$10$$ cm. For each point, there are always going to be two elements or centers that are uniquely connected to that point. play_arrow. Distance between two points. Then the distance formula between the points is given by. The equation of a line ax+by+c=0 in slope-intercept form is given by y=-a/bx-c/b, (1) so the line has slope -a/b. The Distance Formula always act as a useful distance finder tool whenever it comes to finding the distance among any two given points. Example 7: The distance of the middle point of the line joining the points (asinθ, 0)and (0, acosθ) from the origin is _______. Now, a vector from the point to the line is given by … Below is a graph of a straight line, with two different end points A and B. To derive the formula at the beginning of the lesson that helps us to find the distance between a point and a line, we can use the distance formula and follow a procedure similar to the one we followed in the last section when the answer for d was 5.01. The formula for calculatin . To find the distance between the point (x1,y1) and the line with equation ax + bx + c = 0, you can use the formula below. The distance between a point and a line is defined to be the length of the perpendicular line segment connecting the point to the given line. The distance between parallel lines is the shortest distance from any point on one of the lines to the other line. We can just look for the point of intersection between y = 3x + 2 and the line that is perpendicular to y = 3x + 2 and passing through (5, 1)The line that is perpendicular to y = 3x + 2 is given by y = (-1/3)x + b. The distance formula from a point to line is as given below:, P Q PQ P Q = ∣ A x 1 + B y 1 + C ∣ A 2 + B 2 = \frac{\left | Ax_{1} + By_{1} + C \right |}{\sqrt{A^{2} + B^{2}}} = A 2 + B 2 ∣ A x 1 + B y 1 + C ∣ Distance between Two Parallel Lines. The distance (or perpendicular distance) from a point to a line is the shortest distance from a fixed point to any point on a fixed infinite line in Euclidean geometry. You can count the distance either up and down the y-axis or across the x-axis. Solution: Given lines is 5x – 12y + 26 = 0, Equations of any line parallel to given equation is 5x – 12y + m = 0, Distance between both the lines is 4 units (given), Required equations of lines are: 5x – 12y – 26 = 0 and 5x – 12y + 78 = 0. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! Find the perpendicular distance from the point $(5, -1)$ to the line $y = \frac{1}{2}x + 2$ example 3: ex 3: Find the perpendicular distance from the point $(-3, 1)$ to the line $y = 2x + 4$. The distance between any two points is the length of the line segment joining the points. Now consider the distance from a point (x_0,y_0) to the line. We already have (5,1) that is not located on the line y = 3x + 2. To use the distance formula, we need two points. Let P and Q be two given points whose polar coordinates are (r1, θ1) and (r2, θ2) respectively. (Does not work for vertical lines.) Formula for Distance between Two Points . 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Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Applying formula: d=∣C1 − C2∣A2 + B2d=\frac{|C_1 ~- ~C_2|}{\sqrt{A^2~ +~ B^2}}d=A2 + B2∣C1 − C2∣. Distance between a point and a line. How to find the distance between two points if their coordinates are given? Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. If you can solve these problems with no help, you must be a genius! For example, we can find the lengths of sides of a triangle using the distance formula and determine whether the triangle is scalene, isosceles or equilateral. Hang in there tight. Let's learn more about this. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright © 2008-2019. To take us from his Theorem of the relationships among sides of right triangles to coordinate grids, the mathematical world had to wait for René Descartes. Consider Ax + By + C = 0 be an equation of line and P be any point in the cartesian-coordinate plane having coordinates P(x1, y1). the perpendicular should give us the said shortest distance. If we call this distance d, we could say that the distance squared is equal to. Then, plug the coordinates into the distance formula. The distance is found using trigonometry on the angles formed. By formula Given the equation of the line in slope - intercept form, and the coordinates of the point, a formula yields the distance between them. Your email is safe with us. What is the Distance Formula? Once you've done that, just add the numbers that are under the radical sign and solve for d. Example #1Find the distance between a point and a line using the point (5,1) and the line y = 3x + 2.Rewrite y = 3x + 2 as ax + by + c = 0Using y = 3x + 2, subtract y from both sides.y - y = 3x - y + 20 = 3x - y + 23x - y + 2 = 0a = 3, b = -1, and c = 2x1 = 5 and y1 = 1, Example #2Find the distance between a point and a line using the point (-4,2) and the line y = -2x + 5.Rewrite y = -2x + 5 as ax + by + c = 0Using y = -2x + 5, subtract y from both sides.y - y = -2x - y + 50 = -2x - y + 5-2x - y + 5 = 0a = -2, b = -1, and c = 5x1 = -4 and y1 = 2. Since this format always works, it can be turned into a formula: Distance Formula: Given the two points (x1, y1) and (x2, y2), the distance d between these points is given by the formula: d = ( x 2 − x 1) 2 + ( y 2 − y 1) 2. d = \sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2\,} d = (x2. (2) Therefore, the vector [-b; a] (3) is parallel to the line, and the vector v=[a; b] (4) is perpendicular to it. So the distance between these two points is really just the hypotenuse of a right triangle that has sides 6 and 2. Therefore, d = |2(1) + 3(-2) + (-13)|/√(22+32) = 17/√13. The distance between these points is given as: Formula to find Distance Between Two Points in 3d plane: Below formula used to find the distance between two points, Let P(x1, y1, z1) and Q(x2, y2, z2) are the two points in three dimensions plane. To use the distance formula to find the length of a line, start by finding the coordinates of the line segment's endpoints. It is also described as the shortest line segment from a point of line. Solution: Mid-point will be (a sinθ2, a cosθ2)\left( \frac{a\,\sin \theta }{2},\,\frac{a\,\cos \theta }{2} \right)(2asinθ,2acosθ) and distance from origin will be (a sinθ2−0)2+(a cosθ2−0)2=a2\sqrt{{{\left( \frac{a\,\sin \theta }{2}-0 \right)}^{2}}+{{\left( \frac{a\,\cos \theta }{2}-0 \right)}^{2}}}=\frac{a}{2}(2asinθ−0)2+(2acosθ−0)2=2a. So, if we take the normal vector \vec{n} and consider a line parallel t… It is the length of the line segment that is perpendicular to the line and passes through the point. The length of the hypotenuse is the distance between the two points. = (−5−(−3))2+((−1)−(−4))2\sqrt{\left(-5-(-3)\right)^{2}+\left((-1)-(-4)\right)^{2}}(−5−(−3))2+((−1)−(−4))2. Fractions should be entered with a forward such as '3/4' for the fraction $$\frac{3}{4}$$. Here, A = -3, B = 10, C1 = 5 and C2 = 10. Shortest distance from a point to a line. One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. (1/5, 13/5) or (0.25, 2.6)Find the distance using the points (5,1) and (0.25, 2.6). If the two points lie on the same horizontal or same vertical line, the distance can be found by subtracting the coordinates that are not the same. It is the length of the line segment which joins the point to the line and is perpendicular to the line. Pythagoras was a generous and brilliant mathematician, no doubt, but he did not make the great leap to applying the Pythagorean Theorem to coordinate grids. Deriving the distance between a point and a line is among of the toughest things you have ever done in life. 0 = 3x - y + 2. Solution: (a−3)2+(2−4)2=82 ⇒ (a−3)2=60⇒ a−3=± 215 ⇒ a=3 ± 215{{(a-3)}^{2}}+{{(2-4)}^{2}}={{8}^{2}}\,\,\\\Rightarrow \,\,{{(a-3)}^{2}}=60 \\\Rightarrow \,\,a-3=\pm \,2\sqrt{15}\,\,\\\Rightarrow \,\,a=3\,\,\pm \,2\sqrt{15}(a−3)2+(2−4)2=82⇒(a−3)2=60⇒a−3=±215⇒a=3±215. edit close. In the -dimensional case, the distance between and is . Given a point a line and want to find their distance. In a Cartesian grid, a line segment that is either vertical or horizontal. Points on the line have the vector coordinates [x; -a/bx-c/b]=[0; -c/b]-1/b[-b; a]x. d = ∣ a ( x 0) + b ( y 0) + c ∣ a 2 + b 2. All right reserved, $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$, $$d = \frac{|3 \times 5 + -1 \times 1 + 2|}{\sqrt{3^2 + (-1)^2}}$$, $$d = \frac{|15 + -1 + 2|}{\sqrt{9 + 1}}$$, $$d = \frac{|-2 \times -4 + -1 \times 2 + 5|}{\sqrt{(-2)^2 + (-1)^2}}$$, $$d = \frac{|8 + -2 + 5|}{\sqrt{4 + 1}}$$, $$d = \sqrt{(5 - 0.25)^2 + (1-2.6)^2}$$, $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$, $$d = \sqrt{(\frac{-ca + b^2x_1 - y_1ba}{a^2 + b^2} - x_1)^2 + (\frac{-bc + a^2y_1 - abx_1}{a^2 + b^2} - y_1)^2}$$, $$d = \sqrt{(\frac{-ca - y_1ba- a^2x_1 }{a^2 + b^2})^2 + (\frac{-bc - abx_1 - b^2y_1}{a^2 + b^2})^2}$$, $$d = \sqrt{[\frac{-a(c + y_1b + ax_1) }{a^2 + b^2}]^2 + [\frac{-b(c + ax_1 + by_1}{a^2 + b^2}]^2}$$, $$d = \sqrt{\frac{(-a)^2(c + y_1b + ax_1)^2 }{(a^2 + b^2)^2} + \frac{(-b)^2(c + ax_1 + by_1)^2}{(a^2 + b^2)^2}}$$, $$d = \sqrt{\frac{a^2(c + y_1b + ax_1)^2 }{(a^2 + b^2)^2} + \frac{b^2(c + ax_1 + by_1)^2}{(a^2 + b^2)^2}}$$, $$d = \sqrt{\frac{(a^2+ b^2)(c + y_1b + ax_1)^2 }{(a^2 + b^2)^2} }$$, $$d = \sqrt{\frac{(c + y_1b + ax_1)^2 }{(a^2 + b^2)} }$$, $$d = \frac{\left\lvert c + y_1b + ax_1 \right\rvert} {\sqrt{a^2 + b^2}}$$. This distance is actually the length of the perpendicular from the point to the plane. Let (x1,y1) be the point not on the line and let (x2,y2) be the point on the line. Formula to find Distance Between Two Points in 2d plane: Consider two points A(x1,y1) and B(x2,y2) on the given coordinate axis. 1) ax + by + c = 0ax - ax + by + c = -axby + c = -axby + c - c = -ax - cby = -ax - cy = -ax/b - c/by = (-a/b)x - c/b, 2) The line that is perpendicular to y = (-a/b)x - c/b can be written as, y = (b/a)x + y-interceptUse (x1, y1) to find y-intercepty1 = (b/a)x1 + y-intercepty-intercept = y1- (b/a)x1, 3) Set the two equations equal to each other to find expressions for the points of intersection (x2, y2)Set y = (-a/b)x - c/b and y = (b/a)x + y1- (b/a)x1 equal to each other, (b/a)x + (a/b)x = (ba/ba) à [(-c/b + (b/a)x1 - y1], [ (a2 + b2)/ab ] / x = (-ca + b2x1 - y1ba) / ba, x = (-ca + b2x1 - y1ba) / a2 + b2 ( this is x2 ), Now, let us find y2 using the equation y = (-a/b)x - c/b, (-a/b)x = -a/b[ (-ca + b2x1 - bay1) / (a2 + b2) ], (-a/b)x = (ca2 - ab2x1 + ba2y1) / b(a2 + b2)(-a/b)x - c/b = [(ca2 - ab2x1 + ba2y1) / b(a2 + b2)] - c/b(-a/b)x - c/b = (ca2 - ab2x1 + ba2y1 - ca2 - b2c) / b(a2 + b2)(-a/b)x - c/b = (- ab2x1 + ba2y1 - b2c) / b(a2 + b2), (-a/b)x - c/b = b[(- abx1 + a2y1 - bc)] / b(a2 + b2), (-a/b)x - c/b = (- abx1 + a2y1 - bc) / (a2 + b2), y = (- abx1 + a2y1 - bc) / (a2 + b2) (this is y2), Now, find the distance between a point and a line using (x1,y1) and (x2,y2), Top-notch introduction to physics. 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1. ## Word Problem
A boy got 50% of the questions on a test correct. If he had 10 questions correct out of the first 12, and 1/4 of the remaining questions correct, how many questions were on the test?
(1) 16 (3) 26
(2) 24 (4) 28
2. Originally Posted by symmetry
A boy got 50% of the questions on a test correct. If he had 10 questions correct out of the first 12, and 1/4 of the remaining questions correct, how many questions were on the test?
(1) 16 (3) 26
(2) 24 (4) 28
.5*x = (1/2)*x
10 + (1/4)*(x - 12) =(1/2)x
You should be able to solve for x from here.
3. ## ok
Thank you for the equation set up.
I can take it from there.
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# Geometry problem solver
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To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
Any problem, no matter how complex, can be solved if you break it down into smaller, more manageable pieces. The first step is to identify the goal, or what you want to achieve. Once you have a clear goal in mind, you can start to break the problem down into smaller steps that will lead you to your goal. It is important to be as specific as possible when identifying these steps, and to create a timeline for each one. Otherwise, it will be easy to get overwhelmed and lost in the process. Finally, once you have a plan in place, it is important to stick with it and see it through to the end. Only then can you achieve your goal and move on to the next problem.
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Maths answears Geometry photos How to contact problem solvers Solving for a variable Maths no problem
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### 1: Number and Algebra
#### 1.1: Number and place value
1.1.2: Investigate and use square roots of perfect square numbers
1.1.3: Apply the associative, commutative and distributive laws to aid mental and written computation
1.1.4: Compare, order, add and subtract integers
#### 1.2: Real numbers
1.2.2: Solve problems involving addition and subtraction of fractions, including those with unrelated denominators
1.2.3: Multiply and divide fractions and decimals using efficient written strategies and digital technologies
1.2.6: Connect fractions, decimals and percentages and carry out simple conversions
1.2.7: Find percentages of quantities and express one quantity as a percentage of another, with and without digital technologies.
1.2.8: Recognise and solve problems involving simple ratios
#### 1.3: Money and financial mathematics
1.3.1: Investigate and calculate 'best buys', with and without digital technologies
#### 1.4: Patterns and algebra
1.4.1: Introduce the concept of variables as a way of representing numbers using letters
1.4.3: Extend and apply the laws and properties of arithmetic to algebraic terms and expressions
#### 1.5: Linear and non-linear relationships
1.5.1: Given coordinates, plot points on the Cartesian plane, and find coordinates for a given point
1.5.2: Solve simple linear equations
### 2: Measurement and Geometry
#### 2.1: Using units of measurement
2.1.1: Establish the formulas for areas of rectangles, triangles and parallelograms and use these in problem solving
2.1.2: Calculate volumes of rectangular prisms
#### 2.3: Location and transformation
2.3.1: Describe translations, reflections in an axis, and rotations of multiples of 90° on the Cartesian plane using coordinates. Identify line and rotational symmetries
#### 2.4: Geometric reasoning
2.4.1: Classify triangles according to their side and angle properties and describe quadrilaterals
2.4.2: Demonstrate that the angle sum of a triangle is 180° and use this to find the angle sum of a quadrilateral
### 3: Statistics and Probability
#### 3.1: Chance
3.1.1: Construct sample spaces for single-step experiments with equally likely outcomes
3.1.2: Assign probabilities to the outcomes of events and determine probabilities for events
#### 3.2: Data representation and interpretation
3.2.2: Construct and compare a range of data displays including stem-and-leaf plots and dot plots
3.2.3: Calculate mean, median, mode and range for sets of data. Interpret these statistics in the context of data
Correlation last revised: 9/16/2020
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# Main Page
Welcome to the explain xkcd wiki!
We have an explanation for all 2058 xkcd comics, and only 28 (1%) are incomplete. Help us finish them!
## Latest comic
Modified Bayes' Theorem Title text: Don't forget to add another term for "probability that the Modified Bayes' Theorem is correct."
## Explanation
This explanation may be incomplete or incorrect: When using the Math-syntax please also care for a proper layout. Please edit the explanation below and only mention here why it isn't complete. Do NOT delete this tag too soon.
Bayes' Theorem is an equation in statistics that gives the probability of a given hypothesis accounting not only for a single experiment or observation but also for your existing knowledge about the hypothesis, i.e. its prior probability. Randall's modified form of the equation also purports to account for the probability that you are indeed applying Bayes' Theorem itself correctly by including that as a term in the equation.
Bayes' theorem is:
$P(H \mid X) = \frac{P(X \mid H) \, P(H)}{P(X)}$, where
• $P(H \mid X)$ is the probability that $H$, the hypothesis, is true given observation $X$. This is called the posterior probability.
• $P(X \mid H)$ is the probability that observation $X$ will appear given the truth of hypothesis $H$. This term is often called the likelihood.
• $P(H)$ is the probability that hypothesis $H$ is true before any observations. This is called the prior, or belief.
• $P(X)$ is the probability of the observation $X$ regardless of any hypothesis might have produced it. This term is called the marginal likelihood.
The purpose of Bayesian inference is to discover something we want to know (how likely is it that our explanation is correct given the evidence we've seen) by mathematically expressing it in terms of things we can find out: how likely are our observations, how likely is our hypothesis a priori, and how likely are we to see the observations we've seen assuming our hypothesis is true. A Bayesian learning system will iterate over available observations, each time using the likelihood of new observations to update its priors (beliefs) with the hope that, after seeing enough data points, the prior and posterior will converge to a single model.
If $P(C)=1$ the modified theorem reverts to the original Bayes' theorem (which makes sense, as a probability one would mean certainty that you are using Bayes' theorem correctly).
If $P(C)=0$ the modified theorem becomes $P(H \mid X) = P(H)$, which says that the belief in your hypothesis is not affected by the result of the observation (which makes sense because you're certain you're misapplying the theorem so the outcome of the calculation shouldn't affect your belief.)
This happens because, if you apply the original theorem, the modified theorem can be rewritten as: $P(H \mid X) = P(H)(1-P(C)) + P(H \mid X)P(C)$. This is the linear-interpolated weighted average of the belief you had before the calculation and the belief you would have if you applied the theorem correctly. This goes smoothly from the not believing your calculation at all, keeping the same belief as before if $P(C)=0$ to changing your belief exactly as Bayes' theorem suggests when $P(C)=1$.
$1-P(C)$ is the probability that you are using the theorem incorrectly.
As an equation, the rewritten form makes no sense. $P(H \mid X) = P(H)(1-P(C)) + P(H \mid X)P(C)$ is strangely self-referential and reduces to the piecewise equation $\begin{cases}P(H \mid X) = P(H) & P(C) \neq 1 \\ 0 = 0 & P(C) = 1 \end{cases}$. However, the Modified Bayes Theorem includes an extra variable not listed in the conditioning, so a person with an AI background might understand that Randal was trying to write an expression for updating $P(H \mid X)$ with knowledge of $C$ i.e. $P(H \mid X,C)$, the belief in the hypothesis given the observation $X$ and the confidence that you were applying Bayes' theorem correctly $C$, for which the expression $P(H \mid X,C) = P(H)(1-P(C)) + P(H \mid X)P(C)$ makes some intuitive sense.
The title text suggests that an additional term should be added for the probability that the Modified Bayes Theorem is correct. But that's *this* equation, so it would make the formula self-referential. It could also result in an infinite regress -- we'd need another term for the probability that the version with the probability added is correct, and another term for that version, and so on. It's also unclear what the point of using an equation we're not sure of is (although sometimes we can: Newton's Laws are not as correct as Einstein's Theory of Relativity but they're a reasonable approximation in most circumstances. Alternatively, ask any student taking a difficult exam with a formula sheet.).
## Transcript
Modified Bayes' theorem:
P(H|X) = P(H) × (1 + P(C) × ( P(X|H)/P(X) - 1 ))
H: Hypothesis
X: Observation
P(H): Prior probability that H is true
P(X): Prior probability of observing X
P(C): Probability that you're using Bayesian statistics correctly
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1. ## Break even point
So here's the question that i have and i hope someone can help me with this because i am kind of stumped. MY main struggle is determining which is VC and which is FC.
John and Jack have opened a shop in Kelowna to build customized patio furniture. The projected selling price of a patio set will be $3,000. They estimate that the table will use twenty (20) board feet of lumber at a cost of$12 per board foot, $10 per table for hardware, and another$15 for custom finishing materials such as stain, varnish, or painting. Each patio set comes with four (4) chairs. Each chair will use six (6) board feet of lumber, $6 for hardware, and$10 for finishing materials. They estimate the labour time per table to be twenty-five (25) hours, and per chair, ten (10) hours for construction and finishing work. John found warehouse space that they can rent for $2,300 per month including heat. Other utilities will be$275 per month. Other required costs are insurance at $3,500 per year. They will pay themselves$20 per hour (treat this as a variable cost). They will pay a sales commission (based on selling price) of 4% per complete set sold. Jack estimates that they can build 125 sets per year.
1. How many patio sets must they construct to break-even? Calculate the break-even point in:
a. units.
b. sales dollars.
2. Originally Posted by petedam
So here's the question that i have and i hope someone can help me with this because i am kind of stumped. MY main struggle is determining which is VC and which is FC.
John and Jack have opened a shop in Kelowna to build customized patio furniture. The projected selling price of a patio set will be $3,000. They estimate that the table will use twenty (20) board feet of lumber at a cost of$12 per board foot, $10 per table for hardware, and another$15 for custom finishing materials such as stain, varnish, or painting. Each patio set comes with four (4) chairs. Each chair will use six (6) board feet of lumber, $6 for hardware, and$10 for finishing materials. They estimate the labour time per table to be twenty-five (25) hours, and per chair, ten (10) hours for construction and finishing work. John found warehouse space that they can rent for $2,300 per month including heat. Other utilities will be$275 per month. Other required costs are insurance at $3,500 per year. They will pay themselves$20 per hour (treat this as a variable cost). They will pay a sales commission (based on selling price) of 4% per complete set sold. Jack estimates that they can build 125 sets per year.
1. How many patio sets must they construct to break-even? Calculate the break-even point in:
a. units.
b. sales dollars.
Selling Price: 3000 (given)
Labour time (time per table + time per chair):
$25 + 4\cdot 10 = 65 \text{ hours}$
Materials per item (selling price-total cost)
$
20 \cdot 12 + 10 + 15 + 4(12 \cdot 6 +6 + 10) +0.04 \cdot 3000 + 65 \times 20 = 2192$
(edit: 20 should be 12 since each lumber is 12 not 20)
Yearly Production = 125 (given)
Other costs (Insurance + Storage + Rent)
$3500 + 275 \cdot 12 + 2300 \times 12 = 34400$
To break even Revenue = Loss
$3000n - (2192n + 34400) = 0$
Solve for n which is the number of units sold
3. can you tell me why you did this (4(20.6 + 6 +10)
as well you treated wages as fixed cost...i am suppose to add to this Materials per item (selling price-total cost) aren't I?
4. Originally Posted by petedam
can you tell me why you did this (4(20.6 + 6 +10)
as well you treated wages as fixed cost...i am suppose to add to this Materials per item (selling price-total cost) aren't I?
Yes you're right, I changed this in hindsight. The 4(20*6+6+10) is the cost of the chairs. Each chair costs 20*6+6+10 and there are 4 of them
5. Given what you've been asked, essentially all costs are variable except rent, insurance and utilities, since they will be relatively fixed regardless of the units produced -- although in long term, all these costs mentioned are not really fixed (you can cancel insurance, and end your lease, quite easily as compared to owning a facility).
The other costs are variable because they scale with the production. So find the level of production needed to cover the above costs.
6. according to your revenue equation....i got an n value 0f 0.02241...it doesnt make sense.
7. Originally Posted by e^(i*pi)
To break even Revenue = Loss
$3000n - (2192n + 34400) = 0$
Solve for n which is the number of units sold
Originally Posted by petedam
according to your revenue equation....i got an n value 0f 0.02241...it doesnt make sense.
I get an answer of 42.57
$n = \frac{34400}{3000-2192} \approx 42.57$
I attach an excel file for you to look at if you wish (it has the same info on, nothing new). It's just be easier to chop around the numbers
nb: it should be safe but since I use linux I can't be sure that it's totally safe.
8. Originally Posted by e^(i*pi)
I get an answer of 42.57
$n = \frac{34400}{3000-2192} \approx 42.57$
I attach an excel file for you to look at if you wish (it has the same info on, nothing new). It's just be easier to chop around the numbers
nb: it should be safe but since I use linux I can't be sure that it's totally safe.
sorry....i am an idiot and can't solve for n.
9. (edit: 20 should be 12 since each lumber is 12 not 20)
shouldn't it be 2037?
Because there are two of them should it be (65 x 20) x 2? for labour
Again thanks a bunch!
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# 16.28 Significant differences
Before you read this, I suggest that you read posts 16.24 and 16.26.
In this post I want to explore how we make decisions based on experimental results.
Suppose two groups of rats, are identical in all respects, eat the same diet and live in identical conditions. However, one group is given a dietary supplement and the other isn’t.
At the end of a year, all rats that were given the supplement have a mean body mass of 0.420 kg (treated) but those that weren’t have a mean body mass of 0.404 kg (untreated). Can we decide that the supplement is associated with increased body mass? Because these numbers are means of a large range of values, we can’t make this decision confidently. We need more information.
In the graphs below I have plotted the number of rats with each body mass in the range 0.300 to 0.500 kg in steps of 0.005 kg, for the treated rats (red dots) and the untreated rats (blue dots). Although the mean body mass of the treated group is higher, there is a very wide spread of values and the results overlap with those from the untreated group.
So, how are we to decide whether there is a difference? In the graph, the red dots all lie on the bell-shaped curve of a normal distribution (post 16.26) and so do all the blue dots. We can then calculate 95% confidence intervals, as described in posts 16.24 and 16.26. The stages of the calculation are shown in the table below.
Remember that an experiment can’t show that an idea is correct – it can only show that an idea is wrong (see post 16.3). So we start off with the null hypothesis that all the blue dots and all the red dots on the graph come from the same population of results; it’s called a “null hypothesis” because it supposes that there is no difference between the numbers represented by the blue dots and those represented by the red dots.
From the table we see that the 95% confidence interval for the untreated group does not overlap with the 95% confidence interval for the treated group. Therefore, the probability that our null hypothesis is false is at least 95%. Conventionally, we accept that this level of probability is evidence that there is an effect. We can say that there is a significant difference (p < 0.05) between the two groups; this means that there is a probability, p, of less than 0.05 (5%) that the two sets of numbers come from the same population. But be careful – we may need to use statistics to design something – then we need to be more than 95% confident that it won’t fail!
We can also calculate the probability that the numbers represented by the blue dots comes from the same population as the numbers represented by the red dots. We could do this using the properties of the normal distribution. But a more reliable method, especially for a small sample, is to use the ttest. The t-test assumes that our data are normally distributed, so we must check that they lie on the bell-shaped curve before we use this test. If you want to know how the t-test works, see https://en.wikipedia.org/wiki/Student%27s_t-test . However, it’s best to use established computer software to make statistical calculations, as explained in post 16.24. In Microsoft Excel, you can use the T.TEST function. But you could also use the online calculator at http://www.socscistatistics.com/tests/studentttest/ . For the results shown in the graph above, the probability that the red and blue dots come from the same population is only 3 × 10-12 which is zero for all practical purposes. This is a much lower p value than you ever get from a real experiment – because I invented the results to show how decisions can be made from experimental results.
Suppose the results of an experiment are not normally distributed. What can we do? There are statistical tests that don’t assume a normal distribution – they are called nonparametric tests. They are less powerful at detecting significant differences than, for example, the t-test. This is because the t-test uses additional information about the data – that it is normally distributed. If the results from our two groups of rats had not been normally distributed, we could have compared them using a non-parametric test called the Mann-Whitney test (see https://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U_test ). You can’t use Microsoft Excel to do the Mann-Whitney test but there is an online calculator for calculating it at http://www.socscistatistics.com/tests/mannwhitney/ . Alternatively you can use specialist Open Source software: like GNU PSP (https://www.gnu.org/software/pspp/ ) or R (https://www.r-project.org/ ). R is very versatile but difficult to learn; I’ve never used PSP.
The important message is that we can’t trust that one experimental result is bigger than another without investigating whether the difference could have arisen by chance.
Related posts
Follow-up post
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# Statistical dispersion
Article Id: WHEBN0023773247
Reproduction Date:
Title: Statistical dispersion Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Statistical dispersion
In statistics, dispersion (also called variability, scatter, or spread) denotes how stretched or squeezed[1] is a distribution (theoretical or that underlying a statistical sample). Common examples of measures of statistical dispersion are the variance, standard deviation and interquartile range.
Dispersion is contrasted with location or central tendency, and together they are the most used properties of distributions.
## Contents
• Measures of statistical dispersion 1
• Sources of statistical dispersion 2
• A partial ordering of dispersion 3
• References 5
## Measures of statistical dispersion
A measure of statistical dispersion is a nonnegative real number that is zero if all the data are the same and increases as the data become more diverse.
Most measures of dispersion have the same units as the quantity being measured. In other words, if the measurements are in metres or seconds, so is the measure of dispersion. Such measures of dispersion include:
These are frequently used (together with scale factors) as estimators of scale parameters, in which capacity they are called estimates of scale. Robust measures of scale are those unaffected by a small number of outliers, and include the IQR and MAD.
All the above measures of statistical dispersion have the useful property that they are location-invariant, as well as linear in scale. So if a random variable X has a dispersion of SX then a linear transformation Y = aX + b for real a and b should have dispersion SY = |a|SX.
Other measures of dispersion are dimensionless. In other words, they have no units even if the variable itself has units. These include:
There are other measures of dispersion:
Some measures of dispersion have specialized purposes, among them the Allan variance and the Hadamard variance.
For categorical variables, it is less common to measure dispersion by a single number; see qualitative variation. One measure that does so is the discrete entropy.
## Sources of statistical dispersion
In the physical sciences, such variability may result from random measurement errors: instrument measurements are often not perfectly precise, i.e., reproducible, and there is additional inter-rater variability in interpreting and reporting the measured results. One may assume that the quantity being measured is stable, and that the variation between measurements is due to observational error. A system of a large number of particles is characterized by the mean values of a relatively few number of macroscopic quantities such as temperature, energy, and density. The standard deviation is an important measure in Fluctuation theory, which explains many physical phenomena, including why the sky is blue.[2]
In the biological sciences, the quantity being measured is seldom unchanging and stable, and the variation observed might additionally be intrinsic to the phenomenon: It may be due to inter-individual variability, that is, distinct members of a population differing from each other. Also, it may be due to intra-individual variability, that is, one and the same subject differing in tests taken at different times or in other differing conditions. Such types of variability are also seen in the arena of manufactured products; even there, the meticulous scientist finds variation.
In economics, finance, and other disciplines, regression analysis attempts to explain the dispersion of a dependent variable, generally measured by its variance, using one or more independent variables each of which itself has positive dispersion. The fraction of variance explained is called the coefficient of determination.
## A partial ordering of dispersion
A mean-preserving spread (MPS) is a change from one probability distribution A to another probability distribution B, where B is formed by spreading out one or more portions of A's probability density function while leaving the mean (the expected value) unchanged.[3] The concept of a mean-preserving spread provides a partial ordering of probability distributions according to their dispersions: of two probability distributions, one may be ranked as having more dispersion than the other, or alternatively neither may be ranked as having more dispersion.
## References
1. ^ [1]
2. ^ McQuarrie, Donald A. (1976). Statistical Mechanics. NY: Harper & Row.
3. ^ Rothschild, Michael; Stiglitz, Joseph (1970). "Increasing risk I: A definition".
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
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A shopkeeper purchases an article for $6200Rs$ and sells it to customers for $8500Rs$. If sales tax is $8\%$, find the VAT paid by the shopkeeper?
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Hint: Note down all the given values that are given in the question. Find the Tax paid by the shopkeeper and Tax charged by the shopkeeper. The difference between those two would be the VAT paid by the shopkeeper.
It is mentioned in the question that A shopkeeper purchases an article for $6200Rs$ and sells it to customers for $8500Rs$. Sales tax is $8\%$.
So,
Purchase price of the shopkeeper that is the Cost price we can say is $6,200Rs$
Selling price of the article is $8,500Rs$.
Now,
Tax paid by the shopkeeper is calculated on the cost price of the article.
So, Tax paid by the shopkeeper $= 8\%$ of $6,200$ $= \dfrac{8}{{100}} \times 6200$
$= 8 \times 62 \\ = 496Rs \\$
But the Tax that is charged by the shopkeeper will be decided on the selling price of the article.
Similarly, Tax charged by the shopkeeper $= 8\%$ of $6,200$ $= \dfrac{8}{{100}} \times 8500$
$= 8 \times 85 \\ = 680Rs \\$
At last we got two Tax values, Now, to get the VAT value just take the difference between them. Because the VAT refers to the Value Added Tax that should be decided on the selling price too.
Hence VAT paid by the shopkeeper $=$ Tax charged by shopkeeper $-$ Tax paid by the shopkeeper.
We have already derived the above values. Let us substitute them here to get the VAT value.
Hence VAT paid by the shopkeeper $= 680 - 496 = 184Rs$
So, the VAT paid by the shopkeeper is $184Rs$.
Note: When you look at these kinds of problems, be careful in looking whether they have mentioned the cost price or selling price. Because mostly they use terms like purchasing price for the shopkeeper or for the customer. In that situation pay attention in note downing the values.
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# Unit 7: Volume
## Overview
Ever wondered how to teach the concept of volume in an engaging way to your 5th-grade students?
In this unit plan, students will first learn about volume by counting unit cubes and edge lengths. Later on, they will build on this to learn about the volume of right rectangular prisms with whole number edge lengths. Lastly, they will learn about finding the volume of composite rectangular prisms through decomposition and application of the volume formula.
Each lesson helps students master the unit through artistic, interactive guided notes, check for understanding, and culminates with a real-life example that explores how volume is used in practical scenarios. It's an exciting way to teach the concept of volume while comprehensively meeting standards.
## Get the Unit Materials
### Volume of Rectangular Prisms, Composite Solids Guided Notes | 5th Grade Unit
\$8.99 · Extensions sold separately
## Learning Objectives
After this unit, students will be able to:
• Identify unit cubes and understand their role in calculating volume
• Define volume and explain its importance in measuring space
• Calculate the volume of right rectangular prisms with whole number edge lengths using the formula V = l × w × h
• Decompose composite rectangular prisms into simpler shapes in order to calculate their volume
• Apply the volume formula (V = l × w × h) to find the volume of composite rectangular prisms
## Prerequisites
Before starting this unit, students should be familiar with:
• Basic understanding of multiplication and division
• Familiarity with calculating the area of rectangles
• Understanding of the concepts of length, width, and height
• Ability to identify 2D and 3D shapes
• Understanding of basic geometry language (e.g., vertices, edges, faces)
• Proficiency in working with whole numbers
## Key Vocabulary
• Volume
• Unit Cubes
• Rectangular Prism
• Composite
• Decomposing
• Formula
• Edge Lengths
• Solid Figures
## Extensions
#### Volume Pixel Art Unit BUNDLE | 5th Grade CCSS | Cubic Units, Prisms
\$10.99
Students practice volume, including counting cubic units to find volume, calculating volume by multiplying l x w x h, volume of rectangular prisms word problems, and additive volume of composite figures made up of rectangular prisms in this 5th Grade CCSS Volume Pixel Art Bundle. Contains 4 products with 8 pixel art activities total for a complete 5th grade Common Core volume unit.
These self-checking Google Sheets give students instant feedback. Incorrect answers are highlighted in red, and correct answers turn green and unlock parts of a mystery picture. They’re easy to assign, keep students engaged, and free up your time for differentiation and more.
Unit 6
2D Geometry
## Want more ideas and freebies?
Get my free resource library with digital & print activities—plus tips over email.
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# Six Probability Problems
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This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
1) In a survey of 125 college students, it was found that of three newspapers, the Wall Street Journal, New York Times, and Chicago Tribune:
40 read the New York Times
15 read the Wall Street Journal
25 read the Chicago Tribune and New York Times
8 read the New York Times and Wall Street Journal
3 read the Chicago Tribune and Wall Street Journal
a) How many read none of these papers?
b) How many read only the Chicago Tribune?
c) How many read neither the Chicago Tribune nor the New York Times?
2) A survey of families with 2 children is made, and the gender of the children is recorded. Describe the sample space and draw a tree diagram of this experiment.
3) A jar contains 3 white marbles, 2 yellow marbles, 4 red marbles, and 5 blue marbles. Two marbles are picked at random. What is the probability that
a) Both are blue?
b) Exactly 1 is blue?
c) At least 1 is blue?
4) A better coin is such that the probability of heads (H) is ¼ and the probability of tails (T) is 3/4. Show that in flipping this coin twice the events E and F defined below are independent.
E: A head turns up in the first throw
F: A tail turns up in the second throw
5) The records of Midwestern University show that in one semester, 38% of the students failed mathematics, 27% of the students failed physics, and 9% of the students failed mathematics and physics. A student is selected at random.
a) If a student failed physics, what is the probability that he or she failed mathematics?
b) If a student failed mathematics, what is the probability that he or she failed physics?
c) What is probability that he or she failed mathematics or physics?
6) In a certain population of people, 25% are blue-eyed and 75% are brown-eyed. Also, 10% of the blue eyed people are left-handed and 5% of the brown-eyed people are left-handed.
a) What is the probability that a person chosen at random is blue-eyed and left-handed?
b) What is the probability that a person chosen at random is left-handed?
c) What is the probability that a person is blue-eyed, given that the person is left-handed?
https://brainmass.com/math/probability/six-probability-problems-18799
#### Solution Preview
Please see the attached file for the complete solution.
Thanks for using BrainMass.
1) In a survey of 125 college students, it was found that of three newspapers, the Wall Street Journal, New York Times, and Chicago Tribune:
60 read the Chicago Tribune, Denote by A
40 read the New York Times, Denote by B
15 read the Wall Street Journal, Denote by C
25 read the Chicago Tribune and New York Times , Denote by AB
8 read the New York Times and Wall Street Journal, Denote by BC
3 read the Chicago Tribune and Wall Street Journal, Denote by AC
1 read all three ,Denote by ABC
a) How many read none of these papers?
Solution. See Venn graph above, we can get
So,
So, 45 read none of these papers.
b) How many read only the Chicago Tribune?
Solution.
c) How many read neither the Chicago Tribune nor the New York Times?
Solution. See Venn graph above, we have
So,
So 50 read neither the Chicago Tribune nor the New York Times
2) A survey of families with 2 children is made, and the gender of the children is recorded. Describe the sample space and draw a tree diagram of this experiment.
Solution. Assume that A survey of families with 2 children is made. Then the sample space S is ...
#### Solution Summary
Six probability problems are solved using Venn diagrams, tree diagrams and proabability expressions.
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## Solution to 2001 Problem 32
Recall the equation\begin{align*}E^2 = m^2 c^4 + p^2 c^2\end{align*}where $E$ is the total energy of a particle, $m$ is the rest mass of the particle, $c$ is the speed of light, and $p$ is the particle's relativistic momentum. We are given that $E = 2 mc^2$, therefore, this equation becomes\begin{align*}4 m^2 c^4 = m^2 c^4 + p^2 c^2 \Rightarrow p = \boxed{mc \sqrt{3}}\end{align*}Therefore, answer (D) is correct.
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# Higher Maths 2.1.2 - Quadratic Functions
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### Higher Maths 2.1.2 - Quadratic Functions
1. 1. SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART
2. 2. Any function containing an term is called a Quadratic Function . The Graph of a Quadratic Function NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART The graph of a Quadratic Function is a type of symmetrical curve called a parabola . x 2 f ( x ) = ax 2 + bx + c General Equation of a Quadratic Function x a > 0 a < 0 Minimum turning point Maximum turning point with a ≠ 0 turning point ( f ) x
3. 3. Sketching Quadratic Functions NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Before it is possible to sketch the graph of a quadratic function, the following information must be identified: • the nature of the turning point i.e. • the coordinates of the y -intercept • the zeroes or ‘roots’ of the function i.e. the x -intercept(s), if any minimum or maximum • the location of the axis of symmetry and coordinates of the turning point a > 0 or a < 0 substitute x = 0 solve f ( x ) = 0 : ax 2 + bx + c = 0 evaluate f ( x ) at axis of symmetry
4. 4. Perfect Squares NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x 2 + 6 x + 2 x 2 + 6 x + 2 ( ) 2 + 9 – 9 ( ) x + 3 – 7 A perfect square is an expression that can be written in the form Example Complete the square for x 2 + 6 x + 2 = = ( ... ) 2 x 2 + 4 x + 4 ( ) 2 x + 2 = x 2 + 5 x + 9 = ( ) 2 x + ? Step One: Separate number term Step Two: Try to form a perfect square from remaining terms Step Three: Remember to balance the extra number, then write out result
5. 5. Completing the Square NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART y = x 2 – 8 x + 19 = x 2 – 8 x + 19 = ( ) 2 + 16 – 16 ( ) x – 4 + 3 It is impossible for a square number to be negative. The minimum possible value of is zero. ( x – 4 ) 2 The minimum possible value of y is 3 . x = 4 . This happens when The minimum turning point is at ( 4 , 3 ) A turning point can often be found by completing the square. Example y - coordinate x - coordinate Find the minimum turning point of
6. 6. Quadratic Equations can be solved in several different ways: • using a graph to identify roots • factorising • completing the square • using the quadratic formula SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Solving Quadratic Equations NOTE Example f ( x ) = 0 Use the graph to solve x x = -2 - 2 5 x = 5 or Example 2 Solve 6 x 2 + x – 15 = 0 6 x 2 + x – 15 = 0 ( 2 x – 3 )( 3 x + 5 ) = 0 The trinomial can be factorised... or 2 x – 3 = 0 3 x + 5 = 0 2 x = 3 3 x = - 5 x = 3 2 x = - 5 3 ( f ) x
7. 7. If a Quadratic Equation cannot be factorised, it is sometimes possible to solve by completing the square. SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Solving Quadratic Equations by Completing the Square NOTE Example Solve x 2 – 4 x – 1 = 0 The trinomial cannot be factorised so complete the square... x 2 – 4 x – 1 = 0 – 1 = 0 ( ) ( ) 2 + 4 – 4 x 2 – 4 x – 5 = 0 x – 2 ( ) 2 = 5 x – 2 x – 2 = 5 ± x = 2 5 ± Now solve for x ... x ≈ 4.24 - 0.24 or Not accurate! most accurate answer
8. 8. Quadratic Inequations NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART A Quadratic Inequation can be solved by using a sketch to identify where the function is positive or negative. Example Find the values of x for which 12 – 5 x – 2 x 2 > 0 Factorise 12 – 5 x – 2 x 2 = 0 ( 4 + x )( 3 – 2 x ) = 0 The graph has roots x = - 4 and and a y - intercept at 12 x - 4 3 2 Now sketch the graph: The graph is positive for - 4 < x < and negative for x < - 4 3 2 x > and 3 2 3 2 12
9. 9. The Quadratic Formula NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x = - b b 2 – ( 4 ac ) ± 2 a with a ≠ 0 f ( x ) = ax 2 + bx + c If a quadratic function has roots, it is possible to find them using a formula. This is very useful if the roots cannot be found algebraically, i.e. by factorising or completing the square . The roots of are given by x root If roots cannot be found using the quadratic formula, they are impossible to find . x no roots LEARN THIS ( f ) x ( f ) x
10. 10. Real and Imaginary Numbers NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 36 = ± 6 - 36 = It is impossible to find the square root of a negative number. In Mathematics, the square root of a negative number still exists and is called an imaginary number. It is possible for a quadratic equation to have roots which are not real . ? x 1 real root no real roots x x 2 real roots ( f ) x ( f ) x ( f ) x
11. 11. • If • If The Discriminant NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x = - b b 2 – ( 4 ac ) ± 2 a = b 2 – ( 4 ac ) The part of the quadratic formula inside the square root is known as the Discriminant and can be used to find the nature of the roots. The Discriminant • If b 2 – ( 4 ac ) > 0 there are two real roots. b 2 – ( 4 ac ) = 0 there is only one real root. b 2 – ( 4 ac ) < 0 the roots cannot be calculated and are imaginary or non-real . (‘real and unequal’) (‘real and equal’)
12. 12. Using the Discriminant to Find Unknown Coefficients NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Example The quadratic equation 2 x 2 + 4 x + p = 0 Find all possible values of p . has real roots. b 2 – ( 4 ac ) 0 a = 2 b = 4 c = p For real roots, 16 – 8 p 0 16 8 p 8 p 16 p 2 The equation has real roots for . p 2 (for the roots are imaginary or non-real ) p > 2
13. 13. Finding Unknown Coefficients using Quadratic Inequations NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Example Find q given that x 2 + ( q – 3 ) x + q = 0 has non-real roots. a = 1 b = ( q – 3 ) c = q b 2 – ( 4 ac ) < 0 For non-real roots, ( q – 3 ) 2 – 4 q < 0 q 2 – 6 q + 9 – 4 q < 0 q 2 – 10 q + 9 < 0 ( q – 9 ) ( q – 1 ) < 0 Sketch graph of the inequation: q 1 9 q 2 – 10 q + 9 < 0 for 1 < q < 9 The roots of the original equation are non-real when 1 < < 9 q
14. 14. Straight Lines and Quadratic Functions NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART When finding points of intersection between a line and a parabola: • equate the functions • rearrange into a quadratic equation equal to zero • solve for • substitute back to find x Example Find the points of intersection of y = x 2 + 3 x + 2 x 2 + 2 x = 0 y = x + 2 and x 2 + 3 x + 2 = x + 2 x ( x + 2 ) = 0 x = 0 x = - 2 or y = 2 y = 0 or y Points of intersection are ( - 2 , 0 ) and ( 0 , 2 ) .
15. 15. Tangents to Quadratic Functions NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART The discriminant can be used to find the number of points of intersection between a parabola and a straight line. • equate the functions and rearrange into an equation equal to zero • evaluate the discriminant of the new quadratic equation b 2 – ( 4 ac ) > 0 Two points of intersection b 2 – ( 4 ac ) = 0 One point of intersection b 2 – ( 4 ac ) < 0 No points of intersection the line is a tangent
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## Elementary Algebra
$14x^2-10x-8$
We use the rules of exponents to obtain: $\frac{56x^4-40x^3-32x^2}{4x^2}=\frac{56x^4}{4x^2}-\frac{40x^3}{4x^2}-\frac{32x^2}{4x^2}=14x^2-10x-8$
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# Solve for x in Radians sec(x)=- square root of 2
Solve for x in Radians sec(x)=- square root of 2
Take the inverse secant of both sides of the equation to extract from inside the secant.
Simplify the right side.
The exact value of is .
The secant function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from to find the solution in the third quadrant.
Simplify .
To write as a fraction with a common denominator, multiply by .
Combine fractions.
Combine and .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Subtract from .
Find the period of .
The period of the function can be calculated using .
Replace with in the formula for period.
The absolute value is the distance between a number and zero. The distance between and is .
Divide by .
The period of the function is so values will repeat every radians in both directions.
, for any integer
Do you know how to Solve for x in Radians sec(x)=- square root of 2? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name six hundred ninety-three million three hundred thirty-six thousand two hundred thirty-eight
### Interesting facts
• 693336238 has 16 divisors, whose sum is 1195697664
• The reverse of 693336238 is 832633396
• Previous prime number is 167
### Basic properties
• Is Prime? no
• Number parity even
• Number length 9
• Sum of Digits 43
• Digital Root 7
### Name
Name sixty-five million eighty-five thousand forty-six
### Interesting facts
• 65085046 has 16 divisors, whose sum is 105316848
• The reverse of 65085046 is 64058056
• Previous prime number is 701
### Basic properties
• Is Prime? no
• Number parity even
• Number length 8
• Sum of Digits 34
• Digital Root 7
### Name
Name two hundred thirteen million seven hundred ninety-four thousand six hundred sixty-eight
### Interesting facts
• 213794668 has 8 divisors, whose sum is 481038012
• The reverse of 213794668 is 866497312
• Previous prime number is 2
### Basic properties
• Is Prime? no
• Number parity even
• Number length 9
• Sum of Digits 46
• Digital Root 1
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# What is the standard form of y= 2(x-3)^3-x?
Jul 25, 2016
$y = 2 {x}^{3} - 18 {x}^{2} + 53 x - 54$
#### Explanation:
Since ${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$, you can write
$y = 2 \left({x}^{3} - 9 {x}^{2} + 27 x - 27\right) - x$
and then multiply:
$y = 2 {x}^{3} - 18 {x}^{2} + 54 x - 54 - x$
$y = 2 {x}^{3} - 18 {x}^{2} + 53 x - 54$
that's the standard form
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## Someone S Mother Has Four Sons North
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Someone S Mom Has 4 Sons Puzzle With Answer Whatsapppuzzle Puzzle Puzzles Brainteasers Whatsapp What Latest Jokes Funny Brain Teasers Funny Questions
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## Someone S Mother Has Four Sons North South East
Reply to this post with the correct name of the fourth son. The correct answer is someone.
Pin By Yoon Soh On Brain Teaser In 2020 Brain Teasers Teaser Names
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## Someone S Mother Has Four Sons North South East West
North east and west. The riddle goes something like this.
Pin By William Stone On Quotes Funeral Blues Funeral Poems Funeral Readings
### What s the name of the fourth son.
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Such riddles and quizzes facilitate the people to keep their brain cells active during the time of lockdown. Given the name of the three other sons the name of the fourth son can be someone. Someones mom has 4 sons.
What is the name of the fourth son. This someone s mother has 4 sons riddle is one of the trending puzzle which has become viral worldwide. If you lose you have to repost this.
This whatsapp riddle has made its way into several groups. Since someone s mother has four sons north west and south the name of the fourth son is someone. What is the name of the fourth son.
Solving these types of puzzle questions will help you in. Reply to this post with the correct name of the fourth son. The question is a common riddle based on the mathematical topic called sequences.
So that means in common sense we think that fourth son is north west south then fourth son will be east but this is not true. Someone s mother has four sons three are named north east and west. Someone s mother is a riddle that is currently been forwarded by many people on whatsapp.
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https://forums.odforce.net/topic/41575-how-could-i-expand-border-edges/?tab=comments
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Jump to content
# How could I expand border edges..?
## Recommended Posts
How could I expand border edges, without adding more points to my geometry?
As you can see in image bellow, I tried “PolyExtrude SOP” but it adds more points to my geometry.
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After extruding try:
Fuse: Snap -> Distance -> Greatest Point Number
Fuse: Consolidate
expand.hiplc
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Group the edge then Peak SOP and transform by normals?
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29 minutes ago, kleer001 said:
Group the edge then Peak SOP and transform by normals?
that would work for the straight bits, but for the corners you would have to increase the magnitude by which you move them to maintain the angles.
from the top of my head.. I think it was dividing the offset by either the cosine of the angle, or the cosine of half the angle.
So an offset at the corner, at 45 degrees would be 1/cos(45deg) which is 1/0.70710678118 = 1.41421356237 if my memory serves me right.
Which makes sense, as that is the square root of 2, and pythagoras and stuff
Edited by acey195
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8 hours ago, konstantin magnus said:
After extruding try:
Fuse: Snap -> Distance -> Greatest Point Number
Fuse: Consolidate
I don't want to use PolyExtrude because it adds points to my geo.
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So, is there any practical solution?
Edited by Masoud
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You could snap the outer points to polyexpand with nearpoint VOP. In this way, no additional points are created, not even temporarily.
expand_2.hiplc
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You could also loop over the outer points with a detail wrangle:
```int pts_grp[] = expandpointgroup(0, 'outer');
int total = len(pts_grp);
for(int i = 0; i < total; i++){
int pt_prev = pts_grp[ (i - 1) % total ];
int pt_curr = pts_grp[ i ];
int pt_next = pts_grp[ (i + 1) % total ];
vector next = normalize( point(0, 'P', pt_next) - point(0, 'P', pt_curr) );
vector prev = normalize( point(0, 'P', pt_curr) - point(0, 'P', pt_prev) );
vector avg = normalize(next + prev);
vector up = {0, 1, 0};
vector in = cross(avg, up);
float dist = dot(next, avg);
vector dir = in / dist;
vector offset = dir * -chf('extrude');
setpointattrib(0, 'P', pt_curr, offset, 'add');
}```
expand_3.hiplc
• 1
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Yeah Konstatin's version is basically a worked out version of my ramblings.
Forgot to note that the dot product of 2 vectors is equal to the cosine of the angle between those vectors, making the code kind of convenient
@konstantin magnus Just realised you got this technique from the genius himself too (in the other thread)
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### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Hypotenuse Lattice Points
##### Stage: 4 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Building Gnomons
##### Stage: 4 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Seven Squares
##### Stage: 3 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Proximity
##### Stage: 4 Challenge Level:
We are given a regular icosahedron having three red vertices. Show that it has a vertex that has at least two red neighbours.
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### There and Back Again
##### Stage: 3 Challenge Level:
Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives?
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Playground Snapshot
##### Stage: 2 and 3 Challenge Level:
The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone?
##### Stage: 3 Challenge Level:
Can you mark 4 points on a flat surface so that there are only two different distances between them?
### Flight of the Flibbins
##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Travelling Salesman
##### Stage: 3 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### On the Edge
##### Stage: 3 Challenge Level:
If you move the tiles around, can you make squares with different coloured edges?
### Coloured Edges
##### Stage: 3 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Zooming in on the Squares
##### Stage: 2 and 3
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Convex Polygons
##### Stage: 3 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Tilting Triangles
##### Stage: 4 Challenge Level:
A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates?
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Pattern Power
##### Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Cutting a Cube
##### Stage: 3 Challenge Level:
A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical?
### Tetrahedra Tester
##### Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
##### Stage: 3 Challenge Level:
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
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FAQ
# If i leave philadelphia at 5am what time will i get to morocco ?
The fastest direct flight from Philadelphia to Morocco is 7 hours 22 minutes.
Also know, how do flight times work with time zones? Airline departure and arrival times are always given in terms of the local time zone – that is, the time zone at the airport in question for each segment of the trip. So if you’re flying from the West Coast of the United States to the East Coast, your 6:00 p.m. arrival time appears in the Eastern time zone.
Quick Answer, how do you calculate time difference between flights?
Best answer for this question, how do you manually calculate flight times?
1. Time T = D/GS. To find the time (T) in flight, divide the distance (D) by the GS.
2. Distance D = GS X T. To find the distance flown in a given time, multiply GS by time.
3. GS GS = D/T. To find the GS, divide the distance flown by the time required.
Also, does flight time account for time change? The total traveling time is the actual amount of time the trip takes from your departure city to your final destination, according to your itinerary. This calculated time takes into account changes in time zones, time spent in transfer cities and daylight saving time (DST).
## How does traveling across time zones work?
Here’s how it works. When you travel east, the Sun rises higher and higher in the sky. It is as though you are seeing the Sun as it would be at a later time in the day. When you travel west, the Sun gets lower and lower in the sky.
## How does time change affect flights?
It will affect scheduled flight times the same way it will affect your favorite TV show’s air time. It will affect apparent flight duration if the departure and arrival airports switch to DST on different days of the year, but actual time in the air will remain the same.
## How do you calculate time difference?
1. Convert both times to 24 hour format, adding 12 to any pm hours. 8:55am becomes 8:55 hours (start time)
2. If the start minutes are greater than the end minutes…
3. Subtract end time minutes from start time minutes…
4. Subtract the hours…
5. Put(not add) the hours and minutes together – 6:45 (6 hours and 45 minutes)
## What is the formula for calculating travel time?
Estimate how fast you will go on your trip. Then, divide your total distance by your speed. This will give you an estimation of your travel time. For example, if your trip is 240 miles and you are going to be drive 40 miles an hour, your time will be 240/40 = 6 hours.
## How do you find the difference in time between two places?
So if it is 12 noon at Greenwich (0 degree), it would be 12:04 pm at 1 degree meridian and so on. In India, the standard meridian is 82-and-half degree. So the time difference between Greenwich and India is 82.5 x 4, which is 330 minutes (5 hours 30 minutes).
## Are flight arrival times accurate?
An AirTran Airways pilot confessed to Reader’s Digest that “airlines really have adjusted their flight arrival times so they can have a better record of on-time arrivals, so they might say a flight takes two hours when it really takes an hour and 45 minutes.”
## Is departure time take off time?
The departure time is the moment that your plane pushes back from the gate, not the time it takes off. … That’s basically the “wheels up” time on your plane.
## Why do airlines use 24 hour time?
Times for flights are almost always given as local time in a 24-hour format. Most airlines use the 24-hour clock system when telling time. They use this system when assigning trip departures, check-in times and other forms of time designation. The 24-hour clock alleviates communication problems and is more convenient.
## What happens if you travel from east to west across it?
If you are traveling westward, you gain a day, and if you are traveling eastward, you lose a day. For example, if a traveler moves eastward across the Pacific Ocean from Wake Island to the Hawaiian Islands on June 25, they will jump backward to June 24 as soon as they cross the IDL.
## Do you lose or gain time flying to Europe?
Going to Europe you don’t lose a day, you simply fly overnight and arrive in the morning. Flying from Asia to the US you gain a day on the calendar, but will loose it on your way back to China.
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# If an infinite set $S$ of positive integers is equidistributed, is $S+S$ also equidistributed?
By $$S+S$$, I mean $$\{x+y,$$ with $$x,y \in S\}$$. By equidistributed, I mean equidistributed in residue classes, as defined here (the definition is very intuitive, and examples of such equidistributed sets are provided on that page).
My goal here is to prove that if $$S$$ is equidistributed and contains enough elements (see here), then $$S+S$$ covers all the positive integers except a finite number of them. The first step, I think, would be to prove that $$S+S$$ is equidistributed. I d guess the proof is not difficult and this fact has probably been established, but I could not find a reference.
The result seems obvious if you consider "equidistribution modulo 1" instead, so I suppose it can also be derived (probably in a similar way) for "equidistribution in residue classes", as the two concepts are closely related and based on a similar Weyl criterion (a continuous version for modulo 1, a discrete version for residue classes.)
Special case
If the infinite set $$S$$ of positive integers is such that if $$x,y\in S$$ then $$x+y\notin S$$, it is easy to prove that $$S$$ equidistributed implies $$S+S$$ is also equidistributed. Note that $$S$$ is called a Sidon set or sum-free set. Let $$N_S(z)$$ be the number of elements of $$S$$ less or equal to $$z$$, and let's assume that
$$N_S(z)\sim \frac{a z^b}{(\log z)^c} \mbox{ as } z\rightarrow\infty$$
where $$a,b,c$$ are non-negative real numbers with $$b\leq 1$$. The sets that I am interested in all have $$b>\frac{1}{2}$$, for instance, pseudo-primes ($$a=b=c=1$$) or pseudo-superprimes ($$a=b=1, c=2$$). Such sets satisfy this conjecture A (see here) : all but a finite number of positive integers can be written as $$z=x+y$$ with $$x, y \in S$$. That conjecture would imply that $$S+S$$ is de facto equidistributed, since essentially, in this case $$S+S$$ it is the set of all positive integers minus a finite number of them. However, that conjecture A is precisely what I would want to prove, so I can not use it as a justification to establish the much weaker result that $$S+S$$ is supposedly equidistributed if $$b>\frac{1}{2}$$ and $$S$$ is equidistributed. If $$b\leq \frac{1}{2}$$, conjecture A is not true.
Hints to prove the result and win the bounty
In case it is not true, a counter-example will do. Assuming it is correct, a sketch of a proof for a simple case is enough. Here is how it could start.
Let $$T = S+S$$ and
$$S(n,q) = \{x\in S, x=q \bmod{n}\}.$$
We have
$$T(n,q) = \bigcup_{p=0}^{n-1}\Big[S(n,p) + S(n,(q-p) \bmod{n} )\Big]$$ $$T=\bigcup_{q=0}^{n-1}T(n,q)$$
$$T(n,q)$$ is the subset of elements of $$T$$ that are equal to $$q$$ modulo $$n$$. The subsets $$T(n,q)$$ for any given $$n>1$$ form a partition of $$T$$. However the first union for $$T(n,q)$$ consists of potentially overlapping sets, making the problem non-trivial. Proving the result consists in proving that for any integer $$n>1$$ and $$0\leq q,q' we have
$$\frac{N_{T(n,q)}(z)} {N_{T(n,q')}(z)}\rightarrow 1 \mbox{ as } z\rightarrow \infty.$$
Again $$N_T(z)$$ counts the number of elements of $$T$$ less than or equal to $$z$$. We can focus on sets $$S$$ such that
$$N_S(z) \sim \frac{a z^b}{(\log z)^c}$$
with $$0\leq b \leq \frac{1}{2}$$. I will offer the bounty even if the proof is only for the special case $$b=\frac{1}{2}, c=0$$ and $$n=2$$. For computer experiments (generating such a set $$S$$ that is supposed to be equidistributed), proceed as follows: the positive integer $$k$$ belongs to $$S$$ if and only if $$U_k < a/(2\sqrt{z})$$ where the $$U_k$$'s are independent uniform deviates on $$[0, 1]$$.
I did some experiments and here are the results, using $$n=12, b=\frac{1}{2}$$ and looking at all elements of $$S$$ and $$T=S+S$$ up to $$10^6$$:
Equidistribution in $$S$$ (modulo $$n=12$$ in this case) means that the "Ratio_1" tend to be identical and equal to $$\frac{1}{n}$$as you look at more and more elements of $$S$$, while equidistribution in $$T$$ means that the "Ratio_2" tend to be identical also converging to $$\frac{1}{n}$$.
I also did the same test on the set $$S$$ of perfect squares, which is notoriously not equidistributed. The results are below.
Final notes:
• For perfect squares, $$a=1, b=\frac{1}{2}, c=0$$. Even though $$T$$, the set of sums of two perfect squares is not equidistributed, the set $$T+T$$ is the set of all non-negative integers, and is thus equidistributed (that set of course has $$a=0, b=1, c=0$$).
• The last table suggests that the equations $$x^2 + y^2 = 12z+ 3$$, $$x^2 + y^2 = 12z+ 7$$, $$x^2 + y^2 = 12z+ 11$$ do not have integer solutions, while $$x^2 + y^2 = 12z$$, $$x^2 + y^2 = 12z+6$$ and $$x^2 + y^2 = 12z+9$$ might have only finitely many solutions.
• This can give the false impression that we are making little progress towards proving Goldbach and other similar conjectures. I believe this is not the case, because the set $S$ of primes is also notoriously non-equidistributed. Jul 19, 2020 at 17:06
$$S+S$$ need not be equidistributed.
Let $$(a_n)_n$$ be a sequence of even positive integers such that $$a_{n+1} > 3+a_n+a_{n-1}+\dots+a_2+a_1$$ for each $$n \ge 1$$ and such that $$\{\frac{a_n}{2} : n \ge 1\}$$ is equidistributed. It should be clear that such a sequence exists; let me know if you want details. Let $$S = \cup_{n=1}^\infty \{a_n,a_n+1\}.$$ Then $$S$$ is equidistributed: it is clearly equidistributed mod $$2$$, and that $$\{\frac{a_n}{2} : n \ge 1\}$$ is equidistributed implies $$S$$ is equidistributed mod $$n$$, for any $$n \ge 3$$.
Now, $$S+S = \cup_{1 \le n \le m < \infty} \{a_n+a_m,a_n+a_m+1,a_n+a_m+2\}$$. Note the union is a disjoint one, since $$a_{n+1} > 3+a_n+\dots+a_1$$ for each $$n \ge 1$$. Therefore, $$S+S$$ is not equidistributed mod $$2$$, since $$2$$ out of $$3$$ of $$a_n+a_m, a_n+a_m+1,a_n+a_m+2$$ (namely, the first and the third) are even.
• Thank you. I haven't yet looked in detail at your counter-example. Wondering what extra conditions on $S$ might be needed to warrant that $S+S$ is equidistributed. Also wondering if your set $S$ satisfies the same asymptotic class of distributions as discussed in my question. I will look into this. Jul 19, 2020 at 18:21
• @VincentGranville I'm not sure what extra conditions would help. My example stems from the fact (and basically is just the fact) that $\{1,2\}+\{1,2\} = \{2,3,4\}$ is not even equidistributed mod $2$. Jul 19, 2020 at 18:27
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# Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 66
$\frac{3}{8}$
#### Work Step by Step
We calculate the average as: average=$\frac{1}{\frac{2\pi}{3}}\int^{2\pi}_0 \int^{\pi/2}_0 \int^1_0 p^3 cos\phi sin\phi dp$ $d\phi$ $d\theta$ =$\frac{3}{8\pi} \int^{2\pi}_0 \int^{\pi/2}_0 cos\phi \sin \phi d\phi$ $d\theta$ =$\frac{3}{8\pi}\int^{2\pi}_0 [\frac{sin^2\phi}{2}]^{\pi/2}_0d\theta$ =$\frac{3}{16\pi}\int^{2\pi}_0$ =$(\frac{3}{16\pi})(2\pi)$ =$\frac{3}{8}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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You are here » Home » Blog » Geeks Do It Better: Blackjack and Math
Geeks Do It Better: Blackjack and Math
Staff Writer 06/26/2017 Blog 237 views
Despite of what your high school experience may be telling you right now, math is your friend, especially at the Blackjack table where even the basic knowledge of counting can take you a long way. Not being popular and sharing the cafeteria table with the “cool” kids clique may seemed like the end of the world in high-school, but as it turns out, here in the real world geeks actually do it better – if of course by “it” you mean using theoretical knowledge acquired throughout years of formal education and directing it towards a more practical, profit oriented purpose.
And where best to do that than at the Blackjack table, where the skill of counting cards can reduce the house edge and turn it into dust? Employing skill and strategy more than anything else, Blackjack is one card game that is particularly appealing to the so-called geeks, mostly because it requires knowledge of probability and fundamental mathematics.
So, unlock your Memory Palaces as we go over the most important mathematical concepts every true geek has to master before robbing the house blind at the Blackjack table.
Variance and Standard Deviation
Many feel that this is no more than a politically correct term for luck, but true geeks know better and to them, luck is a laymen term for Variance. Defined as a difference between the expected result and the actual outcome, Variance can be easily illustrated with the following example: theoretically, if you play for 100 hours, winning \$25 per hour (known as an Expected Value of the game), you would expect to end up with \$2,500. Any deviation from that particular amount would be considered as a Variance, whether it goes over or below the expected result. This brings as to another important concept in Blackjack – Standard Deviation or SD.
Figuring out the SD of the game of Blackjack as well as its frequency and extent, will allow the players to predict their odds and make an accurate estimate about the winning chance within a particular hand.
N0 or N- Zero
This is something only hard core geeks are familiar with and the concept is usually lost on new players. N0 is the number of the hands necessary for the player to go through in order to get ahead of the SD. To get this number, the player has to stick with the same strategy and game rules as to not disrupt the evaluation accuracy.
Represented by a formula, N0 would look something like this:
N0 = Varience/Expected Value ^ 2
Certainty Equivalence (CE)
Certainty Equivalence is an extremely important factor if you are playing for profit and have a limited budget. CE helps the player determine the element of risk and expected win rate. If the total hand payout shows that the risk exceeds the positive outcome of the game, best course would be to retreat, instead of betting over your budget. Of course, this factor defeats the purpose of gambling, since most of its appeal is not knowing, however, if you are reading this article and devising your Blackjack tactics, chances are you are not much of a risk taker to begin with.
Clearly, there is a lot more to Blackjack than simply waiting for your luck to change. So next time you feel like bullying some unsuspecting nerd and handing out a few wedgies, just remember than befriending them can turn out to be much more profitable in the long run.
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# Thread: not divide
1. ## not divide
Hello, need your help thanks
Prove that if $\displaystyle n\in\mathbb{N}^{*}$ then $\displaystyle 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor$
2. Originally Posted by maria18
Prove that if $\displaystyle n\in\mathbb{N}^{*}$ then $\displaystyle 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor$
If $\displaystyle x^3 = 28$ then $\displaystyle (x-3)(x^2+3x+9) = x^3-27 = 28-27=1$. Therefore $\displaystyle (x-3)^{-1} = 9+3x+x^2$, and
$\displaystyle (x-3)^{-n} = (9+3x+x^2)^n = 9^n + a_1x + a_2x^2 + \ldots + a_{2n}x^{2n}$, a polynomial in x of degree 2n with integer coefficients.
Let $\displaystyle \omega = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt3}2i$, a complex cube root of unity. Note that $\displaystyle 1+\omega+\omega^2 = 0$. Let $\displaystyle X = \sqrt[3]{28}$. Then the complex cube roots of 28 are $\displaystyle X,\ \omega X$ and $\displaystyle \omega^2X$. Therefore
$\displaystyle (X-3)^{-n} + (\omega X-3)^{-n} + (\omega^2X-3)^{-n}$
. . . . .\displaystyle \begin{aligned}= 3*9^n + {}& (1+\omega+\omega^2)a_1X + (1+\omega^2+\omega)a_2X^2 + 3a_3X^3 \\\quad&+ (1+\omega+\omega^2)a_4X^4 + (1+\omega^2+\omega)a_5X^5 + 3a_6X^6 + \ldots\\ = 3*9^n + {}&3*28(a_3 + a_6X^3 + \ldots).\end{aligned}
This is a (real) integer of the form $\displaystyle 3*9^n + 84k$ for some integer k, and is therefore an odd multiple of 3.
Next, notice that $\displaystyle \omega X-3 = -(3+\tfrac12X) + \tfrac{\sqrt3}2Xi$, and $\displaystyle X > 3$. Hence $\displaystyle |\omega X-3|>3\sqrt3$ and so $\displaystyle |(\omega X-3)^{-n}|<3^{-3n/2}$, and similarly $\displaystyle |(\omega^2 X-3)^{-n}|<3^{-3n/2}$. So both these quantities are very small (certainly less than 1/2). It follows from the previous paragraph that $\displaystyle (X-3)^{-n}$ is within distance less than 1 from an odd multiple of 3. So its integer part cannot be a multiple of 6.
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# Momentum Question
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#1
Water of density 1000kgm^-3 flows out of a garden hose of cross sectional area 7.2 x 10^-4 m^2 at a rate of 2 x 10^-4 m^3 per second. How much momentum is carried by the water leaving the hose per second?
Can someone explain how you would work this out plz thanks
0
10 years ago
#2
you can work out the volume, and therefore the mass.. and you have the velocity!
0
#3
(Original post by unamed)
you can work out the volume, and therefore the mass.. and you have the velocity!
sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|
0
10 years ago
#4
0
10 years ago
#5
(Original post by An0nymous)
sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|
it's all right.
It's x-sectional area * speed.
0
#6
(Original post by unamed)
it's all right.
It's x-sectional area * speed.
so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?
0
10 years ago
#7
(Original post by An0nymous)
so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?
2.88 x 10^ -8
although, now I think about it, I might be wrong. >.<
0
#8
(Original post by unamed)
2.88 x 10^ -8
although, now I think about it, I might be wrong. >.<
0
10 years ago
#9
Careful, you don't have the actual speed, but just how much volume comes out each second. By using dimensional analysis (i.e. look at the units) its quite simple to calculate the speed. So first, calculate the mass of water present in that amount of volume, and then find the speed of the water flowing out.
0
10 years ago
#10
The dimensional analysis the way I know it is a bit long..
Spoiler:
Show
We need to know Momentum Per Second (Momentum/Time => p/t => mv/t => mst-2 (kgms-2, thats Newtons by the way)).
So suppose this momentum per second is proportional to the density of water to some power α:
p/t ∝ ρα
and to the volume per second to some power β:
p/t ∝ (V/t)β
and to the area to some power λ:
p/t ∝ Aλ
So we get:
p/t = (ρ )α×(V/t)β×(A)λ
And in terms of the units:
kgms-2 = (kgm-3)α×(m3s-1)β×(m2)λ
OK then.. we need s-2 on the right side, as it is on the left.
So β = 2 as there is no other second on the right side:
kgms-2 = (kgm-3)α×(m3s-1)2×(m2)λ
kgms-2 = (kgm-3)α×(m6s-2)×(m2)λ
But then we get m6 and we need m1 as on the left. We also need kg1. So α = 1.
kgms-2 = (kgm-3)1×(m6s-2)×(m2)λ
kgms-2 = (kgm-3)×(m6s-2)×(m2)λ
kgms-2 = (kgm3s-2)×(m2)λ
Lets see what we got. The units of mass are OK, the units of time too.
We only need to get m1 from m3. And we have m there.
1 = 3+2λ
-2 = 2λ
λ = -1
kgms-2 = (kgm3s-2)×(m2)-1
kgms-2 = kgm3s-2×m-2
kgms-2 = kgms-2
So we got:
p/t = ρ1×(V/t)2×A-1
or
p/t = ρ×(V/t)2/A
That gives 5.6×10-2.
A shorter way, but the same idea.
We need momentum per second, that's kgms-2.
We have seconds in Volume/Time given only, so I'll square that to get s-2.
We get m6s-2.
OK.. Multiply with the density, we're left with kgm3s-2.
Divide by the area and we get kgms-2.
I dont really know what (V/T)2 means, but that gives the correct answer.
0
#11
The dimensional analysis the way I know it is a bit long..
Spoiler:
Show
We need to know Momentum Per Second (Momentum/Time => p/t => mv/t => mst-2 (kgms-2, thats Newtons by the way)).
So suppose this momentum per second is proportional to the density of water to some power α:
p/t ∝ ρα
and to the volume per second to some power β:
p/t ∝ V/tβ
and to the area to some power λ:
p/t ∝ Aλ
So we get:
p/t = (ρ )α×(V/t)β×(A)λ
And in terms of the units:
kgms-2 = (kgm-3)α×(m3s-1)β×(m2)λ
OK then.. we need s-2 on the right side, as it is on the left.
So β = 2 as there is no other second on the right side:
kgms-2 = (kgm-3)α×(m3s-1)2×(m2)λ
kgms-2 = (kgm-3)α×(m6s-2)×(m2)λ
But then we get m6 and we need m1 as on the left. We also need kg1. So α = 1.
kgms-2 = (kgm-3)1×(m6s-2)×(m2)λ
kgms-2 = (kgm-3)×(m6s-2)×(m2)λ
kgms-2 = (kgm3s-2)×(m2)λ
Lets see what we got. The units of mass are OK, the units of time too.
We only need to get m1 from m3. And we have m there.
1 = 3+2λ
-2 = 2λ
λ = -1
kgms-2 = (kgm3s-2)×(m2)-1
kgms-2 = kgm3s-2×m-2
kgms-2 = kgms-2
So we got:
p/t = ρ1×(V/t)2×A-1
or
p/t = ρ×(V/t)2/A
That gives 5.6×10-2.
A shorter way, but the same idea.
We need momentum per second, that's kgms-2.
We have seconds in Volume/Time given only, so I'll square that to get s-2.
We get m6.
OK.. Multiply with the desity, we're left with kgm3s-2.
Divide by the area and we get kgms-2.
I dont really know what (V/T)2 means, but that gives the correct answer.
lol thanks this was a multiple choice question worth 1 mark!
1
4 years ago
#12
p = mv
m = density x volume.
= 1000 x 2.0 x 10^-4
v = d/t. d being the 'length component' of the water coming out the house
volume of cylinder = area x length(d)
(d) = volume / area
= 2.0 x 10^-4 / 7.2 x 10^-4
= 2 /7.2
p = mv, p = 1000 x 2.0 x 10^-4 x 2 /7.2
= 5.5(recurring) x 10^-2
= 5.6 x 10^-2 (approximately)
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# What are the values of x and y?
• 863 views
a) 3x + 5y = 26
b) 2x + 2y = 12
Step One: We need to find a way to equate either the x terms of the y terms in each equation. Multiply equation a) by 2 and equation b) by 3 to form the following equations.
a) 6x + 10y = 52
b) 6x + 6y = 36
Step Two: Take equation b) from equation a) to eliminate the x component.
a) 6x + 10y = 52
- b) 6x + 6y = 36
0x + 4y = 16
y = 4
Step Three: substitute the value of y into either equation to find the value of x.
b) 2x + 2y = 12
2x + (2x4) =12
2x + 8 = 12
2x = 4
x = 2
x=2 y=4
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This op-amp configuration is giving me trouble
I came across this configuration earlier today and can't for the life of me figure out the equation for Vout:
It looks like a unity-gain differential amplifier but the ground resistor on V2 is before the series resistor.
• The resistor on V2 can thus be ignored, so the voltage on the non-inverting input is V2. Does that help? – Spehro Pefhany Jul 3 '15 at 4:08
• Could you post a link to the page where you saw this schematic? Maybe we can glean more from the context. – Nick Alexeev Jul 3 '15 at 5:08
• @SpehroPefhany Could this be a badly drawn difference amplifier? – Nick Alexeev Jul 3 '15 at 5:08
• To me it looks like a standard inverting amplifier, where an extra resistor (vertically positioned) is added to define the input impedance of the circuit and the other extra resistor (horizontally positioned) is for symmetry for offset currents. – jippie Jul 3 '15 at 5:49
• @NickAlexeev looks like h/w to me, but I guess I could come up with a practical application in signal conditioning. – Spehro Pefhany Jul 3 '15 at 12:15
Re drawing the schematic to I can add component references etc.
simulate this circuit – Schematic created using CircuitLab
I don't recognise this as a standard circuit but its got negative feedback so will try to keep both the inverting and non-inverting inputs the same
The non-inverting input is simply $V_2$
And the inverting input:
$\dfrac{\dfrac{V_1}{R_3}+\dfrac{V_{out}}{R_4}}{\dfrac{1}{R_3}+\dfrac{1}{R_4}}$
Noting all resistors are the same value and equating the 2 together we get
$\dfrac{\dfrac{V_1}{R}+\dfrac{V_{out}}{R}}{\dfrac{1}{R}+\dfrac{1}{R}} = V_2$
Solving for $V_{out}$
$\dfrac{V_1}{R} + \dfrac{V_{out}}{R} = V_2 \cdot \dfrac{2}{R}$
$V_{out} = 2 \cdot V_2 - V_1$
But as others have pointed out the circuit may be drawn incorrectly with $R_1$ in the wrong position and a differential amplifier could be what is intended.
Apply the superposition principle.
When V1 is zero you have Vout that depends on V2 (simple non-inverting amplifier). When V2 is zero you have an inverting amplifier whose output depends on V1.
Finally, for the said principle, you sum these two results and obtain the Vout in function of both V1 and V2.
As Warren says, its gain is 2*V2 - V1.
The resistor from V2 to ground serves a couple of purposes:
1. If V2 is disconnected, it provides a DC path to ground, preventing spurious outputs
2. It presents the same impedance as input V1 to differential signals, which may be important if it's connected to the output of a passive filter, or a sensor wired as a bridge.
The series resistor from V2 to In+ matches the V1 input resistor on V1, which may improve both DC performance (if the opamp has input bias currents) and high frequency performance (by ensuring both inputs, which have capacitance) are fed from the same source impedance).
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# the boy who
Essay by thisisfaHigh School, 12th gradeC, October 2014
MEI STRUCTURED MATHEMATICS
Marking C3 Coursework
10 tips to ensure that the right mark is
awarded
MEI Conference 2013 Marking C3 Coursework Page 2
Marking C3 Coursework
C3 coursework is very prescriptive. Providing assessors follow the criteria carefully there is no problem with the assessment. However, there are difficulties for the External moderator.
⢠Errors are made in the marking ⢠Work is not checked but assumed to be correct. ⢠Credit is given for work that is not evident.
It is not a question of "what is a good piece of coursework?" but "how can I ensure that I give an appropriate mark?" 1 Terminology This task is all about solving equations. Therefore, candidates should write equations. Persistent errors should be penalised in domain 5. Examples which should be penalised: I am going to solve the equation x3 â 4x â 1. I am going to solve the equation y = x3 â 4x â 1. I am going to solve the equation f(x) = x3 â 4x â 1. Correct terminology: I am going to solve the equation x3 â 4x â 1 = 0. Or I am going to solve the equation f(x) = 0 where f(x) = x3 â 4x â 1. 2 Illustrations All three methods require a graph and an illustration for both success and failure. A graph of the function is not an illustration of the method. Example I am going to solve the equation x3 â 4x â 1 = 0 Here is a graph of y = x3 â 4x â 1.
MEI Conference 2013 Marking C3 Coursework Page 3
You can see from the graph that there is a root of the equation in the range [1,2] which I shall find. This is not enough - only the graph has been drawn.
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Just For Fun
## 5 more maths magic tricks
Following on from the previous blogs about the number 6174 and the (not-so) magic number 10, here are 5 more cool maths tricks that you can use to impress your friends.Trick 1For this trick, pick a three digit number with repeating digits (eg. 222 or 999)Add up the digits of… Continue reading 5 more maths magic tricks
Just For Fun
## The Magic Number 10 – A mathematical trick
Here's another maths magic trick that you can use to impress your friends. This trick works for ANY number that you choose. Follow these instructions and you will always end up with 10. Pick a number.Multiply it by 3.Then add 30.Then multiply by 2.Then divide by 6.Subtract your original number.… Continue reading The Magic Number 10 – A mathematical trick
Younger Years
## Times Table Practice
We've all been there at some stage - 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12… and so on. It's not fun, but it is effective. Unfortunately, knowing your times tables is incredibly important for maths students of any age and ability. That… Continue reading Times Table Practice
Just For Fun
## 6174 – Kaprekar’s Constant
Here's a random 4-digit number that I bet you didn't know was special. 6174. Follow these instructions and you will (if you have done things correctly), ALWAYS end up with 6174. 1. Pick any four numbers between 0 - 9. The only rule here is you must pick at least… Continue reading 6174 – Kaprekar’s Constant
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## The Toymaker's Mysterious Math Carnival - To Amuse & Delight
Funded! This project successfully raised its funding goal on July 15, 2012.
A printable eBook set of ten mathematical paper toys that amuse and delight. Impossible? Improbable you say? Wait and see!
# The Toymaker's Mysterious Math Carnival
When I was in first grade, my first math workbook had color pictures of cowboys, candy, puppies and toys. I decided that this arithmetic thing was very cool and I couldn't wait for more. The next math book didn't have pictures at all and I remember feeling pretty much cheated. In second grade I got in trouble for illuminating the margins of my math homework. My love affair with numbers waned after that.
My first Kickstarter project is to design a way to make learning and reviewing math actually interesting, beautiful and colorful. If I can help even one kid get better at multiplication or discover something new, then I'll be a happy Toymaker.
The Toymaker's Mysterious Math Carnival will be a printable PDF eBook of ten printable paper toys designed to make learning and reviewing simple math a delight. The addition, subtraction, multiplication tables, division and fractions games will reinforce basic skills in a fun and entertaining way.
Here is a rough list of toy ideas. I can't wait to get started designing these!
Goal = Ten amusing paper toys that reinforce basic math facts, addition, subtraction multiplication, division and fractions.
1. A full set of multi-pies 1 through 12. (Right now what I have on my website only goes up to 6) These flash cards fold up into triangles. Cover the answer with your thumb and work your way around. = Multiplication and Division
2. A super simple number sorting game match the word three with the picture of three apples and the numeral 3. = Basic numeracy
3. A “shooting gallery” game where you blow over funny monster targets with a straw. Then add up the points. = Addition and possibly fractions
4. A memory concentration game where you have match up a card 9 -3 with a card that says six. - = Subtraction
5. A “Feed the Carnival Animals” game where you have to put together plates of animal chow. ½, ¾ 5/6, etc. Fractions
6. A numbered fish game using a paper clip hook string and stick fishing pole. Addition and possibly subtraction Your “catch” might be a surprise. Be careful! Some of the fish will be stinky and cause the player to lose points.
7. A bowling arcade game with marbles. Addition and Subtraction.
8. A “Fortune Teller” guess the next number, skip counter game. Can you guess the next number that will appear in the crystal ball? 3,6,9,12,? Skip Counting
9. A Division game = A side show game.
10. A Multiplication Ferris wheel or possibly a bareback rider Fraction game.
# Rewards
A special note from the Toymaker...
I will be offering signed 8.5 x 11 laser jet prints suitable for framing as rewards. You pick which one that you would like.
Once the project funds I will be able to dedicate the time to finish the toys and convert them to printable PDF files. Part of the money will go to pay taxes, pay for editing and fund the gifts.
Thank you for your support of my big idea. My goal is to help grownups and kids spend time together making things. It is my wish to amuse and delight,
# My Dream Team
Dr. Marcia S. Popp is helping me as curriculum advisor. She is the author of three college level textbooks on elementary teaching methods and a master teacher.
Ronn Waters is also advising me on content. Besides being an amazing elementary school teacher with fifteen years of experience and an educational technology specialist, he is a very fine husband.
Thank you to Kevin MacLeod for the lovely piano music in the video. You can hear more of his work at http://incompetech.com/m/c/royalty-free/
--
"If one advances confidently
in the direction of his dreams
and endeavors to live the life
which he has imagined
he will meet with a success
unexpected in common hours."
Henry David Thoreau
### FAQ
Have a question? If the info above doesn't help, you can ask the project creator directly.
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Funding period
Jun 20, 2012 - Jul 15, 2012 (25 days)
• ##### Pledge \$5 or more
7 backers
My sincere thanks and I will add you to my mailing list so that you can keep up on my progress. Plus free paper toys in your inbox each month.
Estimated delivery: Aug 2012
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# Unlocking Solutions: A Step-by-Step Guide to How to Solve Non-linear Equations by Substitution
Solving systems of non-linear equations by substitution involves isolating one variable in one of the equations and then substituting the resulting expression into the other equation. This method can be particularly useful when one of the equations in the system can be easily solved for one of the variables. Here’s a step-by-step guide to tackle these systems:
## A Step-by-Step Guide to How to Solve Non-linear Equations by Substitution
### Step 1: Understand the Equations
• Begin by carefully examining your system of non-linear equations. Identify which equation and which variable might be easiest to isolate.
### Step 2: Isolate a Variable
• Choose one of the equations and solve it for one of the variables. This means you will express one variable in terms of the other variable(s).
### Step 3: Substitute the Expression
• Take the expression you obtained for the isolated variable and substitute it into the other equation in the system. This means you replace every instance of that variable in the second equation with the expression you found.
### Step 4: Solve the Resulting Equation
• After substitution, you will have an equation with one variable. Solve this equation to find the value of that variable. This may involve solving a quadratic equation, or in some cases, higher-degree equations.
### Step 5: Back-Substitute to Find Other Variable(s)
• Once you have found the value of one variable, substitute it back into any of the original equations to find the value of the other variable.
### Step 6: Check the Solution
• Substitute the values of the variables back into the original equations to ensure they satisfy both equations. This is an important step to verify the correctness of your solution.
### Step 7: Consider All Solutions
• Be aware that non-linear systems can have multiple solutions, a single solution, or no solution at all. Make sure to consider all possible solutions, especially when dealing with equations that could have more than one solution (e.g., quadratic equations).
## Tips for Success
• Be meticulous with algebraic manipulations to avoid errors.
• When solving quadratic or higher-degree equations, remember to consider all possible roots.
• Checking your solution is crucial to ensure accuracy.
This method is effective for systems where one of the equations can be conveniently solved for one variable, making it easier to substitute into the other equation(s).
### What people say about "Unlocking Solutions: A Step-by-Step Guide to How to Solve Non-linear Equations by Substitution - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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# The probability that iid draws from a mean zero random variable sum to zero
Suppose we have a probability distribution $$p(\cdot)$$ supported on the integers between $$-m$$ and $$m$$ for some positive integer $$m$$, with $$\sum_k kp(k) = 0$$. Suppose furthermore that all $$p(k)$$ are rational. Let $$q_n$$ (for $$n \geq 0$$) be the probability that $$X_1 + \dots + X_n = 0$$ where $$X_1,\dots,X_n$$ are iid draws from $$p(\cdot)$$ (so $$q_0 = 1$$ trivially).
Example: $$m=1$$, $$p(1)=p(-1)=1/2$$, $$p(0)=0$$. Then $$q_n$$ equals 0 for $$n$$ odd and equals $${n \choose n/2} 2^{-n}$$ for $$n$$ even.
Question: Must the generating function $$\sum_{n=0}^{\infty} q_n x^n$$ be algebraic? $$D$$-finite?
• This is $\int \frac{1}{ 1- x \sum_k p(k) y^k} \frac{dy}{y}$ with the integral taken over the unit circle and expressing that integral as a sum of residues at poles in the unit disc should give algebraicity. Feb 24 at 23:18
I guess I can convert my comment to an answer.
Recall that for a bivariate power series $$F(x,y) = \sum_{i,j \geq 0} f(i,j) x^i y^j$$, its diagonal is the univariate power series $$\operatorname{diag} F(x,y) = \sum_{n \geq 0} f(n,n) x^{n}$$. In other words, we extract the coefficients of $$x^ny^n$$ and make them into a new power series.
Since your $$q_n$$ is the coefficient of $$x^0$$ in $$(\sum_{k=-m}^{m} p(k) x^k)^n$$, it follows that your generating function $$\sum_{n \geq 0} q_n x^n = \operatorname{diag} F(x,y)$$ where $$F(x,y) = 1/(1-(xy\sum_{k=-m}^{m}p(k) x^k)$$).
Now, it is well known that if $$F(x,y)$$ is rational, then $$\operatorname{diag} F(x,y)$$ is algebraic: see for example Theorem 6.3.3 of Stanley's EC 2. Since the above $$F(x,y)$$ is clearly rational, it follows that your $$\sum_{n \geq 0} q_n x^n$$ is algebraic. And notice that we never used the fact that the mean of your random variable is zero, or that the $$p(k)$$ are rational.
(In fact, there is a very strong connection between diagonals of bivariate rational generating functions and contour integrals - see the relevant section of Stanley - so this approach ends up being more-or-less the same as Will Sawin's.)
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Question
# In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70ms–1 and 63ms–1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.
Solution
## Speed of wind on the upper surface of the wing, V1=70m/s Speed of wind on the lower surface of the wing, V2=63m/s Area of the wing, A=2.5m2 Density of air, ρ=1.3kgm–3 According to Bernoulli’s theorem, we have the relation: P1+12ρV21=P2+12ρV22P2−P1=12ρ(V21−V22) Where, P1 = Pressure on the upper surface of the wing P2= Pressure on the lower surface of the wing The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane. Lift on the wing =(P2−P1)A =12ρ(V21−V22)A=121.3((70)3−(63)2)×2.5 =1512.87 =1.51×103N Therefore, the lift on the wing of the aeroplane is 1.51×103N.
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hw3 - formulas for R i R j R i F j F i R j and F i F j For...
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Math 113 Homework # 3, due 9/23/9 at 2:10 PM 1. Fraleigh section 4, exercise 41. 2. Fraleigh section 5, exercise 13. 3. (a) Fraleigh section 5, exercise 54. (b) Is this still true if one replaces intersection by union? Prove or give a counterexample. 4. Let n > 1 be an integer and let θ = 2 π/n . Let P be the regular n -gon with vertices (cos iθ, sin ) for i Z n . The dihedral group D n is the symmetry group of P , which consists of rotations R i and reflections F i for i Z n . Here R i is the counterclockwise rotation around the origin by angle , and F i is the reflection across the line through the origin and (cos iθ/ 2 , sin iθ/ 2). Your problem: find (and give at least some justification for) general
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Unformatted text preview: formulas for R i R j , R i F j , F i R j , and F i F j . For example, R i R j = R i + j , where the addition of indices is mod n . 5. Find all subgroups of D 4 . 6. If G is a group, the center of G is defined to be Z ( G ) = { x ∈ G | xy = yx for all y ∈ G } . (a) Show that Z ( G ) is a subgroup of G . (b) For n > 2, what is the center of D n ? (Use the multiplication rules you found above. The answer depends on whether n is even or odd.) 7. Fraleigh section 6, exercise 32 (justify as always). 8. How challenging did you find this assignment? How long did it take?...
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The Harmonic Series.
Murat Aygen from Turkey sent this solution to part (a) and Dapeng Wang from Claremont Fan Court School, Guildford sent in essentially the same solution.
Let us partition the terms of our series, starting from every $k= 2^{i-1}+ 1$, as follows: $\sum_{k=1}^{n} \left({1\over{k}}\right) = 1 + \left({1\over{2}}\right) + \left({1\over{3}} + {1\over{4}}\right) + \left({1\over{5}} + {1\over{6}} + {1\over{7}} + {1\over{8}}\right) + \sum_{i=4}^{n} \left(\frac{1}{2^{i-1}+1} + \dots + {1\over2^{i}}\right)$
How many terms are there in each partition? We see that there are 1, 2, 4, 8, 16... terms in the successive partitions given by $2^i - 2^{i -1}$ where $2^i - 2^{i -1} = 2^{i -1}(2-1) = 2^{i -1}$.
The smallest term in a partition is clearly the rightmost term ${1\over 2^i}$. This smallest term multiplied by the number of terms in the partition is equal to $${1\over 2^i}\times 2^{i -1} = {1\over 2}$$ which is always less than the sum of the terms of the partition. Anyway it is a positive constant! Since the number of terms in the series is as many as one wishes, we can form as many partitions as we wish whose partial sums are not less than 1/2. For reaching a sum of 100 only 200 partitions are needed.
Noah and Ariya from The British School of Boston, USA and Aled from King Edward VI Camp Hill School for Boys sent excellent solutions to part (b) .
Each of the areas of the pink rectangles is representative of one fraction in the sum
$$S_n = 1 +{1\over 2} + {1\over 3} + {1\over 4} + ... + {1\over n}.$$This is verified by knowing that each rectangle x has a base of length 1 and a height of length 1/x. Every rectangle has an area greater than that under the curve $y=1/x$ it overlaps, as illustrated above (note that this will remain so because the function $y= 1/x$ is monotonic for positive numbers).
Consider the area under the graph $y = 1/x$ between $x=a$ and $x=b$. This area lies between two rectangles and so we get $${b-a \over b} < \int_a^b{ 1\over x }dx = \ln b - \ln a < { b-a\over a}$$ If we evaluate the expression between $a = {1\over n}$ and $b= {1 \over {n-1}}$ we get:
$${1\over n} < \ln {1 \over {n-1}} - \ln {1 \over n} = \ln n - \ln (n-1) < { 1\over n-1}$$
and this gives:
$${1 \over 2} < \ln 2- \ln 1< { 1\over 1}$$
$${1 \over 3} < \ln 3- \ln 2< { 1\over 2}$$
$${1 \over 4} < \ln 4- \ln 3< { 1\over 3}$$
...and so on
$${1 \over n} < \ln n- \ln (n-1)< { 1\over n-1}$$
Summing these expressions (noting that $\ln1 = 0$) we get:
$${1\over 2} + {1\over 3} + {1\over 4} + ... + {1\over n} < \ln n < 1 + {1\over 2} + {1\over 3} + {1\over 4} + ... + {1\over n-1} .$$
The series on each side of this inequality grow infinitely large and differ by less than 1 so the series grows like $\ln n$.
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# LSAT and Law School Admissions Forum
Get expert LSAT preparation and law school admissions advice from PowerScore Test Preparation.
## #21 - Most of the students who took Spanish 101 at the
• PowerScore Staff
• Posts: 8224
• Joined: Feb 02, 2011
#33829
Complete Question Explanation
Must Be True—FL. The correct answer choice is (E)
This stimulus contains facts involving the use of Formal Logic. The first statement is that most of the university’s Spanish 101 students last semester attended every class session. We can diagram this statement as:
• StudentSpanish 101 attended every class session
Next, we are told that each student with a grade lower than a B- missed at least one session. We can diagram this just like we do with any conditional relationship, recognizing that missing at least one session is logically identical to “did not attend every class session”:
• Studentlower than B- attended every class session
and the contrapositive:
• attend every class session Studentlower than B-
We can connect the first, “most” relationship to the contrapositive of the second relationship across the common term “attend every class session”:
• StudentSpanish 101 attended every class session Studentlower than B-
Based on this connection, we can infer that most students who took Spanish 101 at the university last semester received a grade of B- or higher (i.e., not lower than B-).
The question stem identifies this as a Must Be True question. Our prephrase is the inference made above, that most students who took Spanish 101 at the university last semester received a B- grade or higher.
Answer choice (A): Since the stimulus did not tell us about students who received a grade of A- or higher, we cannot make this inference.
Answer choice (B): This answer choice contains a reversal of what we know from the stimulus. We know that every student who received a grade lower than a B minus missed at least one class session, but we cannot simply reverse that statement to say that most people who missed at least one class session received a grade lower than a B minus. At most, we can say that some people who missed at least one class session received a grade lower than a B minus.
Answer choice (C): This is a tricky answer choice, because it is very close to our actual prephrase. However, this answer choice is incorrect because it leaves out the possibility of those students receiving a B minus, stating that most of the students received a grade “higher than B minus.”
Answer choice (D): The only rule we have that associates grades with attendance is the second rule, which told us about students who received a grade lower than a B minus. We cannot infer anything about the attendance of those students who received a grade of B minus or higher.
Answer choice (E): This is the correct answer choice, and it restates our prephrase, that most of the Spanish 101 students received a grade of B minus or higher.
mokkyukkyu
• Posts: 97
• Joined: Aug 17, 2016
#29163
Hi,
Is A wrong because
It could be true that A students might not exist? Or they might skip a class too?
In real life, usually there are some % As stundets Bs students etc but from the stimuls we only know about B minus students so we don't really know about all other students...right?
Is Be wrong because it is illegal negation? (illegal revese)
And...
At first, I thought in real life, grades consist of test, attendance, and other factors too so I thought E may not be true too...but in this case we only care about class attendance right?
• PowerScore Staff
• Posts: 3813
• Joined: Apr 14, 2011
#29342
A is wrong here because we just cannot know, based on this info, about any A- students. B is wrong because we just don't have enough info about the students that missed some classes.
When I face the word "most", I start by putting numbers into the equation. I imagine that there are 100 students, and at least 51 of them attended every class and also got at least a B-. They had to have gotten at least that grade, because any lower grade would mean they had missed a class. This proves answer E to be correct - more than half the class got at least a B-.
If every student with less than a B- skipped a class, I imagine just one student fitting that description.
Now, I have 48 students about whom I know nothing at all. What grade did they get? Did they miss any classes? It's all up in the air.
It looks like you are trying to bring some causality into this equation, but the author did not do that so you shouldn't either. This isn't about why they got a certain grade, or what effect attending class had - it's just about numbers.
ShannonOh22
• Posts: 70
• Joined: Aug 15, 2019
#71067
Why wouldn't B work as an answer for this question? It says each student who got a B- or lower missed at least one class. Can we not then infer that most students who missed at least one class received a grade lower than B-? I feel like I'm missing how this answer is a Mistaken Reversal...please help
Paul Marsh
• PowerScore Staff
• Posts: 290
• Joined: Oct 15, 2019
#71537
Hi Shannon! Answer Choice (B) is incorrect because the stimulus leaves open the possibility that there could be many students who missed at least one class but still received a grade of a B minus or higher.
The last sentence of our stimulus tells us:
B minus (or lower) Missed a class (or more)
To say that if you missed a class, then you probably got a B minus, (as answer choice (B) does), would be a Mistaken Reversal! We can only read a conditional from left (sufficient condition) to right (necessary condition). So if a student got a B minus, then we know something (in this case, that he or she missed a class). But if a student missed a class, we don't know anything about that student's performance.
For example, say that 3 students in the class received a grade lower than a B minus. From our stimulus, we know that all 3 of those students must have missed at least one class. However, there could be 20 other students that missed a class but all got an A plus! The stimulus never rules out that possibility. So we don't know whether most students who missed a class got a B minus or lower, which means answer choice (B) is incorrect. Hope that helps!
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# Algebraic Fractions: Addition and Subtraction
So what do I have to multiply to 11 to make 66, 6 isn’t it? So we have to multiply this denominator by 6.
But guys, remember you can’t just multiply the denominator by six if you’re doing something to the bottom, you must do the same thing to the numerator to the top.
So I’m going to multiply six to bottom and top for this fraction. Now, what do I have to multiply to six to get sixty-six? This time we have to multiply 11, don’t we?
I’m going to multiply 11 to 6. But again you have to do the same thing to the top.
This is what I’m going to do multiply 6 to top and bottom for this fraction and multiply 11 to both top and bottom for this fraction. Okay?
And then just simplify so what’s 6 times 8x 48 and 6 times 11 is 66 same for this one and 11 times 5 is 55. So 55x. Okay? So can you see that now we have the same denominator?
Now keep that denominator and you merge the numerators. So you do 48x minus 55x. What is that? All of you should be able to do this. It’s negative 7.
So negative 7x sorry so negative 7x over the same denominator 66 is your answer. So you leave the denominator as 66, Okay? Let’s do the same thing.
What is common what’s the common multiple the lowest common multiple of 3 and 5? It should be 15, isn’t it? So I need to multiply 3 by 5 and 5 by 3 to get that same 15.
Make sure when you’re multiplying by five you do it to the top as well as the bottom and when you’re multiplying by three you do it to both top and bottom like that.
Here we have 5x over 15 and here we have 6x over 15. So we have the same denominator. All you need to do is merge the numerator. So 5x plus 6x is 11x, so it’s 11x over 15, Okay?
So you keep the denominator the same all the time all right? Easy peasy.
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# I am a prime number. Double of myself is equal to the square if myself. Which number am I ?
453 views
I am a prime number. Double of myself is equal to the square if myself. Which number am I ?
posted Mar 2, 2018
2X = X^2
X^2 - 2X = 0
X(X - 2)=0
X = 0, 2
So the answer is 2 (because its a prime) and not 0.
answer Mar 2, 2018
+1 vote
2 is also prime number
when, we double this is equal to 4
if, we square its is equal to 4
so, required answer is 2
answer Mar 3, 2018
+1 vote
answe----------------- two
answer Mar 8, 2018
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# MULTIPLE CHOICE QUESTIONS CLASS VII TOPIC - Maths
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WWW.MATHSTIMES.COMMULTIPLE CHOICE QUESTIONSCLASS VIITOPIC: INTEGERSTIME: 30 MINUTESTOTAL QUESTIONS 40Q1. In addition and subtraction of two integers , sign of the answer depends upon(a) Smaller number (b) Their difference (c) Their sum(d) Greater numerical valueQ2. Sum of two negative number is always(a) Positive(b) Negative(c) 0(d) 1(b) Positive(c) 1(d) 0(b) 65(c) ̶ 7(d) 7(b) 40(c) 2(d) ̶ 2Q3. Sum of two Positive number is always(a) NegativeQ4. Sum of – 36 and 29 is(a) ̶ 65Q5. Sum of ̶ 19 and ̶ 21 is(a) ̶ 40Q6. Which of the following statement is false:(a) ̶ 7 (̶ 6 ) ̶ 13(b) ̶ 5 1 4(c) 2 (̶ 1 ) 1(d) 8 (̶ 9 ) ̶ 1Q7. The pair of integers whose sum is ̶ 5(a) 1, ̶ 4(b) ̶ 1 , 6(c) ̶ 3 , ̶ 2(d) 5, 0Q8. What integers or number should be added to ̶ 5 to get 4(a) 1(b) ̶ 1(c) ̶ 9(d) 9(c) 3(d) 5Q9 . What will be the additive inverse of ̶ 5(a) ̶ 6(b) ̶ 4Q10. What will be the additive inverse of 7(a) ̶ 7(b) ̶ 6(c) ̶ 5(d) ̶ 4Q11. Predecessor of ̶ 9 is(a) ̶ 8(b) 8(c) ̶ 10(d) 10(b) 0(c) 1(d) 2Q12. Successor of ̶ 1 is(a) ̶ 2WWW.MATHSTIMES.COM1
WWW.MATHSTIMES.COMQ13. The value of 6 ̶ ( ̶ 3 ) is(a) 3(b) ̶ 9(c) ̶ 3(d) 9(c) ̶ 56(d) 56(c) 7 ̶ 4 4 ̶ 7(d) 7 ̶ 4 ̶ 3Q14. The value of 26 ̶ 30 is equal to(a) 4(b) ̶ 4Q15. Which of the following statement is true(a) 7 ̶ 4 4 ̶ 7(b) 7 ̶ 4 4 ̶ 7Q16. Choose appropriate number for blank: ̶ 7 ̶ ( ) 2(a) 5(b) ̶ 5(c) 9(d) ̶ 9(c) ̶ 12(d) 7(c) ̶ 140(d) 90(c) ̶ 10(d) 7Q17. Multiplication of 3 and ̶ 4(a) ̶ 7(b) 12Q18. Multiplication of ̶ 2 , ̶ 7 and ̶ 10 gives(a) ̶ 34(b) 19Q19. Multiplication of 2 , ̶ 5 and 0 gives( a) 10(b) 0Q20. Identify the property used in the following:(a) Commutative(b) Closure213 813 ( 2 8 )(c) Associative13(d) DistributiveQ21. Which number is multiplicative identity for the whole numbers(a) 0(b) 1( c) 2(d) 3Q22. What will be multiplicative inverse of – 8(a) 8(b)(c) ̶Q23. Which property is reflected in the following:(a) ClosureQ24.(d) 075 57(b) Commutative(c) Associative(d) Distributive(b) 9(c) ̶ 9(d) ̶ 16(b) 2(c) ̶ 2(d) 3(c) ̶ 5(d) 5̶ 18 2 gives(a) 36Q25. – 6 ( ̶ 3 ) gives(a) ̶ 9Q26.15 divided by ̶ 3 is equal to(a) 12(b) ̶ 12WWW.MATHSTIMES.COM2
WWW.MATHSTIMES.COMQ27. 0 10 gives(a) 0(b) 10(c) 1(d) ̶ 10(c) 12 0 12(d) 4 1 4Q28. Which of the following is not true(a) 0 2 0(b) – 25 5 ̶ 5Q29. Which of the following is true(a) 5 7 7 5(b) 0 3 0 5(c) 2 ( 3 ̶ 1) 2 3 ̶ 2 1(d) 4 1 1 4Q30 . Which of the following does not represent pair of integer (a, b) such that a b 2(a) ( ̶ 6 , ̶ 3 )(b) ( ̶ 2 , 1)(c)( ̶ 10 , ̶ 5 )(d)(8 , 4 )Q31 . On dividing a negative integers by other negative integer the quotient will be(a) Always negative (b) Always positive (c) Either positive or negative(d) 1Q32. Which of the following statement is true(a) 7 0 7(b) 7 0 0(c) 7 0 0 7(d) 0 7 0Q33. Product of two negative integers is always(a) Always negative (b) Always positive (c) Either positive or negativeQ34. The integer whose product with(b) ̶ 20(a) 20(d) 0̶ 1 is ̶ 40 is(c) ̶ 40(d) 40(c) 11(d) 0(c) ̶ 720(d) 720(c) 52(d) 320(c) 16(d) ̶ 16(c) ̶ 5(d) 100Q35. Absolute value of ̶ 11 isQ36.(a)̶ 10̶81010 2 is equal to(a) 162Q38.̶ 16(b) 17125 ( ̶ 25 ) is equal to(a) 1Q40.(b) 192( ̶ 1) is equal to(a) ̶ 17Q39.9 is equal to(b) ̶ 27(a) 27Q37. 16(b) 10(b) 5( ̶ 50 ) ̶ 1 , number in the blank will be(a) 49(b) 50(c) ̶ 50(d) 51WWW.MATHSTIMES.COM3
WWW.MATHSTIMES.COMMULTIPLE CHOICE QUESTIONSCLASS VIIFRACTIONS AND DECIMALSTOPIC:TIME: 30 MINUTESTOTAL QUESTIONS 401. Which of the following fraction has numerator 5a)25b)57(c) 157(d) 783(d)152. Which of the following fraction has denominator8.a)83b) 138(c) 1383. What fraction does the shaded portion in the adjoining fig. represents.a)5235b)(c)25(d)53(c)47(d)434. Which one of the following is proper fraction?a)75b)325. Which one of the following is improper fraction?a)236. What is the value ofa)5147. What is the value ofa)512b)57(c)74(d)1257(c)67(d)35142935(c)31352 3 7 7b)3 2 5 7b)(d)535WWW.MATHSTIMES.COM5
WWW.MATHSTIMES.COM8. What is the value of2 1 7 3 3 3103b)9. What is the value of2 1 7 3 3 3a)a)103b)10. What is the value )438(c)215(d) none of these5 38 8b)11. What is the value of1094 2̶5 3b)141512. Which of the following drawing shows 215a)b)c)d)13. Which of the following drawing shows 3 a) c)b) 14. The value ofa) 121231 244 d) None of theseof 24 isb)112(c) 48(d)148WWW.MATHSTIMES.COM6
WWW.MATHSTIMES.COM15. The product of(a)14gives320(b)512(c)16. Which of the following product gives the value15(d) None of these(a) 32517. The product of(a)(b)5512125(d)20378513(c) 35261gives4(b)2110(c)1110(d)1510(b) 132(c)53(d)35(c) 9(d)116218. The reciprocal of 1 is3(a)3219. The value of 12(a) 16(b) 120. The value of 312(a) 421. The value of 4(a) 4283(c)94(d)2116(b) 13(c)139(d)913(c)12(d)21(b)133 is22. Which of the following is the least form of(a)36(b)918183623. What is the sum of 5.300 and 3.250(a) 8.550(b) 85.50(c) 5.6250(d) 8550WWW.MATHSTIMES.COM7
WWW.MATHSTIMES.COM24. What is the value of 29.35 ̶ 04.56(a) 23.75(b) 16.35(c) 16.25(d) 24.79(c) 0.005(d) 0.05(c) 0.2301(d) 23.01(c) Rs0.707(d) Rs 770(c) 0.0005(d) 0.05(c) Tenth(d) Hundredth25. Which one of the following is greater(a) 5.0(b) 0.526. Which one of the following is smaller( a) 2.031(b) 2.30127. 7 Rupees 7 paisa can be written in rupees as(a) Rs7.07(b) Rs7.7028. 5 cm in Km can be written as(a) 0.0005(b) 0.0000529. The place value of 2 in 21.38 is(a) Ones(b) Tens30. Which one of the following represent the expansion2 0 0 3(a) 20.0331. The value of 2.71(b) 2.03(c) 200.03(d) 2.034(b) 1355(c) 13.55(d) 1.355(c) 153.7(d) 1537(c) 43.07(d) 430.7(c) 0.003(d) 30(b) 0.403(c) 4.03(d) 0.0403(b) 0.16(c)is(a) 135.532. The product of 153.7 and 10 is(a) 1.537(b) 15.3733. The value of 43.07is(a) 4.307(b) 430734. The value of 0.03is(a) 0.0000335. The value of 1.3(b) 3is(a) 40336. The value of 0.80(a) 1637. The value of 52.5(a) 5.25is116(d) 1.6is(b) 0.525(c) 525(d) 5250WWW.MATHSTIMES.COM8
WWW.MATHSTIMES.COM38. The value of 0.78(a) 780039. The value of 26.3(a) 0.026340. The value of 7.75(a) 31is(b) 0.0078(c) 0.78(d) 7.8(c) 26300(d) 26.300(c) 0.31(d) 3.1is(b) 0.2630is(b) 0.0031WWW.MATHSTIMES.COM9
WWW.MATHSTIMES.COMTopic: Fractions and decimalsANSWER KEYClass – ISConceptsBasic concepts of fractionsAddition and subtraction of fractionsMultiplication of fractionsDivision of fractionsBasic concepts, addition and subtraction ofdecimalsMultiplication of decimalsDivision of decimalsTally ,,17,1819,20,21,2223 to 3031 to 3536 to 40WWW.MATHSTIMES.COM10
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WWW.MATHSTIMES.COMMULTIPLE CHOICE QUESTIONSCLASS VIITOPIC: PERIMETER AND AREATIME: 30 MINUTESQ.1.The perimeter of regular polygon is(a)(c)Q.2.Q.5.(b)decrease(c)remains same(d) none of these1 m2(b)4 m2(c)2 m2(d)3 m29000 cm227 cm2(b)90 cm2(c)9 cm2(d) 900 cm2is isosceles AE 6 cm , BC 9 cm, the area of(b)54 cm2(c)22.5 cm2is(d) 45 cm2The area of parallelogram is(a)Q.8.increaseIn the figure(a)Q.7.no. of sides lengths of one sideno. of sideslengths of one sideWhich figure encloses more area : a square of side 2 cm ; a rectangle of side 3 cm &2 cm ;An equilateral triangle of side 4 cm(a) rectangle(b) square(c) triangle (d) same of rectangle & squareThe area of rectangle whose length is 15 cm & breadth is 6 m(a)Q.6.(b)(d)The area of a square whose perimeter is 4 m(a)Q.4.no. of sides lengths of one sideno. of sides – lengths of one sideIf the area of rectangle increases from 2 cm2 to 4 cm2 the perimeter will(a)Q.3.TOTAL QUESTIONS 40base height (b) baseheight (c) basebase(d)heightbase(d)(d)areaheightThe base in the area of parallelogram is(a)(b)(c)areaareaheightQ.9.The height in the area of parallelogram is(a)(b)(c)Q.10. Which of the following has the formula Base(a)(c)area of parallelogramarea of triangleareabaseheightHeight(b) area of quadrilateral(d) area of trapeziumWWW.MATHSTIMES.COM15
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WWW.MATHSTIMES.COMANSWERKEYMENSURATIONCLASS a)(d)(b)(c)(a)(b)(b)(a)(a)QUESTIONSCORRECT )(c)(c)(b)(b)(a)ANALYSE YOUR PERFORMANCEQUESTIONS1, 3, 7, 8, 9, 11, 12,14, 27, 31, 3415, 26, 28, 32, 36, 382, 4, 6, 10, 21, 23, 24,29, 33, 35, 39, 405, 13, 16, 17, 18, 19,20, 22, 24, 30, 37TALLY MARKSREVISE THESE CONCEPTSKnowledge of formulaeConcept of termsUnderstanding of conceptsApplicationsCLASS VIIWWW.MATHSTIMES.COM19
WWW.MATHSTIMES.COMTIME: 30 MINUTESMULTIPLE CHOICE QUESTIONSCLASS VIITOPIC: ALGEBRAIC EXPRESSIONTOTAL QUESTIONS 40Q1: Write the expression for the statement: the sum of three times x and 11(a) x 3 11( b) 3x 11(c) 3 11x(d) 3x-11Q2 : Write an expression : Raju s father s age is 5 years more than 3 times Raju s age . If Raju sage is x years , then father’s age is(a) 3x 5(b) 5-3x(c) 3x-5(d) 15xQ3 : Identify the coefficient of x in expression 8-x y(a ) 0(b) 8(c) -1(d) 1Q4: The number of terms in 4p2q -3pq2 5 is(a) 7(b) 3(c) 1(d) 4Q5: The expression for sum of numbers a and b subtracted from their product is(a) a b –ab(b) ab-a b(c) ab-(a b)(d) ab a - b.Q6: The sum of mn 5-2 and mn 3 is(a) 2mn 3(b)6( c) 2mn 8(d) 2mn 6.Q7: What is the statement for the expression 3mn 51of product of m and n3(b) number 5 added to product of number m and n(c) number 5 added to 3 times the product of m and n.(d) 5 more than 3 times the product of the numbers m and n(a) 5 more thanQ8 : The constant term in the expression 1 x2 x is(a) 1(b) 2(c) x(d) x2Q9: The coefficient of y3 in the expression y – y3 y2 is(a) 1(b) y(c) -y3(d) -1Q10: The number of terms in the expression 1.2ab – 2.4 b 3.6a is(a)1.2(b) -2.4(c) 3.6a(d) 3Q11: What is the numerical coefficient of y2 in the expression 2x2y – 15xy2 7y(a) -15x(b) -15(c) 2(d) 7WWW.MATHSTIMES.COM20
WWW.MATHSTIMES.COMQ12: The expression x y – xy is(a) Monomial(b) Binomial(c) Trinomial(d) Quadrinomial(b) Binomial(c) Trinomial(d) Zero polynomialQ13: The expression xyz is(a) MonomialQ14: From the following expressions 10pq, 7p, 8q, -p2q2, -7pq, -23, ab,3a,b.The like terms are(a) 3,7p(b) 10 pq, -7pq(c) ab,3a,b(d)10pq,7p,8qQ15:From the following expressing 3ab,a2,b2,a,5ab,-2ab,2a2 the three terms are(a) 3ab,5ab,-2ab(b) a2,a,2a2(c)3ab,a2,b2(d)2a2,a2, a(b) 3m 2n(c) 5m(d) 5n(c) 2xy x y(d) 2xy x 2y(c)8b - 32(d)28b -52(c) 2b(d) 2a-2b(c) 4 y2(d) -6 y2Q16: Sum of 3m and 2n is(a)5mnQ17: Sum of xy, x y and y xy is(a) 2xy 2x y(b) 3xy 2yQ18:The value of 21b-32 7b – 20b is(a) 48b- 32(b) -8b - 32Q19:Subtract a –b from a b the result is(a) 2a 2b(b) 2aQ20:Subtracting -5y2 from y2 ,the result is(a) -4y2(b) 6y2Q21: The value of expression 5n – 2, when n -2 is(a) -12(b) 8(c) 1(d) -8(c) 21a-8b(d) 29Q22:The value of expression 7a-4b for a 3,b 2 is(a) 13(b) 7a-6bQ23: When x-0,y -1,then the value of expression 2x 2y is(a) 4(b) 0(c) -2(d)2Q24:Factors of the term 15x2 in the expression 15x2 -13x are(a) 15,x,x(b) 15,-13(c) 15x2,-13x(d)15Q25:Factors of the terms -4pq2 in the expression 9p2q2 -4 pq2 are(a) 9 p2q2 ,-4 pq2(b)9,-4(c)-4,p,q,q(d) -4Q26:If the length of each side of the equilateral triangle is l, then the perimeter of theequilateral triangle isWWW.MATHSTIMES.COM21
WWW.MATHSTIMES.COM(a) 3l(b) 3 l(c) 3-l(d) l/3(c) 4x 2y 3(d) 4y 5x z ̶ 1(c) 5a ̶ 7(d) a b c ̶ 3Q27:Which of the following is monomial(a) 2x 3(b) 2xQ28:Which of the following is trinomial(a) 2a 6b-1(b) 1Q29: Terms with factors y in the expression 8 xy xyz are(a) xy, xyz(b) x, xz(c) 8, xy, xyz(d) y, xzQ30: Identify the terms in the expression x y 1 which are not constant(a) x,y,1(b) x, y(c) x,1(d) y,1(c) 4(d) 2Q31:The value of expression 4x ̶ 3 at x 2 is(a) -4(b) 5Q32: The value of expression 5n2 5n ̶ 2 for n ̶ 2 is(a) 13(b) 3(c) 8(d) 12Q33: The value of expression 2a2 2b2 ̶ ab for a 2, b 1is(a) 2(b) 8(c) 6(d) 10(c) 12(d) 37Q34: The value of x 7 4(x ̶ 5) for x 2(a) ̶ 3(b) 31Q35: The value of expression 2a ̶ 2b ̶ 4 ̶ 5 a at a 1, b -2(a) 10(b) ̶ 2(c) 12(d) ̶ 4Q36: What must be subtracted from 2a b to get 2a ̶ b(a) 2b(b) 4a(c) 0(d) 4a 4b(c) x ̶ 2y(d) x 2y(c) 2b(d) ̶ 2a 2bQ37: What must be added to 3x y to get 2x 3y(a) 5x 4y(b) ̶ x 2yQ38: Subtract a 2b from sum of a ̶ b and 2a b(a) 2a ̶ 2b(b) 4a 2bQ39: On simplifying (a b-3) – (b –a 3) (a ̶ b 3) the result is(a) a ̶ b 3(b) a ̶ b ̶ 3(c) 3a ̶ b ̶ 3(d) 3a b 3Q40:What should be value of ‘a’ if y2 y – a equals to 3 for y 1(a) ̶ 1(b) ̶ 5(c) 5(d) 0WWW.MATHSTIMES.COM22
WWW.MATHSTIMES.COMANSWER CT .Q.40.aacacabcabbcbababacaKEY FOR ANALYSISBasic ConceptsFormation of algebraic expressionConcept of terms , factors and coefficientAddition and subtractionvalue of an expressionQuestion no.1, 2, 7, 26,3, 4, 5, 8, 9, 10, 11, 12, 13, 14, 15,24, 25, 27,28, 29, 306, 16, 17, 18, 19, 20, 36, 37, 38, 3921, 22, 23, 31, 32, 33, 34, 35, 40WWW.MATHSTIMES.COM23
WWW.MATHSTIMES.COMMULTIPLE CHOICE QUESTIONSCLASS VIITOPIC: GEOMETRYTIME: 30 MINUTESTOTAL QUESTIONS 401. A line that intersects two or more lines at distinct points is called(a) Parallel(b) transversal(c) intersecting(d) none of these2. If two adjacent angles are supplementary, then they form .(a) Corresponding angles (b) vertically opposite angles(c) a linear pair of angles (d) a ray3. If two angles are supplementary then the sum of their measures is .(a) 90o(b) 180o(c) 360o(d) 45o4.If two angles are complementary, then the sum of their measures is .(a) 45o(b) 180o(c) 90o(d) 360o5.If
CLASS VII TOPIC: FRACTIONS AND DECIMALS TIME: 30 MINUTES TOTAL QUESTIONS 40 1. Which of the following fraction has numerator 5 a) 5 2 b) 7 5 (c) 7 5 1 (d) 5 1 7 2. Which of the following fraction has denominator8. a) 3 8 b) 8 3 1 (c) 3 8 1 (d) 8 3 3. What fraction does the shaded portion in
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# Older blog entries for amatus (starting at number 95)
I'm still waiting for the awesome language that will finally kill C.
I'm still waiting for the awesome language that will finally kill C.
Damien Katz: The Unreasonable Effectiveness of C
The Unreasonable Effectiveness of C. For years I've tried my damnedest to get away from C. Too simple, too many details to manage, too old and crufty, too low level. I've had intense and torrid love a...
Syndicated 2013-01-10 17:54:53 from David Barksdale - Google+ Posts
A solution to the 12 coins problem: 1 L-- 2 LL- 3 LR- 4 L-L 5 R-L 6 RRR 7 RLR 8 RLL 9 --R 10 L 11 -RR...
A solution to the 12 coins problem:
1 L--
2 LL-
3 LR-
4 L-L
5 R-L
6 RRR
7 RLR
8 RLL
9 --R
10 L
11 -RR
12 -RL
#cleaningscrapsofpaperoffmydesk
Syndicated 2012-12-06 03:20:14 from David Barksdale - Google+ Posts
If you have a unit interval and you want to know how many ways you can divide it such that division marks...
If you have a unit interval and you want to know how many ways you can divide it such that division marks are at m/2^n and the divisions are sorted from smallest to largest, use this recurrence relation:
a[0] = 1
a[n] = 1 + sum_{i=0}^{n-1} {2^(i(n-i-1))a[i]}
#cleaningscrapsofpaperoffmydesk
Syndicated 2012-12-06 03:01:38 from David Barksdale - Google+ Posts
If you have an n by m grid of points, you can connect 4 points to make a square in (2mn^3-2mn-2n^4+n^...
If you have an n by m grid of points, you can connect 4 points to make a square in (2mn^3-2mn-2n^4+n^2)/12 different ways.
#cleaningscrapsofpaperoffmydesk
Syndicated 2012-12-06 02:52:15 from David Barksdale - Google+ Posts
Apparently Rosencrantz was flipping a coin with Bose-Einstein statistics.
Apparently Rosencrantz was flipping a coin with Bose-Einstein statistics.
This Quantum World | Quantum coins and quantum diceA comparison of the behavior of ordinary coins and dice with that of quantum coins and dice (bosonic as well as fermionic).
Syndicated 2012-11-29 18:35:29 from David Barksdale - Google+ Posts
I remember seeing this comet at the McDonald Observatory.
I remember seeing this comet at the McDonald Observatory.
Comet Hyakutake - Wikipedia, the free encyclopediaHyakutake became visible to the naked eye in early March 1996. By mid-March, the comet was still fairly unremarkable, shining at 4th magnitude with a tail about 5 degrees long. As it neared its closes...
Syndicated 2012-11-28 19:46:45 from David Barksdale - Google+ Posts
I believe his point about just-in-time delivery making our industry very fragile.
I believe his point about just-in-time delivery making our industry very fragile.
Syndicated 2012-11-26 04:05:23 from David Barksdale - Google+ Posts
Fabbing a chip that could encode data in a twisted vortex of light
Fabbing a chip that could encode data in a twisted vortex of light
Tiny silicon-based device shapes twists photons for communication purposes
Syndicated 2012-10-30 19:38:11 from Google+ List of Activities for Collection PUBLIC
86 older entries...
New Advogato Features
New HTML Parser: The long-awaited libxml2 based HTML parser code is live. It needs further work but already handles most markup better than the original parser.
Keep up with the latest Advogato features by reading the Advogato status blog.
If you're a C programmer with some spare time, take a look at the mod_virgule project page and help us with one of the tasks on the ToDo list!
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Algebra
Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5.88 × 10*12 miles. Suppose a star is 9.8 × 10^1 light-years from Earth. In scientific notation, approximately how many miles is it?
A. 5.88 × 10^13 miles
B. 5.76 × 10^14 miles***
C. 5.88 × 10^12 miles
D. 9.8 × 10^12 miles
1. correct .
posted by Steve
2. So the answer is B?
posted by Alyssa
3. I really hope so cause im on the same question
posted by Tatum
Similar Questions
1. Algebra
Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5.88*10^12 miles. Suppose a star is 3.2*10^12 light-years from Earth. In scientific notation,
2. algebra
Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5.88 *10^12 miles. Suppose a star is 3.2 * 10^2 light-years from Earth. In scientific notation,
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Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5.88 *10^12 miles. Suppose a star is 3.2 * 10^2 light-years from Earth. In scientific notation,
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Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5.88 *10^12 miles. Suppose a star is 3.2 * 10^2 light-years from Earth. In scientific notation,
5. Math
Simplify the expression. (-4)^-6(-4)^-7 My answer: -4 Simplify and give the answer in scientific notation. (5 X 10^6((3 X 10^5) My answer: 1.5 X 10^12 (3 X 10^6)(8 X 10^-4) My answer: 2.4 X 10^3 Astronomers measure large distances
6. math
simplify (8*10^7)(7*10^4) i think it's 5.6*10^10 simplify (3*10^6)(8*10^-4) i think it's 2.4*10^3 Astronomers measure large distances in light-years. One light-year 5.88*10^12 mi Star is 9.8*10^1 light years from Earth in
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61_PartUniversity Physics Solution
# 61_PartUniversity Physics Solution - Motion Along a...
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Motion Along a Straight Line 2-31 2.76. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let y + be downward. The egg has 0 0 y v = and 2 9.80 m/s y a = . At the height of the professor&s head, the egg has 0 44.2 m yy −= . EXECUTE: 2 1 00 2 y y yy v t a t −= + gives 0 2 2( ) 2(44.2 m) 3.00 s 9.80 m/s y t a == = . The professor walks a distance (1.20 m/s)(3.00 s) 3.60 m x xx v t −= = = . Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is 2 (9.80 m/s )(3.00s) 29.4 m/s = . It is traveling much faster than the professor. 2.77. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let y + be upward. At the maximum height, 0 y v = . When the rock returns to the surface, 0 0 = . EXECUTE: (a) 22 2( ) y vv a =+ gives 2 1 0 2 y y aH v =− , which is constant, so EE MM a H = . 2 E ME 2 M 9.80 m/s 2.64 3.71 m/s a H HH H a ⎛⎞ = ⎜⎟ ⎝⎠ . (b) 2 1 2 y y t with 0 0 gives 0 2 y y at v , which is constant, so aT a T = . E M 2.64 a TT T a ⎡⎤ ⎢⎥ ⎣⎦ . EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. 2.78. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let y + be upward. 2 9.80 m/s y a . 0 0 = when the ball returns to its original position. EXECUTE: (a) It takes her 5.50 m 2.20 s 2.50 m/s = to reach the table and an equal time to return. For the ball, 0 0 , 4.40 s t = and 2 9.80 m/s y a . 2 1 2 y y t gives 2 11 0 ( 9.80 m/s )(4.40 s) 21.6 m/s va t =− − = . (b) Find 0 when 2.20 s t = . 2 (21.6 m/s)(2.20 s) ( 9.80 m/s )(2.20 s) 23.8 m t = +− = EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. 2.79. (a) IDENTIFY: Use constant acceleration equations, with , y ag = downward, to calculate the speed of the diver when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the -direction y + to be downward. 0 21.3 m, + 2 9.80 m/s , y a 0 0 y v = (since diver just steps off), ? y v = ) y EXECUTE: 2 0 2 ( ) 2(9.80 m/s )(31.3 m) 20.4 m/s. y y We know that y v is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use 2 1 2 y y t to find 2.085 s. t = The diver gains 9.80 m/s of speed each second, so has 2 (9.80 m/s )(2.085 s) 20.4 m/s y v when she reaches the water, which checks.
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https://testbook.com/question-answer/inabc-o-is-the-circumcentre-of-the-t--5fe99c16dd06b989429eaf38
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# In ΔABC, O is the circumcentre of the triangle and ∠BOC = 60°, then what will be the value of ∠BAC?
1. 90°
2. 30°
3. 60°
4. 45°
Option 2 : 30°
## Detailed Solution
Given:
O is circumcentre
∠BOC = 60°
Formula used:
The angle subtended at the center is double the angle subtended at the circle
Calculation:
∠BOC = 2 × ∠BAC
⇒ 60° = 2 × ∠BAC
⇒ ∠BAC = 30°
∴ ∠BAC is 30°
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# Chapter 5 - Section 5.1 - Introduction to Decimals - Exercise Set - Page 338: 50
$-0.59\lt-0.52$
#### Work Step by Step
Working left to right, compare the digits with the same place value. The first value that differs is the hundredths position. $-9\lt-2$, so $-0.59\lt-0.52$. Note that when comparing negative numbers, the number with the larger digit is the smaller number. The larger digit is "more negative".
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Parking Lot Puzzle [closed]
I have to answer someone urgently but I am not able to find out the answer.
• Hollup, this is not a duplicate, the numbers are different. Commented Jun 29, 2017 at 4:58
• But probably the same concept. Commented Jun 29, 2017 at 5:07
• It doesn't look to me like it's the same. With the other one, once you see what's going on the answer is obvious. With this one, at least for me, applying the idea from the other one doesn't make anything obvious. Commented Jun 29, 2017 at 9:22
• Is there actually a clear defined answer? For sequence puzzles like these it is possible to make a case for practically any number being correct. How do we know which one is the intended solution? Commented Jun 29, 2017 at 18:43
• The "I have to answer urgently" line lets me assume that the OP does not know the answer, which makes this puzzle rather... bad. Also, the tagging lets one assume it is really just about finding the answer to "18,114,158,X,338" which makes it really bad and very differnt form the "original" which had some out-of-the-box-but-obvious solution. I'm therefore voting to close unless the post is somewhat improved. Commented Jun 30, 2017 at 8:44
The car is parked in spot
274. The difference between 18 and 158 is 140, 158 and 338 is 180. Therefore 114 + 160 = 274
• How do you infer "160" of just two numbers (140 and 180) ? What makes you assume that "+20" is the way to go? It could as well be $* Sqrt(9/7)$.. Commented Jun 29, 2017 at 17:52
• (Not saying that your solution isn't "beautiful", but if it is the correct solution, the puzzle is really bad in my eyes.) Commented Jun 29, 2017 at 17:53
• If it's the solution it's also not a very good sequence, as the first two numbers aren't really related, but it got stuck in my head because it was so oddly "neat"
– Emma
Commented Jun 29, 2017 at 18:56
That number might be 294.
It starts with number 18.
18 x 6 + 6 = 114
18 x 8 + 14 = 158
18 x 16 + 6 = 294
18 x 18 + 14 = 338
I here use some sequences. Also the last digit 4 8 4 8 4 also recursive with my answer.
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The MCQ: The example of linear equation involving two variables is; "Linear Equations: Math" App Download (Free) with answers: 7x+3y+4z = 20; 6x+2y = 10; 8x = 2+10; 7a+8b+9c = 10+5; for online bachelors degree. Practice Linear Equations Math Quiz Questions, download Google eBook (Free Sample) for business management degree online.
Linear Equations Math MCQs: Questions and Answers
MCQ 1:
The variables of linear equation is implicitly raised to
1. first power
2. second power
3. third power
4. four power
MCQ 2:
The example of linear equation involving two variables is
1. 7x+3y+4z = 20
2. 6x+2y = 10
3. 8x = 2+10
4. 7a+8b+9c = 10+5
MCQ 3:
In the linear equation 'ax+by = c' the a,b and c are considered as
1. variable
2. constants
3. zero
4. real numbers
MCQ 4:
In the linear equation ax+by = c, the a and b cannot be equal
1. to rational numbers
2. to one
3. to zero
4. set of even numbers
MCQ 5:
The two variables x and y if involved in linear equation then the equation is
1. ax+by = c
2. ab+xy = c
3. ac+bx = y
4. ax+bc = y
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# Curve building for a swap
I'm a student learning how to build a swap curve, I've deposit 6m rate = 5%, fra 6-12m rate = 5.8% and 12m-18m rate= 6% and swap 2y =7% and 3y swap rate = 7.5%. I get the correct discount factors for the deposit and fras but for swaps I get wrong values..
DF6m =0.97547758 DF6-12m =0.94709626 DF12-18m =0.91921345 DF 2y swap =?? DF 3y swap =??
For this example starting date is 10/09/2010 using day conv act/360. Can you help me get the right discount factors for 2y and 3y swaps. or direct me to book with examples of curve buildings ? thank you
Try Howard Corb's book ["Interest Rate Swaps and Other Derivatives", Columbia Business School Publishing, 2012].
And your 2y rate looks odd which should be intuitively obvious - simple arithmetic mean of your 6m, 6mf6m and 12mf6m rates is 5.6%. What does that suggest about your 18mf6m rate for the 2y rate to be 7%?!
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Optimization Metal Tank Problem
1. Jan 4, 2006
Xcron
Ok, well the problem states:
A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?
The first step that I took was to draw a picture. I just drew a semicircle with a right circular cylinder right in the middle with the base on the bottom of the hemisphere and the top touching at two points to the rounding of the semicircle. I've attached a crude representation of my drawing.
Then, I defined the volume of this storage tank, V = (1/2)(4/3*pi*r(hemisphere)^3) - (pi*r(cylinder)^2*h(cylinder)). I kept in mind the fact that V is a constant that was given in the problem and that I would use it in my final equation that I would optimize.
I am not sure of this is correct but it seems to me that the metal storage tank would be a hemisphere with a hole that is a right circular cylinder...
Next, I tried to eliminate one of the three variables. I did this by drawing a triangle from the top-right corner of the cylinder to the center of the cylinder and connected it to the right side of the cylinder. I have attached a representation of this too.
I used the pythagorean theorem from there and said that (h/2)^2 = (r(hemi))^2 - (r(cyli))^2. Then I continued to square root both sides to solve for h/2.
I thought that I would need to solve for the surface area of the tank to see how much material would be needed and I write an equation for that as A = (1/2)(4*pi*r(hemi)^2) - (2*pi*r(cyli)^2 + 2*pi*r(cyli)*h). After that I tried to get r(cyli) onto one side so that I could solve for it but I ended up with a rather large and extremely convoluted mess from which I was not able to do so.
Any help would be appreciated. I'm stuck as to the actual representation of the problem...*sigh*...
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2. Jan 4, 2006
Staff: Mentor
I think you have the geometry wrong. It sounds to me like the problem is describing an old grain-silo type shape. Or a long cylinder with a rounded end kind of like a condom if you haven't seen old grain silos. Sorry I can't think of a better description. I think that the radius of the hemisphere is equal to the radius of the cylinder and that the volumes and surface areas are simply added.
So you have a cylinder with a hemisphere stuck on top. I think there are only two design variables, the radius and the height of the cylinder portion. You should then be able to solve for e.g. r in terms of V and h and then substitute into the equation for the surface area to get A as a function of h and V.
-Dale
Last edited: Jan 4, 2006
3. Jan 4, 2006
Mindscrape
Why did you subtract the cylinder's volume? And yes, I think it's fair to assume the radius is the same for the two, which now makes it an easy/straightforward optimization problem.
4. Jan 5, 2006
Xcron
Hhahahahahahah! If you are correct then it will truly become an easy and straightforward problem. I must admit that I did not understand the visualization part of the problem correctly...perhaps because I have not been trained to deal with images of such words as surmounted.
Is that what surmounted really means by any chance?....Because I did a problem where the image was once given and it said a rectangle that is surmounted by a semicircle and it looked just like what I was trying to draw out except there was a rectangle instead of a cylinder.
If only the said had stated that it was mounted on top of a standing cylinder instead...*sigh*...
I subtracted the cylinder's volume because I thought that is where something would be stored...this idea was produced from my confusion after trying to visualize the problem...I'm sorry.
Would anyone confirm as to the explicit meaning of the imagery/language of the problem?
I believe that DaleSpam has quite a good set-up but I am not sure if that is the correct one that I should be using...
Also, would anyone explain how I may use the mathematical pictures of pi and such things that I have seen people use instead of my crude typing of i.e. 2*pi*r(hemi)...
Last edited: Jan 5, 2006
5. Jan 5, 2006
Mindscrape
But you wouldn't subtract it for storage because then you would have a rod with hemisphere dug into it. I think they want the volume of the whole thing anyway, for possible storage. So you probably had it right, except add this time.
$$V = \frac{2}{3} \pi r^3 + \pi r^2 h$$
Oh, the typesetting is TeX. Ever used LaTeX before? Pretty much what this is. If you ever get a math project where you have to type something up, then you should use LaTeX.
Last edited: Jan 5, 2006
6. Jan 5, 2006
Staff: Mentor
Am I allowed to confirm that I am correct.
Check out the LaTeX thread in the tutorials section (https://www.physicsforums.com/showthread.php?t=8997) It has a bunch of examples and "how to" links and you are allowed to practice there. If you practice there please be sure to delete your post.
-Dale
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# Math for Computer science Questions and Answers 201 to 210
## Math for Computer science
### Questions 201 to 210
201.
(p Ú q) → q is a
(a) Contingency (b) Contradiction (c) Tautology (d) Contrapositive (e) Inverse.
202.
The fallacy of affirming the consequent is denoted by
(a) (b) (c) (d) (e) .
203.
From the following which is not the method of proof ?
(a) Vacuous proof (b) Trivial proof (c) Indirect proof (d) Non-trivial proof (e) Direct proof.
204.
Let F(x) : x is a bird, P(x): x can fly.
The statement “All birds can fly” equivalent is
(a) (b) (c) (d) (e) .
205.
Which of the following statements is false
(a) ÆÎ power set of {0} (b) 0Î power set of {Æ} (c) {0}Î power set of {0} (d) {Æ}Ì power set of {0} (e) Æ = power set of Æ.
206.
Which of the following statements is false?
(a) (b) (c) (d) (e) .
207.
Which of the following is Absorption law
(a) (b) (c) (d) B Ç (A È B) = A (e) .
208.
If E and F be the events with P(F) > 0. The conditional probability P(E | F) =
(a) P( E Ç F) × P( F) (b) P( E Ç F) + P(F) (c) P( E Ç F) – P(F) (d) P( E Ç F) / P(F) (e) P(E È F) / P(F).
209.
Determine which of the following functions from {a, b, c, d} to itself is one-to-one.
(a) f(a) = b, f(b) = a, f(c) = c, f(d) = b (b) f(a) = a, f(b) = a, f(c) = c, f(d) = d (c) f(a) = a, f(b) = b, f(c) = c, f(d) = d (d) f(a) = b, f(b) = c, f(c) = c, f(d) = d (e) f(a) = c, f(b) = d, f(c) = c, f(d) = d.
210.
What is the probability of getting a tail when the coin is biased so that heads comes up twice as often as tails?
(a) 1/2 (b) 2/3 (c) 3/2 (d) 2/5 (e) 1/3.
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# Show that there do not exist any distinct natural numbers a,b,c,d such that
Show that there do not exist any distinct natural numbers a,b,c,d such that
$$a^3+b^3=c^3+d^3$$ and $$a+b=c+d$$
Suppose $$a+b=c+d$$ and $$a^3+b^3=c^3+d^3$$. $$a+b=c+d$$ $$(a+b)^3=(c+d)^3$$ $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ $$3ab(a+b)=3cd(c+d)$$ $$ab=cd$$
Let $$a+b=c+d=m$$ and $$ab=cd=n$$
a and b are the roots of the quadratic equation $$x^2-mx+n=0$$ by Vieta's relations because a+b=m and ab=n. But c and d are also roots of the equation for similar reasons. But a quadratic equation can have at most two distinct roots.
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## Ranking / Arrangement Concept in Reasoning for SSC Exams
Ranking / Arrangement Concept
In this type of questions, a set of information pertaining to persons, objects or some other entities alongwith their qualities, which can be compared is provided. Candidates are required to arrange the given entities in either ascending or descending order on the basis of relative quality. At first look, such questions appear to be very simple but sometimes these are made tricky by infusing intricacies and complicated data. However, one can organize the given data quickly and easily by comprehending the given information in systematic manner. Sometimes, the given information may seem to be insufficient but a cogent and co-problem turns to be very simple. You may also find some superfluous statements, which are given to confuse you. You can recognise such statements in no time if you have gained command over such types of questions. If you haven't checked some of our latest quizzes, use below given links.
• Maths Quizzes - Set 1 Set 2 Set 3
• English Quiz on Spotting Errors
• English Quiz on One-Word Substitution
• Quiz on Indian History
• Quiz on General Science
• Sometimes you are required to ascertain only the rank of a person either from top or bottom in a class or a group of persons.In determining the rank, the information about the total number of persons in the class or a group is a prerequisite otherwise you cannot determine the rank. Sometime the rank, of a person from either end is given and you are required to calculate the total number of persons.
In another type of question on RANKING, you are required to ascertain some other qualities such as height, weight, marks, age etc. In this type of question on Ranking, the two extreme ends are occupied by the persons having the lowest and the highest degrees of quality that has to be compared. While solving such type of questions, you should employ certain symbols and notations to organize the given information so that you can use the data more conveniently and quickly too.
You Might Also Like::
# Ranking / Arrangement Quiz
• #### 1.Five men A,B,C,D and E read a newspaper. The one who reads first give it to C, the one who reads last had taken it from A, E was not the first or last to read. There were two readers between B and A. Find the Person who read the newspaper last.
1. E
2. B
3. D
4. A
Explanation:
Note:
# Ranking / Arrangement Quiz
• #### 2.If you are eleventh in a queue starting either end. How many are there in the queue?
1. Eleven
2. Twenty
3. Twenty One
4. Twenty Two
Explanation:
Note:
# Ranking / Arrangement Quiz
• #### 3.In a row of boys, if A who is 10th from the left and B who is 9th from the right interchange their positions, A becomes 15th from left. How many boys are there in the row?
1. 23
2. 27
3. 28
4. 31
Explanation:
Note:According to question, B is 9th from the right end and 15th from the left end. Therefore, total number of boys in the row = 9+15- 1 = 23
# Ranking / Arrangement Quiz
• #### 4.A is older than B but younger than C. D is younger than E but older than A. If C is younger than D, who is the oldest of all?
1. A
2. C
3. D
4. E
Explanation:
Note:According to question
C > A >B --------------(i)
E >D > A----------------(ii)
D > C-------------------(iii)
From all the three statements
E > D > C > A > B
Therefore, E is the oldest among them.
# Ranking / Arrangement Quiz
• #### 5.Heavier coins are costlier. Ram's coin is heavier than Mohan's and costlier than Ramesh's. Naresh's coin is costlier than Ram's but lighter than Yogesh's. Ramesh's coin is costlier than Mohan's. So who is the owner of the costliest coin?
1. Ram
2. Ramesh
3. Yogesh
4. Naresh
Explanation:
Note: Ram > Mohan ---------------(i)
Ram > Ramesh---------------(ii)
Yogesh > Naresh > Ram ---------(iii)
(Y) (N) (R)
Ramesh > Mohan ------------------(iv)
(Rm) (M)
From all the statements
Y > N > Rm > M
Therefore, the owner of the costliest coin is Yogesh
# Ranking / Arrangement Quiz
• #### 6.Among five friends, A is shorter than B but taller than E, C is slightly taller than B but D is slightly shorter than B and slighthy taller than A. Who is the shortest?
1. A
2. E
3. C
4. D
Explanation:
Note: B > A > E .
C > B,
B > D > A
Thus C > B > D > A > E
# Ranking / Arrangement Quiz
• #### 7.There are five friends - S ,K , M, A , R, S is shorter than K, but taller than R. M is the tallest. A is a little shorter than K and little taller than S. Who has two persons taller and two persons shorter than him?
1. R
2. S
3. K
4. A
Explanation:
Note: K > S > R--------------(i)
K > A > S--------------(ii)
M is the tallest-------(iii)
From statements (i), (ii) and (iii)
M > K > A > S > R
# Ranking / Arrangement Quiz
• #### 8.Of the five members of a panel sitting in a row, A is to the left of B, but n the right of C, D is on the right of B but is on the left of E. Find the member who is sitting in the middle?
1. B
2. D
3. A
4. C
Explanation:
Note:
# Ranking / Arrangement Quiz
• #### 9.A, B , C ,D and E are sitting on a bench. A is sitting next to B, C is sitting next to D, D is not sitting with E who is on the left end of the bench. C is on the second position from the right. A is on the right of B and E. A and C are sitting together. In which position is A sitting?
1. Between B and D
2. Between B and C
3. Between E and D
4. Between C and E
Explanation:
Note:
# Ranking / Arrangement Quiz
• #### A, P, R, X, S and Z are sitting in a row. S and Z are in the centre, and A and P are at the ends. R is sitting on the left of A. Then who is on the right of P?
1. A
2. X
3. S
4. Z
Explanation:
Note:
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# Is the sum of six consecutive integers even?
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Hey, guys, look here:
Is the sum of six consecutive integers even?
1. The first integer is odd
2. The average of six integers is odd
1. As far as I did some calculations, the sum of 6 consectituve integers is ALWAYS odd, no matter the first integer is even or odd.
2. According to the number properties theory the SUM must be even here (even number divided by another even (6) may give either even or odd (which is actually odd here as mentioned));
What I see is some inconsistency here. According to DS logic the answer is D, but according to common sense this question is irrelevant, since the stimulus and both pieces of information contradict each other.
Can anybody help?
Thank you
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### Show Tags
17 Sep 2003, 03:17
vaka wrote:
Hey, guys, look here:
Is the sum of six consecutive integers even?
1. The first integer is odd
2. The average of six integers is odd
1. As far as I did some calculations, the sum of 6 consectituve integers is ALWAYS odd, no matter the first integer is even or odd.
2. According to the number properties theory the SUM must be even here (even number divided by another even (6) may give either even or odd (which is actually odd here as mentioned));
What I see is some inconsistency here. According to DS logic the answer is D, but according to common sense this question is irrelevant, since the stimulus and both pieces of information contradict each other.
Can anybody help?
Thank you
This question only makes sense with an odd number of integers, say 5, not six, hence there must be a typo in your source. As you noticed, the sum of six consecutive integers is ALWAYS odd. Hence, there is no need for conditions. Were the number of integers odd, it would matter whether the starting number was odd or even (or equivalently, whether the middle number (which is the average) is odd or even.)
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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17 Sep 2003, 04:46
Thank you, AkamaiBrah,
I took this question from this site (Qantitive Section Lessons) so I thought there could be no mistake. Wrong. A person, who uploaded the lesson must have made a typo.
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17 Sep 2003, 04:46
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#### Strassen Matrix Multiplication
T
p
O
o
A
a
and 8 more contributors
```from __future__ import annotations
import math
def default_matrix_multiplication(a: list, b: list) -> list:
"""
Multiplication only for 2x2 matrices
"""
if len(a) != 2 or len(a[0]) != 2 or len(b) != 2 or len(b[0]) != 2:
raise Exception("Matrices are not 2x2")
new_matrix = [
[a[0][0] * b[0][0] + a[0][1] * b[1][0], a[0][0] * b[0][1] + a[0][1] * b[1][1]],
[a[1][0] * b[0][0] + a[1][1] * b[1][0], a[1][0] * b[0][1] + a[1][1] * b[1][1]],
]
return new_matrix
def matrix_addition(matrix_a: list, matrix_b: list):
return [
[matrix_a[row][col] + matrix_b[row][col] for col in range(len(matrix_a[row]))]
for row in range(len(matrix_a))
]
def matrix_subtraction(matrix_a: list, matrix_b: list):
return [
[matrix_a[row][col] - matrix_b[row][col] for col in range(len(matrix_a[row]))]
for row in range(len(matrix_a))
]
def split_matrix(a: list) -> tuple[list, list, list, list]:
"""
Given an even length matrix, returns the top_left, top_right, bot_left, bot_right
quadrant.
>>> split_matrix([[4,3,2,4],[2,3,1,1],[6,5,4,3],[8,4,1,6]])
([[4, 3], [2, 3]], [[2, 4], [1, 1]], [[6, 5], [8, 4]], [[4, 3], [1, 6]])
>>> split_matrix([
... [4,3,2,4,4,3,2,4],[2,3,1,1,2,3,1,1],[6,5,4,3,6,5,4,3],[8,4,1,6,8,4,1,6],
... [4,3,2,4,4,3,2,4],[2,3,1,1,2,3,1,1],[6,5,4,3,6,5,4,3],[8,4,1,6,8,4,1,6]
... ]) # doctest: +NORMALIZE_WHITESPACE
([[4, 3, 2, 4], [2, 3, 1, 1], [6, 5, 4, 3], [8, 4, 1, 6]], [[4, 3, 2, 4],
[2, 3, 1, 1], [6, 5, 4, 3], [8, 4, 1, 6]], [[4, 3, 2, 4], [2, 3, 1, 1],
[6, 5, 4, 3], [8, 4, 1, 6]], [[4, 3, 2, 4], [2, 3, 1, 1], [6, 5, 4, 3],
[8, 4, 1, 6]])
"""
if len(a) % 2 != 0 or len(a[0]) % 2 != 0:
raise Exception("Odd matrices are not supported!")
matrix_length = len(a)
mid = matrix_length // 2
top_right = [[a[i][j] for j in range(mid, matrix_length)] for i in range(mid)]
bot_right = [
[a[i][j] for j in range(mid, matrix_length)] for i in range(mid, matrix_length)
]
top_left = [[a[i][j] for j in range(mid)] for i in range(mid)]
bot_left = [[a[i][j] for j in range(mid)] for i in range(mid, matrix_length)]
return top_left, top_right, bot_left, bot_right
def matrix_dimensions(matrix: list) -> tuple[int, int]:
return len(matrix), len(matrix[0])
def print_matrix(matrix: list) -> None:
print("\n".join(str(line) for line in matrix))
def actual_strassen(matrix_a: list, matrix_b: list) -> list:
"""
Recursive function to calculate the product of two matrices, using the Strassen
Algorithm. It only supports even length matrices.
"""
if matrix_dimensions(matrix_a) == (2, 2):
return default_matrix_multiplication(matrix_a, matrix_b)
a, b, c, d = split_matrix(matrix_a)
e, f, g, h = split_matrix(matrix_b)
t1 = actual_strassen(a, matrix_subtraction(f, h))
t2 = actual_strassen(matrix_addition(a, b), h)
t3 = actual_strassen(matrix_addition(c, d), e)
t4 = actual_strassen(d, matrix_subtraction(g, e))
t5 = actual_strassen(matrix_addition(a, d), matrix_addition(e, h))
t6 = actual_strassen(matrix_subtraction(b, d), matrix_addition(g, h))
t7 = actual_strassen(matrix_subtraction(a, c), matrix_addition(e, f))
top_left = matrix_addition(matrix_subtraction(matrix_addition(t5, t4), t2), t6)
top_right = matrix_addition(t1, t2)
bot_left = matrix_addition(t3, t4)
bot_right = matrix_subtraction(matrix_subtraction(matrix_addition(t1, t5), t3), t7)
# construct the new matrix from our 4 quadrants
new_matrix = []
for i in range(len(top_right)):
new_matrix.append(top_left[i] + top_right[i])
for i in range(len(bot_right)):
new_matrix.append(bot_left[i] + bot_right[i])
return new_matrix
def strassen(matrix1: list, matrix2: list) -> list:
"""
>>> strassen([[2,1,3],[3,4,6],[1,4,2],[7,6,7]], [[4,2,3,4],[2,1,1,1],[8,6,4,2]])
[[34, 23, 19, 15], [68, 46, 37, 28], [28, 18, 15, 12], [96, 62, 55, 48]]
>>> strassen([[3,7,5,6,9],[1,5,3,7,8],[1,4,4,5,7]], [[2,4],[5,2],[1,7],[5,5],[7,8]])
[[139, 163], [121, 134], [100, 121]]
"""
if matrix_dimensions(matrix1)[1] != matrix_dimensions(matrix2)[0]:
msg = (
"Unable to multiply these matrices, please check the dimensions.\n"
f"Matrix A: {matrix1}\n"
f"Matrix B: {matrix2}"
)
raise Exception(msg)
dimension1 = matrix_dimensions(matrix1)
dimension2 = matrix_dimensions(matrix2)
if dimension1[0] == dimension1[1] and dimension2[0] == dimension2[1]:
return [matrix1, matrix2]
maximum = max(*dimension1, *dimension2)
maxim = int(math.pow(2, math.ceil(math.log2(maximum))))
new_matrix1 = matrix1
new_matrix2 = matrix2
# Adding zeros to the matrices so that the arrays dimensions are the same and also
# power of 2
for i in range(maxim):
if i < dimension1[0]:
for _ in range(dimension1[1], maxim):
new_matrix1[i].append(0)
else:
new_matrix1.append([0] * maxim)
if i < dimension2[0]:
for _ in range(dimension2[1], maxim):
new_matrix2[i].append(0)
else:
new_matrix2.append([0] * maxim)
final_matrix = actual_strassen(new_matrix1, new_matrix2)
# Removing the additional zeros
for i in range(maxim):
if i < dimension1[0]:
for _ in range(dimension2[1], maxim):
final_matrix[i].pop()
else:
final_matrix.pop()
return final_matrix
if __name__ == "__main__":
matrix1 = [
[2, 3, 4, 5],
[6, 4, 3, 1],
[2, 3, 6, 7],
[3, 1, 2, 4],
[2, 3, 4, 5],
[6, 4, 3, 1],
[2, 3, 6, 7],
[3, 1, 2, 4],
[2, 3, 4, 5],
[6, 2, 3, 1],
]
matrix2 = [[0, 2, 1, 1], [16, 2, 3, 3], [2, 2, 7, 7], [13, 11, 22, 4]]
print(strassen(matrix1, matrix2))
```
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Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao.
Presentation on theme: "Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao."— Presentation transcript:
Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao
Overview Introduction of the problem. A brief discussion of one-dimensional sequence. Two approaches of finding the maximum- density path in a tree.
Input & Output Input ： A weighted tree with n edges and a lower bound L. The lower bound is necessary. Output ： A maximum-density path with length at least L.
The Density of a Path The density of a path is defined as follow ： Given a path with k edges E={e 1, e 2, …, e k }, and w(e) is an edge weight function. The density of such path is defined as
Lower Bound L = 4 3 2 3 3 5 6 8 7 5 6 3 1 5 1 4 2 1 8 4 2 7 4 8 Max-Density = ( 7+4+6+5+8 ) / 5 = 6
The Efficiency We propose two efficient algorithms reach O(nL) time. One of them is further modified to solve some special cases such as full m-ary trees in linear time.
One-Dimensional Sequence The maximum-average segment problem arises naturally in several areas of sequence analysis. For example, given a DNA sequence, which segment of the sequence of length at least L has the highest GC ratio. The efficiency of the naïve algorithm of one- dimensional sequence is O(n 2 ).
Lower Bound L There exists an optimal segment of length at most 2L-1. It can be proved by a counter argument. The naïve algorithm is O(nL) now. AB
The Relation between Two Overlapped Sequences Let P[x] denotes the maximum-density of those segments that start from x. For a given j { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4183782/slides/slide_9.jpg", "name": "The Relation between Two Overlapped Sequences Let P[x] denotes the maximum-density of those segments that start from x.", "description": " For a given j
Right-Skew Segment A sequence S={s 1, s 2, …, s k } is right-skew if and only if the average of any prefix {s 1, s 2, …, s i ) is never larger than the average of the remaining subsequence {s i+1, …, s k ). According to the tricks mentioned above, Goldwasser et al. proposed a linear time algorithm for one-dimensional sequence.
Right-skew Decomposition 5397738918635 4 5 6 7 Decreasing right-skew decomposition is unique
Finding Maximum-Density Path In a Tree The path in a tree is similar to the segment of an one-dimensional sequence. That is, there exists a maximum-density path of length at most 2L-1. We propose two approaches reach the time complexity of O(nL).
Downward & Upward Paths We classify the paths that start from node K into two types called downward and upward paths. One is to stretch downward to its children only, called downward paths. And the other is to include at least its parent, called upward paths.
An upward path of node K K A downward path of node K A general tree
Notation Let denotes the maximum-density downward path of node K of length i. Let denotes the second-best one. If there is a tie, choose an arbitrary path. Similarly, represent the maximum upward path of node K of length i.
3 5 3 4 3 6 9 2 7 6 3 5 8 4 2 1 8 4 2 7 4 8 8 A
Contributor The node which determines its parent’s maximum-density downward path is called contributor. Each maximum-density downward path of a given length has its own contributor. The upward path has no contributor because each node has at most one parent.
3 5 3 4 3 6 9 2 7 6 3 5 8 4 2 1 8 4 2 7 4 8 8 A B C D node D node C node D node B
Downward & Upward Table The downward table of node K is composed of and, where 1 ≦ i ≦ 2L-1. Note that there exists an optimal path with length at most 2L-1 Similarly, the upward table of node K is composed of, where 1 ≦ i ≦ 2L-1.
Constructing Downward Table For a given internal node K with m children {K 1, K 2, …, K m }. Let e j denotes the edge (K, K j ). We can construct the downward table of node K by bottom-up dynamic programming. All contributors should be recorded.
Constructing Upward Table Suppose K’ is the parent of node K, and e’ denote the edge (K’, K). Upward Table of node K can be constructed by top-down dynamic programming.
Time Complexity Deciding a given length of downward and upward path of any node is O(1). Thus, constructing downward and upward table of a node takes O(L). There are totally n nodes in a tree, so it take O(nL) time to complete all downward and upward tables.
Approach I ： Finding the Path from its End Node Once the downward and upward table of each node are constructed, then we can determine the maximum-density path of a given node with length from L to 2L-1 in O(L). Therefore, it takes O(nL) time to check all nodes since there are n nodes in such tree.
Approach II ： Finding the Path from its LCA Node LCA stands for least common ancestor. We are now combining two downward paths together.
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# Molecular Formula
Molecular Formula - represent the actual numbers of atoms of the different elements in one molecule of a compound. An example is Ca(NO3)2
Ca = Ca (only one Ca) , (1*2)N = 2N (two nitrogens) , (3*2)O = 6O (six oxygens)
It is not written CaN2O6 but Ca(NO3)2 - why is that?
That is because there is more information in Ca(NO3)2 than CaN2O6. By the molecular formula we can see that Ca(NO3)2 contain two NO3 - groups. Other groups used in a typical moelcular formula: NO2 (nitrite) , SO4 (sulfate) , SO3 (sulfite) , OH (hydroxyl)...
Determination of the molecular formula by the empirical formula and mass percent composition
Example: A combustion analysis gives the following mass %: H= 9.15% C = 54.53% O = 36.32% . Determine the molecular formula knowing that : molecular mass = 132.16 , empirical formula = C2H4O
Solution
Assume a 100 gram sample which will convert the given percentages to gram amounts
(9.15 gram H)/(1 gram/mole) = 9.15 moles
(54.53 gram C)/(12 gram/mole) = 4.54 moles
(36.32 gram O)/(16 gram/mole) = 2.27 moles
Determination of the simplest moles ratio:
Divide each of the moles figures by the lowest of the three (in this case 2.27).
2.27 moles O /2.27 =1 , 9.15 moles H/2.27 = 4 , 4.54 moles C/2.27 = 2
2*12.0 + 4*1.0 + 1*16.0 = 44
Calculating the common factor:
The common factor defines the ratio (molecular formula)/(empirical formula):
132.16/ 44 = 3
Determination of the molecular formula :
The solution: Multiply C2H4O by the common factor --> C(2*3)H(4*3)O(1*3) = C6H12O3
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lab_solution6
# lab_solution6 - ECE 101 Linear Systems Fundamentals Fall...
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ECE 101 – Linear Systems Fundamentals, Fall 2009 Lab # 6 Solutions December 1, 2009 (send questions/comments to [email protected]) NOTE: These solutions demonstrate the decoding of the telephone numbers: 491-5877 (Toscanini’s Ice Cream Shop in Cambridge, MA) and 253-1000 (MIT, also in Cambridge, MA). These places were frequented, no doubt, by the authors of the “Computer Explorations” book. The phone numbers that you decoded were local ones: 534-2230 (UCSD) and 550-0406 (Regents Pizzeria). These are places frequented by your Professor! In this lab assignment, you’ll use the DTFT to analyze touch-tone telephone sounds, play your own phone number with MATLAB, and decode a couple phone numbers. When you push a key on your phone, the sound you hear is the sum of two sinusoids, y[n] = sin( ω C n) + sin( ω R n) where ω C and ω R are pair of frequencies given by the table below: The corresponding frequency-domain spectrum, then, should be sinusoidal peaks (delta-functions) centered at plus/minus the two frequency values. (a) To listen to, say, the sound of the digit 2, we can use the commands: n = 0:999; d2 = sin(0.5346*n) + sin(1.0247*n); sound(d2,8192); The digit 2 has the frequencies ω R = 0.5346 and ω C = 1.0247 associated with it, as seen from the table. A plot of the tone looks like this:
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(b) In MATLAB, we can use the Fast Fourier Transform, fft( ) , to approximate the DTFT of the touch- tone signal. We’ll examine the digits 2 and 9. According to the table, the digit 2 should have sinusoidal peaks at ω R = 0.5346 and ω C = 1.0247, and the digit 9 should have peaks at ω R = 0.6535 and ω C = 1.1328. Below are plots of the magnitude responses of the DTFTs of d
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# What is the value of x in the equation below?
Students were asked over to answer a question at academics and to indicate what is most important for them to succeed. The one that response stood out from the rest was practice. Successful persons definitely not born successful; they become successful from hard work and commitment. This is how you can complete your goals. Below are one of the answer and question example that you could definitely utilise to practice and elevate your practical knowledge and also give you insights that might just guide you to preserve your study in school.
## Question:
What is the value of x in the equation below?1+2e^x+1=9
a). x=log4-1
b) x=log4
c). x=Ln4-1
d). x=ln4
Step-by-step explanation:
1+2e^x+1=9
2e^x+1=9-1
e^x+1= 8/2
e^x+1 = 4
Here we can applicate the Ln:
ln e^x+1 = ln 4
Applicating the log property for exponent:
(x+1).ln e = ln 4
Ln e = 1, so:
x+1 = ln 4
x = ln 4-1
From the answer and question examples above, hopefully, they can assist the student resolve the question they had been looking for and take note of every thing stated in the answer above. You could definitely then have a discussion with your classmate and continue the school learning by studying the topic altogether.
READ MORE which sentence uses the passive voice for the main verb? if we use our cell phones during class, the teacher confiscates them. when they can, students slide down the banister in the science wing. when we can pick our reading, mythology is most often chosen in my class. if they choose to participate, students perform their poems at the poetry slam.
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## Content
### Factoring
Factoring is the reverse process to expanding. Writing the number 3128 as $$8\times 17 \times 23$$ gives us useful information about the number. For example, it tells us that 3128 is divisible by 8 and 17 and 23. It also allows us to calculate $$3128 \div 23$$ without using a calculator. Similarly, in algebra, factoring an algebraic expression is a very useful and important skill. Indeed, it is very unwise in algebra to expand a complicated collection of brackets unless you have no other option. Factoring is often much the better thing to do. Factoring also enables us to solve certain types of equations and will reappear when we look at quadratic equations and again in the TIMES module Polynomials (Years 9–10) .
#### Basic factoring
Students should always be on the lookout for simple common factors.
For example, the terms in the expression $$3x^2+12x$$ have a common factor of $$3x$$ which can be factored out. Thus, $$3x^2+12x=3x(x+4)$$.
Sometimes after doing this we can spot a further common factor. For example,
\begin{align*} 12xy - 15x - 16y + 20 &= 3x(4y - 5) - 4(4y - 5)\\ &= (3x - 4)(4y - 5). \end{align*}
##### Exercise 4
Factorise
$4x^3-12x^2-x+3.$
Hence state the values of $$x$$ which make this expression equal to 0.
#### Quadratics
A quadratic expression has the form $$ax^2+bx+c$$, where $$a,b,c$$ are given numbers with $$a \ne 0$$. In some cases a quadratic expression can be factored. For example, starting with the expression $$x^2+5x+6$$, we can, rather cleverly, write this as
\begin{align*} x^2+2x+3x+6 &= x(x+2)+3(x+2)\\ &= (x+3)(x+2). \end{align*}
Now how did we know to split the middle term in this way? The answer lies in looking at what happens when we expand:
$(x+a)(x+b) = x^2+(a+b)x+ab.$
Clearly the coefficient of $$x$$ is the sum of $$a$$ and $$b$$ and the constant term is their product. This is why we split the $$5x$$ term into $$2x+3x$$. In practice, when factoring $$x^2+5x+6$$, we seek two integers whose sum is 5 and whose product is 6. Clearly the numbers are 2 and 3, and so we obtain the desired factorisation.
#### Example
Factor
$x^2+23x-50.$
#### Solution
We seek two numbers whose sum is 23 and whose product is $$-50$$. The numbers 25 and $$-2$$ will do. Hence,
$x^2+23x-50 = (x+25)(x-2).$
Provided we can find the appropriate numbers, this method works well for monic quadratics, that is, for quadratics whose leading coefficient is 1.
In the case of a non-monic quadratic, the following algorithm is employed. It is best seen by example.
To factor $$3x^2+14x-5$$, we first multiply the 3 and the $$-5$$ to obtain $$-15$$. Now we find two numbers whose sum is 14 and whose product is $$-15$$. Clearly, these are 15 and $$-1$$. We use these numbers to split the middle term and write
$3x^2+14x-5= 3x^2+15x-x-5.$
Now we factor in pairs, as above. Thus
\begin{align*} 3x^2+14x-5 &= 3x^2+15x-x-5\\ &= 3x(x+5)-1(x+5)\\ &= (3x-1)(x+5). \end{align*}
Of course there are quadratic expressions such as $$x^2+x+1$$ which cannot be factored in this way.
#### Special factorisations
The following three special factorisations are exactly the same as the three special expansions discussed in Expanding (Special expansions) in this module, but used in reverse:
\begin{align*} x^2+2xy+y^2 &= (x+y)^2\\ x^2-2xy+y^2 &= (x-y)^2\\ x^2-y^2 &= (x+y)(x-y). \end{align*}
Of these the last is probably the most important, and students need to spot this factorisation whenever it occurs.
#### Example
Factor
$9x^2-25y^2.$
#### Solution
$$9x^2-25y^2=(3x+5y)(3x-5y)$$.
##### Exercise 5
Factorise fully
$(a^2+ab)^2-(ab+b^2)^2.$
Next page - Content - Algebraic fractions
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Home Projects Experiments Circuits Theory BLOG PIC Tutorials Time for Science
Logic Function Instructions
The next instructions set is the Logic Function instructions. Within this set, there are instructions to perform the typical logic functions such as OR, AND, XOR etc. Here is the list of them:
Instruction Description ANDLW k Logic AND between the contents of W and the literal value k. The results is stored back to W. ANDWF f , d Logic AND between the content of W register with the content of a file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register 'f'. IORLW k Logic OR between the contents of W and the literal value k. The results is stored back to W. IORWF f , d Logic OR between the content of W register from the content of a file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register 'f'. XORLW k Logic EXCLUSIVE OR between the contents of W and the literal value k. The results is stored back to W. XORWF f , d Logic EXCLUSIVE OR between the content of W register from the content of a file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register 'f'.
Now, let's see the instructions one by one:
ANDLW k
With this instructions you perform an AND logic function between the W register and the literal value k. The result is stored back to the W register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• W = W +.AND. f
Where 0 <= k <= 255
Example
``` movlw b'10000100' ;The W register has now the binary value '10000100'
andlw b'00001111' ;The binary value '00001111' AND with the W register.
;Now the W register has the binary value '00000100'
```
ANDWF f , d
With this instructions you perform an AND logic function between the W register and the file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• Destination (d) = W .AND. f
Where 0 <= f <= 127
Example
```;This example requires that you have declare a file register for example
;TempRegister equ 0x20
movlw b'10000100' ;The W register has the value '10000100'
movwf TempRegister ;The TempRegister has now the value '10000100'
movlw d'00001111' ;The W register has the value '00001111'
andwf TempRegister,1 ;The W register AND with the contents of TempRegister
;and the result ('00000100') is stored to TempRegister
```
IORLW k
With this instructions you perform an OR logic function between the W register and the literal value k. The result is stored back to the W register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• W = W +.OR. f
Where 0 <= k <= 255
Example
``` movlw b'10000100' ;The W register has now the binary value '10000100'
iorlw d'00001111' ;The binary value '00000011' OR with the W register.
;Now the W register has the binary value '10001111'
```
IORWF f , d
With this instructions you perform an OR logic function between the W register and the file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• Destination (d) = W .OR. f
Where 0 <= f <= 127
Example
```;This example requires that you have declare a file register for example
;TempRegister equ 0x20
movlw b'10000100' ;The W register has the value '10000100'
movwf TempRegister ;The TempRegister has now the value '10000100'
movlw d'00001111' ;The W register has the value '00001111'
iorwf TempRegister,1 ;The W register OR with the contents of TempRegister
;and the result ('10001111') is stored to TempRegister
```
XORLW k
With this instructions you perform an EXCLUSIVE OR logic function between the W register and the literal value k. The result is stored back to the W register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• W = W +.XOR. f
Where 0 <= k <= 255
Example
``` movlw b'10000100' ;The W register has now the binary value '10000100'
xorlw d'00001111' ;The binary value '00000011' XOR with the W register.
;Now the W register has the binary value '10001011'
```
XORWF f , d
With this instructions you perform an EXCLUSIVE OR logic function between the W register and the file register 'f'. If d is 0 the result is stored in the W register, if d is 1 the result is stored back to the file register. The status affected from this function are:
• Z - Zero
The logic operation performed is:
• Destination (d) = W .XOR. f
Where 0 <= f <= 127
Example
```;This example requires that you have declare a file register for example
;TempRegister equ 0x20
movlw b'10000100' ;The W register has the value '10000100'
movwf TempRegister ;The TempRegister has now the value '10000100'
movlw d'00001111' ;The W register has the value '00001111'
xorwf TempRegister,1 ;The W register OR with the contents of TempRegister
;and the result ('10001011') is stored to TempRegister
```
The Logic Functions Table
Here is a final tip for you. The following table demonstrates the result for the above functions between two bits.
Bit 1 Bit 2 AND OR XOR 0 0 0 0 0 0 1 0 1 1 1 0 0 1 1 1 1 1 1 0
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At 23 February 2011, 20:06:32 user Kammenos wrote: [reply @ Kammenos]capn than you for the correction. I pressed 'd' instead of 'b'. 'd' stands for decimal but i declare a binary...At 23 February 2011, 17:04:03 user capn wrote: [reply @ capn]I think you have a problem in your ANDlw code example. movlw b'10000100' andlw d'00001111' ;this should be b'00001111' should it not? because b'10000100' AND b'00001111' will give you b'00000100'.
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Mat137 study sheet for T1
# Mat137 study sheet for T1 - MATH 137(U of T Study Sheet...
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Unformatted text preview: MATH 137 (U of T) Study Sheet Properties of Real Numbers Properties of real numbers: associativity, commutativity, existence of identity, existence of inverses Subsets of real numbers: natural numbers (e.g. 1,2,3,…), integers (e.g. -2,-1,0,1,2,…), rational numbers (i.e. xx= p q for p, q integers, q ≠ 0 ). Strong (or complete) induction: assume that the statement holds for n < k , and then prove it true for n=k. EXAMPLE: Prove that 1 + 2 + ... + n = n(n + 1) (i.e. the sum of 2 the first n positive integers is equal to n(n + 1) .) 2 SOLUTION: Properties of Functions Domain is the set of values of x for which f(x) is defined Range is the set of all possible values f(x) Even functions are functions that have the property f − x = f x for all x () () Odd functions are functions that have the property f (− x ) = − f ( x ) for all x Invertible function: a function f is invertible if there exists a function g such that y = f (x ) if and only if g ( y ) = x Increasing function: a function is increasing on an interval I if for all x, f ( x1 ) < f ( x 2 ) whenever x1 < x 2 on I Decreasing function: a function is decreasing on an interval I if for all x, f ( x1 ) > f ( x 2 ) whenever x1 < x 2 on I EXAMPLE: Suppose f(x) is even and x2 f (x ) − 5 f x 3 g (x ) = () (− x ) 3 f (− x ) − 5 f ([− x] ) is even. Therefore, since k ⋅ (k + 1) 1 + 2 + ... + k = 2 1 + 2 + ... + k = b such that for all k ⋅ (k + 1) , we can re-write it as 2 2 Collecting the (k + 1) terms, we find: ) since k (k + 1 ) + (k + 1 ) 2 ⎛k ⎞ ( k + 1 ) ⋅ (k + 2 ) = (k + 1 ) ⋅ ⎜ + 1 ⎟ = 2 ⎝2 ⎠ (( k + 1) + 1) = (k + 1 ) 2 (1 + 2 + ... + k ) + k + 1 = x∈ A, x ≤ b. Least Upper Bound: b is the least upper bound of A iff b is an upper bound of A and if for all other upper bounds a of A, b ≤ a . Greatest Lower Bound: b is the greatest lower bound of A iff b is a lower bound of A and if for all lower bounds a of A, a ≤ b . Completeness of real numbers: If A is a non empty set of real numbers and A has an upper bound then A has a least upper bound If A is a non empty set of real numbers and A has a lower bound then A has a greatest lower bound EXAMPLE: Solve the inequality x2 + 7 x − 8 < 0 . SOLUTION: First, lets factor the expression x 2 + 7x − 8 as x + 7 x − 8 = ( x + 8)( x − 1) . (x + 8)(x − 1) < 0 . For x < − 8 , both x + 8 and x − 1 are negative and, therefore F ( x ) = ( x + 8)( x − 1) is positive. follows: We now need to show the statement is true for n = k + 1. That means, we are concerned with the sum of the first k + 1 integers, i.e. with the sum 1 + 2 + ... + k + k + 1 . But we can write this as follows: 1 + 2 + ... + k + k + 1 = (1 + 2 + ... + k ) + k + 1 . (1 + 2 + ... + k ) + k + 1 = k (k + 1) + (k + 1) 2 x2 x2 = = 3 f (− x ) − 5 f − x f ( x ) − 5 f (x 3 ) f (x ) Thus, the left side is equal to the right side and the expression is valid for the case n = 1. Induction Step: Our induction hypothesis is that the expression holds for the case n = k, and thus the following expression holds: a≤x 2 This gives us: When we pass through x = −8, x + 8 changes sign and, therefore, so does F(x). But when we later pass through x = 1, x − 1 changes sign and F(x) changes back to being positive. Thus, F(x) is negative for −8 < x < 1. follows: SOLUTION: ( 1⋅ (1 + 1) 2 = = 1. 2 2 But since our induction hypothesis is . Determine whether g(x) is even, odd, or neither. g (− x ) = We’ll use induction to prove the above equality. Base case: When n = 1, the left side of the equality is equal to 1. When n = 1, the right side of the equality is equal to has a lower bound if there is a number a such that for all x ∈ A , . Bounded: A is bounded iff A has either an upper bound and a lower bound; iff there exists a number Logarithmic Functions Properties of logarithms: log b (mn) = log b m + log b n , ⎛ m⎞ log b ⎜ ⎟ = log b m − log b n , ⎝ n⎠ 1 log b mr = r log b m , log b = − log b m , m log b 1 = 0 , log b b = 1 , log b b r = r , More free study sheet and practice tests at: g ( x ) = g (− x ) , g ( x ) is Therefore, we know: even. Mathematical Induction INDUCTION STEP BY STEP Step 1—Base case. Show that the statement holds in the simplest case (normally for n = 1 ). n = 0 and/or Step 2—Induction Hypothesis. Assume the statement holds for an arbitrary k. Step 3. Prove that it holds for above induction hypothesis. k + 1 given the (1 + 2 + ... + k ) + k + 1 = (k + 1) ((k + 1) + 1) This is 2 what we need to show. Therefore, the statement is true for all positive integers n. b logb m = m , 10 = x , e ln x = x , log a m . log b m = log a b log x EXAMPLE: Inequalities and Absolute Values Absolute Value Rules: a−b ≥ a − b a+b ≤ a + b , a ⋅b = a ⋅ b Solve for x: ln(x2 + x) = 1 + ln x SOLUTION: ln(x2 + x) = 1 + ln x ln(x2 + x) − ln x = 1 ⎛ x2 + x ⎞ ⎟ =1 ⎜x⎟ ⎠ ⎝ , Upper/lower bound: A has an upper bound if there ln ⎜ is a number a such that for all ln(x + 1) = 1⇒ x + 1 = e ⇒ x = e − 1. x∈ A, x ≤ a . A More free Study Sheets and Practice Tests at: www.prep101.com www.prep101.com More free study sheets and practice tests at Limits and Continuity L’HOPITAL’S RULE: Formal Definition: The function f(x) has limit L, as x approaches a, denoted lim f x = L if given any () x→a ε > 0 , there exists a δ > 0 such that |f(x) - L| < ε for all x satisfying 0 < |x – a|< δ. lim[ f ( x ) ± g ( x )] = lim f ( x ) ± lim g (x ) , x→a x→a x→a lim x→a [ n x→a x→a ε >0. Since ε Then x − a < δ 1 , then f ( x ) − L < ε Also, since such . Let x−a < δ 2 , then g (x ) − K < δ = minimum (δ 1 , δ 2 ) . Hence, if g (x ) − K < Therefore, if then 5x 3 − 4 x + 2 lim = lim x→∞ 3 + x 2 − 6 x 3 x→∞ 4 2⎞ ⎛ x3 ⎜ 5 − 2 + 3 ⎟ x x⎠ ⎝ 1 3⎞ ⎛ x3⎜− 6 + + 3 ⎟ x x⎠ ⎝ 5−0+0 5 =− . x→∞ − 6 + 0 + 0 6 Continuity: a function is continuous at point a if and only if the left hand and right hand limits exist and left hand limit = right hand limit = f(a) (i.e. if lim f ( x) = f (a) ). = lim . ε 2 =ε (by the triangle x →a as required. and lim g ( x ) = K x→a lim[ f (x ) + g ( x )] = L + K . x→a By the I.V.T, there is a ⎛ π⎞ ⎜ 0, ⎟ ⎝ 2⎠ such that f(c) = 1, SQUEEZE THEOREM: If f (x ) ≤ g (x ) ≤ h( x ) near a and lim x ) = lim h( = L More free study sheet and practicef (testsx)at: 2 lim f ( x ) = L x →a π ). 2 i.e. such that c is the solution to the expression x + sin x = 1. , then ≤ f (x ) − L + g (x ) − K 2 Thus, f(0) < 1 < f( π )= π + sin π = π +1. 22 22 5x − 4 x + 2 3 + x 2 − 6 x3 . SOLUTION: x →∞ 2 + f(0) = 0 + sin 0 = 0 and f( ⎡ π⎤ . ⎢0, 2 ⎥ ⎣ ⎦ number c in the interval lim 2 [ f (x ) + g (x )] − (L + K ) = [ f (x ) − L] + [g (x ) − K ] ε Let f(x) = x + sin x. Then f is continuous on 3 . So, inequality) < f (c ) = k Determine the following limit: f (x ) − L < ε and Use the Intermediate theorem to show that the equation x + sin x = 1 has a solution in the interval EXAMPLE: x − a < δ 1 and ε and , and therefore, x−a <δ x − a < δ2 ε f (a ) and f (b ) then there exists a c such that ⎡ π⎤ . ⎢0, 2 ⎥ ⎦ ⎣ SOLUTION: First try substitution, factoring If you’re taking the limit as x → ∞ of a quotient, and substitution yields an invalid answer, then try first to divide both the numerator and the denominator of the limit expression by its highest power of x If the limit expression contains absolute values, then try breaking the limit up into two one-sided limits. If the limit expression is a quotient, and if factoring doesn’t work, then try l’Hopital’s Rule x →a such that if a<c<b TECHNIQUES FOR FINDING LIMITS: 2 lim g ( x ) = K , there exists δ 2 > 0 , ⇒ EXAMPLE: x →a that if ) 3 x + 6 x − 10 14 = =7 x→2 3x 2 − 6 x + 2 2 > 0. 2 lim f ( x ) = L , there exists δ 1 > 0 x →1 is continuous on [a, b ] and if k is between If = lim x→a Let f (x ) ) 2 lim[ f ( x ) + g ( x )] = L + K . = lim+ 2 x = 2 Therefore: 1 + eK = 2 K = 0. Thus, if f ( x) is continuous at x = 1, then K = 0. ( ε − δ proof of the addition property of If lim f ( x ) = L and lim g ( x ) = K , then x →a x →a 2x 2 − 2x x−1 x−1 ( n ) x →1+ INTERMEDIATE VALUE THEOREM: d3 x + 3 x 2 − 10 x x 3 + 3 x 2 − 10 x lim dx = lim x → 2 x 3 − 3x 2 + 2 x x→2 d x 3 − 3x 2 + 2 x dx Give an SOLUTION: x →1 (x − 1) ⋅ 2 x x →1 0 ( x →1 K = 1 + e K = f (1) = lim+ . Note first that this limit is in 0 form. Therefore, use EXAMPLE: limits: x 3 + 3 x 2 − 10 x L’Hopital’s theorem: We find lim[ f ( x )] = lim f ( x ) x →1− lim f (x ) = lim− x + e x →1 − x →1+ SOLUTION: x→a lim f (x ) = lim f ( x ) = f (1) We require: lim f (x ) = lim+ 3 2 Evaluate x → 2 x − 3x + 2 x lim[ f ( x ) ⋅ g ( x )] = lim f ( x ) ⋅ lim g ( x ) , ⎡ f ( x ) ⎤ lim f ( x ) , lim⎢ = x →a x →a g ( x ) ⎥ ⎣ ⎦ lim g ( x ) x →a f (x ) and g ( x ) are of types 0 or ∞ , and if g ′(a ) 0 ∞ is not 0, then lim f ( x ) = lim f ′( x ) x→a g ( x) x → a g ′( x ) EXAMPLE: PROPERTIES OF LIMITS: x→a SOLUTION: If the limit of the quotient of differentiable functions EXAMPLE: , If ⎧x + eK if x ≤ 1 is continuous at x = 1, ⎪ f (x ) = ⎨ 2 x 2 − 2 x if x > 1 ⎪ ⎩ x −1 then determine the value of K. x →a then x→a lim g ( x ) x →a exists and is equal to L. EXAMPLE: If, for all values of x in the interval (0, 5), 1 + x ≤ f (x) ≤ 3 + sin πx, find lim f x . x →2 () SOLUTION: lim 1 + x =1+2=3 x→2 lim 3 + sin πx x→2 = 3 + sin 2π = 3. ⇒ lim f ( x ) = 3 by the squeeze theorem. x →2 Helping students since 1999 ...
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# Difference between revisions of "1996 AHSME Problems"
## Problem 1
The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{r}&\ \texttt{6 4 1}\\ \texttt{8 5 2} &+\texttt{9 7 3}\\ \hline \texttt{2 4 5 6}\end{tabular}$ (Error compiling LaTeX. ! Extra alignment tab has been changed to \cr.)
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$
## Problem 2
Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$
## Problem 3
$\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$
## Problem 4
Six numbers from a list of nine integers are $7,8,3,5, 9$ and $5$. The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ (Error compiling LaTeX. ! Undefined control sequence.)
## Problem 5
Given that $0 < a < b < c < d$, which of the following is the largest?
$\text{(A)}\ \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\ \frac{b+c}{a+d} \qquad\text{(D)}\ \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b}$
## Problem 6
If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$, then $f(0)+f(-1)+f(-2)+f(-3) =$
$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$
## Problem 7
A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$, which of the following could be the age of the youngest child?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$
## Problem 8
If $3 = k\cdot 2^r$ and $15 = k\cdot 4^r$, then $r =$
$\text{(A)}\ -\log_{2}5\qquad\text{(B)}\ \log_{5}2\qquad\text{(C)}\ \log_{10}5\qquad\text{(D)}\ \log_{2}5\qquad\text{(E)}\ \frac{5}{2}$
## Problem 9
Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$? $[asy] real r=sqrt(2)/2; draw(origin--(8,0)--(8,-1)--(0,-1)--cycle); draw(origin--(8,0)--(8+r, r)--(r,r)--cycle); filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black); filldraw(origin--(8,0)--(8,6)--(0,6)--cycle, white, black); pair A=(6,0), B=(2,0), C=(2,4), D=(6,4), P=B+1*dir(-65); draw(A--P--B--C--D--cycle); dot(A^^B^^C^^D^^P); label("A", A, dir((4,2)--A)); label("B", B, dir((4,2)--B)); label("C", C, dir((4,2)--C)); label("D", D, dir((4,2)--D)); label("P", P, dir((4,2)--P));[/asy]$
$\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$
## Problem 10
How many line segments have both their endpoints located at the vertices of a given cube?
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$
## Problem 11
Given a circle of raidus $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.
$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$
## Problem 12
A function $f$ from the integers to the integers is defined as follows:
$$f(x) =\begin{cases}n+3 &\text{if n is odd}\\ n/2 &\text{if n is even}\end{cases}$$
Suppose $k$ is odd and $f(f(f(k))) = 27$. What is the sum of the digits of $k$?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$
## Problem 13
Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?
$\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1}$
## Problem 14
Let $E(n)$ denote the sum of the even digits of $n$. For example, $E(5681) = 6+8 = 14$. Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$
## Problem 15
Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See figure for $n=5, m=7$.] What is the ratio of the area of triangle to the area of triangle ?
$[asy] int i; for(i=0; i<8; i=i+1) { dot((i,0)^^(i,5)); } for(i=1; i<5; i=i+1) { dot((0,i)^^(7,i)); } draw(origin--(7,0)--(7,5)--(0,5)--cycle, linewidth(0.8)); pair P=(3.5, 2.5); draw((0,4)--P--(0,3)^^(2,0)--P--(3,0)); label("B", (2.3,0), NE); label("A", (0,3.7), SE); [/asy]$
$\text{(A)}\ 1\qquad\text{(B)}\ m/n\qquad\text{(C)}\ n/m\qquad\text{(D)}\ 2m/n\qquad\text{(E)}\ 2n/m$
## Problem 16
A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?
$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$
## Problem 17
In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$? $[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("A", A, dir((5, 3.5)--A)); label("B", B, dir((5, 3.5)--B)); label("C", C, dir((5, 3.5)--C)); label("D", D, dir((5, 3.5)--D)); label("E", E, dir((5, 3.5)--E)); label("F", F, dir((5, 3.5)--F)); label("2", (0,1), dir(0)); label("6", (7.5,0), N);[/asy]$ $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$
## Problem 18
A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?
$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3$
## Problem 19
The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?
$[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label("A", A, dir(origin--A)); label("B", B, dir(origin--B)); label("C", C, dir(origin--C)); label("D", D, dir(origin--D)); label("E", E, dir(origin--E)); label("F", F, dir(origin--F)); [/asy]$
$\text{(A)}\ \frac{1}{2}\qquad\text{(B)}\ \frac{\sqrt 3}{3}\qquad\text{(C)}\ \frac{2}{3}\qquad\text{(D)}\ \frac{3}{4}\qquad\text{(E)}\ \frac{\sqrt 3}{2}$
## Problem 20
In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$?
$\text{(A)}\ 10\sqrt 3\qquad\text{(B)}\ 10\sqrt 5\qquad\text{(C)}\ 10\sqrt 3+\frac{ 5\pi}{3}\qquad\text{(D)}\ 40\frac{\sqrt{3}}{3}\qquad\text{(E)}\ 10+5\pi$
## Problem 21
Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC, then$ \angle C+\angle D \$ is
$[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); [/asy]$
$\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$
## Problem 22
Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $AB$ intersects the chord $CD$?
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{1}{3}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$
## Problem 23
The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$
## Problem 24
The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$
## Problem 25
Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
## Problem 26
An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$
## Problem 27
Consider two solid spherical balls, one centered at $(0, 0,\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$
## Problem 28
$[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("A", A, NE); label("B", B, NW); label("C", C, S); label("D", D, E); label("4", (4,2,0), SW); label("4", (2,4,0), SE); label("3", (0, 4, 1.5), E); [/asy]$ Solution
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# Probabilistic Inventory Models
Pages: 5 (1270 words) Published: June 11, 2011
Probabilistic Inventory Models
1. CONTINUOUS REVIEW MODELS
1.1 "Probabilitized" EOQ Model
Some practitioners have sought to adapt the deterministic EOQ model to reflect the probabilistic nature of demand by using an approximation that superimposes a constant buffer stock on the inventory level throughout the entire planning horizon. The size of the buffer is determined such that the probability of running out of stock during lead time (the period between placing and receiving an order) does not exceed a prespecified value. Let
L = Lead time between placing and receiving an order
[pic] = Random variable representing demand during lead time [pic] = Average demand during lead time
[pic]= Standard deviation of demand during lead time
B = Buffer stock size
a = Maximum allowable probability of running out of stock during lead time The main assumption of the model is that the demand,[pic] ，during lead time L is normally distributed with mean [pic] and standard deviation [pic]-that is, N([pic], [pic]) FIGURE 14.1
Buffer stock imposed on the classical EOQ model
[pic]
Figure 14.1 depicts the relationship between the buffer stock, B, and the parameters of the deterministic EOQ model that include the lead time L, the average demand during lead time, [pic] , and the EOQ, y*. Note that L must equal the effective lead time. The probability statement used to determine B can be written as P {[pic] [pic] B +[pic]} [pic] [pic]
We can convert [pic] into a standard N (O, 1) random variable by using the following substitution [pic]
Thus, we have
[pic]
Figure 14.2 defines [pic] such that
[pic]
Hence, the buffer size must satisfy
[pic]
The demand during the lead time L usually is described by a probability density function per unit time (e.g., per day or week), from which the distribution of the demand during L can be determined. Given that the demand per unit time is normal with mean D and standard deviation [pic], the mean and standard deviation, [pic] and [pic], of demand during lead time, L, are computed as [pic]
The formula for [pic] requires L to be (rounded to) an integer value. [pic]
Example
EOQ = 1000 units. If the daily demand is normal with mean D = 100 lights and standard deviation [pic] = 10 lights-that is, N (100, 10)-determine the buffer size so that the probability of running out of stock is below a = .05. We know, the effective lead time is L = 2 days. Thus,
[pic]
Given [pic]= 1.645, the buffer size is computed as
B [pic] 14.14 * 1.645 [pic] 23 neon lights
Thus, the optimal inventory policy with buffer B calls for ordering 1000 units whenever the inventory level drops to 223 (= B + [pic]= 23 + 2 * 100) units.
1.2 Probabilistic EOQ Model
There is no reason to believe that the "probabilitized" EOQ model will produce an optimal inventory policy. The fact that pertinent information regarding the probabilistic nature of demand is initially ignored, only to be "revived" in a totally independent manner at a later stage of the calculations, is sufficient to refute optimality. To remedy the situation, a more accurate model is presented in which the probabilistic nature of the demand is included directly in the formulation of the model. Unlike the case in Section 1.1, the new model allows shortage of demand, as Figure 14.3 demonstrates. The policy calls for ordering the quantity y whenever the inventory drops to level R. As in the deterministic case, the reorder level R is a function of the lead time between placing and receiving an order. The optimal values of y and R are determined by minimizing the expected cost per unit time that includes the sum of the setup, holding, and shortage costs. [pic]
The model has three assumptions.
1. Unfilled demand during lead time is backlogged.
2. No more than one outstanding order is allowed.
3. The distribution of demand during lead time remains stationary (unchanged) with time.
To develop the total cost function per unit time, let
f(x) = pdf of demand, x,...
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30 November, 13:15
# The sum of two consecutive integers is - 27. If n represents one of the integers, what equation can you write to find the value of the two integers and what are the integers?
+2
1. 30 November, 14:44
0
-14,-13
Step-by-step explanation:
n+n+1=-27
2n = -27-1
2 n=-28
n=-14
n+1=-14+1=-13
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# Higher order partial derivatives
1. Jun 8, 2010
### magnifik
Consider the partial di erential equation, (y4-x2)uxx - 2xyuxy - y2uyy = 1. We will make the substitution x = s2 - t2 and y = s - t, to simplify
(a) Solve for s and t as functions of x and y
the farthest point i got to was
x = s^2 - t^2 = (s+t)(s-t) = y(s+t)
y = s - t
s+t = x/y
i don't know what to do after that.. i have the solution, but i have no idea how to get to it.
the solution is
s = x + y^2 / 2y
t = x - y^2 / 2x
2. Jun 8, 2010
### tiny-tim
Hi magnifik!
No, you're mssing the point …
that isn't the solution, it's just writing s and t in terms of x and y …
s+t = x/y, s-t = y, so 2s = x/y + y = (x + y2)/y, 2t = (x - y2)/y.
To get the solution, you need to find uss ust and utt (and I expect ust will be zero, which will make the solution fairly easy ).
3. Jun 8, 2010
### magnifik
errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!
4. Jun 8, 2010
### magnifik
but i don't understand how you got from s+t = x/y, s-t = y to 2s and 2t...how did you combine the previous two equations to get that?
5. Jun 8, 2010
### HallsofIvy
Excuse me, tiny-tim, but the only question asked was "(a) Solve for s and t as functions of x and y".
Presumably, there is a "b" but so far all we want to do is solve for s and t.
magnifik, you have $x= s^2- t^2$ and $y= s- t$.
From $y= s- t$ you can say that $s= y+ t$ and, putting that into $x=s^2- t^2$, $x= y^2+2yt+ t^2- t^2= y^2+ 2yt$. Solve that equation for t, then replace t by that expression in $s= y+ t$.
6. Jun 8, 2010
### magnifik
THANK YOU!! makes so much sense now :)
7. Jun 8, 2010
### tiny-tim
erm … I added, and subtracted!
Add s+t to s-t, you get 2s; subtract, you get 2t …
this is a common transformation, and you should be familiar with it.
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# Meridian altitude
Meridian Altitude is an astronomical navigation method of calculating an observer's latitude.
## Principle
This is the simplest calculation of astronomical navigation and is when an observer determines his latitude by measuring the altitude of a heavenly object at the time of its meridian passage. Meridian passage is when the object passes the observer's meridian, i.e. passes through south or north. This is usually done with the sun for a noon latitude but can be done with any heavenly object. Noon is when the sun passes through the meridian.
Let us imagine that the sun is overhead (at the zenith) at a point on the equator (latitude 0°) and observer A is standing at this point, the geographical point of the sun. If he were to measure the height of the sun above the horizon with a sextant he would find that the altitude of the sun was 90°. By subtracting this figure from 90° he would find the zenith distance of the sun which is 0°, which is the same as his latitude. Observer B is standing at one of the poles (latitude 90°N or 90°S) he would see the sun on the horizon at an altitude of 0°. By subtracting this from 90° he would find that the zenith distance is 90°, which is his latitude. Observer C at the same time is at latitude 20°N on the same meridian, i.e. on the same longitude as observer A. His measured altitude would be 70° and by subtracting this from 90° gives a 20° zenith distance which in turn is his latitude. In short, the zenith distance of a heavenly object at meridian altitude is the difference in latitude between it and the observer.
## Methodology
The estimated time of meridian altitude of the heavenly object is extracted from the nautical almanac. A few minutes before this time the observer starts observing the altitude of the object with a sextant. The altitude of the object will be increasing and the observer will continually adjust the sextant to keep the reflected image of the object on the horizon. As the object passes the meridian a maximum altitude will be observed. The time in UTC of this is observed.
The altitude obtained is corrected for dip (the error caused by the observers height above the sea) and refraction to obtain the true altitude of the object above the horizon. This is then subtracted from 90° to obtain the angular distance from the position directly above to obtain the zenith distance.
A further correction must then be taken into account to counter the "wobble" of the earth's spin and rotation relative to the sun and planets. This is given in the declination for the body on a particular day in the year (also taken from the nautical almanac). If the declination of the body is in the opposite hemisphere (ie if you are in the northern hemisphere and the declination is in the southern hemisphere) then the declination must be subtracted from your true zenith distance, otherwise the declination is added.
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# Thread: Missundestanding of how complex numbers work not helping with fourier analysis
1. ## Missundestanding of how complex numbers work not helping with fourier analysis
I am studying electrical engineering at university level in the UK. As part of a course on Circuit and Signal processing I have tutorial sheets on Fourier Analysis.
I am fine with fourier analysis except when I start to look at exponential fourier series, I have the question, I even have the answers and how the tutor reached the answer (the workings). But I am not understanding how he is jumping from one line to another. I have attached the whole solution but the bit that I am misunderstanding is:
$\displaystyle r_{m}=\frac{-V}{4\pi^2}\displaystyle\{\frac{-1}{jm}[2\pi]\displaystyle\}$
$\displaystyle \\r_{m}=\frac{-jV}{4m\pi^2}\displaystyle\{[2\pi]\displaystyle\}$
where j = complex number $\displaystyle j^2=-1$, how is j moving from the denominator to the numerator?
surely the next line should be
$\displaystyle \\r_{m}=\frac{V}{j4m\pi^2}\displaystyle\{[2\pi]\displaystyle\}$
Any help would be greatly appreciated. Thanks Ed.
2. Originally Posted by emmv
I am studying electrical engineering at university level in the UK. As part of a course on Circuit and Signal processing I have tutorial sheets on Fourier Analysis.
I am fine with fourier analysis except when I start to look at exponential fourier series, I have the question, I even have the answers and how the tutor reached the answer (the workings). But I am not understanding how he is jumping from one line to another. I have attached the whole solution but the bit that I am misunderstanding is:
$\displaystyle r_{m}=\frac{-V}{4\pi^2}\displaystyle\{\frac{-1}{jm}[2\pi]\displaystyle\}$
$\displaystyle \\r_{m}=\frac{-jV}{4m\pi^2}\{[2\pi]\}$
where j = complex number $\displaystyle j^2=-1$, how is j moving from the denominator to the numerator?
surely the next line should be
$\displaystyle \\r_{m}=\frac{V}{j4m\pi^2}\displaystyle\{[2\pi]\displaystyle\}$
Any help would be greatly appreciated. Thanks Ed.
Now multiply top and bottom by $\displaystyle j$, this gives a $\displaystyle j$ on the top and on the bottom $\displaystyle j^2=-1$, or just $\displaystyle -j$ on the top:
$\displaystyle r_{m}=\frac{V}{j4m\pi^2}(2\pi)=\frac{Vj}{j^24m\pi^ 2}(2\pi)=\frac{-jV}{4m\pi^2}(2\pi)$
CB
3. Thanks Captain Black, to clarify my understanding:
If I had $\displaystyle \frac{1}{5}$ and I multiplied top and bottom by 5 then the answer would be $\displaystyle \frac{5}{25}$ which is still $\displaystyle \frac{1}{5}$ the actual values haven't changed just the figures representing them. So is it the case that with $\displaystyle j$ that $\displaystyle j^2$ is $\displaystyle -1$ which effectively gets lost within the other numbers?
So actually the steps are as below, but then how are you moving the minus from the denominator to the numerator?
$\displaystyle r_{m}=\frac{V}{j4m\pi^2}(2\pi)=\frac{Vj}{j^24m\pi^ 2}(2\pi)=\frac{jV}{(-1)4m\pi^2}(2\pi)$
Thanks, Ed.
4. Sorry Captain Black, I didn't think before I asked my question. If the fraction was $\displaystyle \frac{1}{-5}$ this would in fact be $\displaystyle -\frac{1}{5}$ as the minus relates to the whole of the fraction. Thank you for helping me with this. Regards Ed.
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Question
# Which constant must be added and subtracted to solve the quadratic equation 9x2+34x−2=0 by the method of completing the square?
A
18
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B
164
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C
14
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D
964
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Solution
## The correct option is B 164 Given equation is 9x2+34x−2=0 ⇒(3x)2+14(3x)−2=0 On putting 3x = y, we have y2+14y−2=0 ⇒y2+14y+(18)2−(18)2−2=0⇒(y+18)2=164+2⇒(y+18)2=12964 ⇒y=±√1298−18 ⇒y=√129−18 or −(√129+1)8 Thus, (18)2=164 must be added and subtracted to solve the given equation.
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A polygon is a 2-dimensional many sided shape. If all the sides are the same length it is a regular polygon, if not it is an irregular polygon.
A polygon is usually thought of as a convex shape (roughly circular), but they don’t have to be, they can fold in on themselves and have intersecting lines. The only requirements are that they must have straight sides and form a closed 2D shape. So all these shapes are valid polygons:
Lets forget about these though and focus on convex polygons. The most visually satisfying are regular polygons. Here are the first nine:
An n-sided polygon has n interior angles, as shown below. If it is a regular polygon, each interior angle is the same, look at the regular polygons above.
Each interior angle has an exterior angle ‘partner’.
The sum of each interior angle and exterior angle is 180°, so:
e1 + i1 = 180°
e2 + i2 = 180°
Again, if it is a regular polygon, all the exterior angles are equal.
The sum of the exterior angles is 360°. Imagine walking round the polygon say anti-clockwise. Each time you reach a corner you have to turn anti-clockwise through the exterior angle. By the time you get back to the starting point, you have made one full turn, 360°, and turned through all of the exterior angles added together. Hence we can say for any polygon:
e1 + e2 + e3 + …. + en = 360°
where en is the nth exterior angle and …. denotes all other exterior angles between 3 and n. It can be written more compactly as:
∑e = 360°
where the ∑ symbol means sum of. For a regular polygon each e is the same so:
e + e + e + …. + e = 360°
which can be written:
n × e = 360°
rearranging to make e the subject gives:
e = 360°/n
This immediately enables us to write down a formula for the interior angle, i, of a regular polygon starting from the fact that the sum of the interior and exterior angles is 180°:
i + e = 180°
rearranging to make i the subject:
i = 180° – e
substituting for e:
i = 180° – 360°/n
factorising by pulling out the common factor of 180°:
i = 180°(1-2/n)
pulling 1/n out of the bracket:
i = 180°×(n-2)/n
which is a formula that enables us to calculate the interior angle for any regular polygon. So for example the interior angle of a regular pentagon is:
i = 180°×(5-2)/5 = 36° × 3 = 108°
Hiding in this formula is the more general result for the sum of the interior angles (multiplying both sides by n):
n×i = ∑i = 180°×(n-2)
This result holds for any polygon, regular or irregular.
Both versions of this formula need to be learnt and applied for GCSE! It is much much easier to do this if you have a deep understanding of where the formula come from.
Here is a sample question (EdExcel Nov 2014 1MAO/1H):
Here is my worked solution.
There are other ways to get the answer, such as considering the polygon CKFED, or by calculating the external angle of the octagon. So long as you lay the answer out neatly, explaining what you are doing it doesn’t matter which approach you use.
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## Reflection: Student Ownership Going Deeper with Interior and Exterior Angles - Section 3: Think-Group-Share: Three Polygons Meet at Point B
This is the kind of problem that I love to showcase in a whole-class discussion as it provokes interest by posing a question students actually want to answer—this is important. Equally important is the fact that this problem can be solved in multiple interesting ways, which reinforces the message that the answer is not nearly as valuable as the thinking and problem solving efforts that lead students to be able to arrive at the same, logical conclusion.
As students first shared out initial thoughts and ideas in their small groups, I circulated the room, listening carefully to understand and differentiate between students’ solution paths. I chose student presenters in an order that I thought would promote the deepest understanding of the problem amongst the greatest number of students. I essentially looked for a student who would start the discussion by sharing a widely used and fairly traditional approach (by using the interior angle measure of the unknown polygon and the interior angle sum formula) to start the share-out discussion; this also meant that I looked for different and perhaps novel approaches (sketching an exterior angle to the unknown polygon and visualizing the number of times that exterior angle would fit into 360 degrees, for example). The order, I think, matters in that it allows the greatest number of students the most access and opportunity to deeply understand important relationships.
Using a Simple Yet Engaging Problem to Publicly Value Multiple Approaches
Student Ownership: Using a Simple Yet Engaging Problem to Publicly Value Multiple Approaches
# Going Deeper with Interior and Exterior Angles
Unit 8: Discovering and Proving Polygon Properties
Lesson 3 of 9
## Big Idea: By working on a Hands-On Activity in groups students develop collaboration skills and practice justifying their conjectures.
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1 teacher likes this lesson
Standards:
Subject(s):
Math, Geometry, polygons (Determining Measurements), problem solving, properties of polygons, angles and angle measurement
55 minutes
### Jessica Uy
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The Geometry of Circles Definitions and formulas for the radius of a circle, the diameter of a circle, the circumference (perimeter) of a circle, the area of a circle, the chord of a circle, arc and the arc length of a circle, sector and the area of the sector of a circle
The radius of a circle: The radius of a circle is the distance from the center of the circle to the outside edge.
The diameter of a circle: The diameter of a circle is longest distance across a circle. (The diameter cuts through the center of the circle. This is what makes it the longest distance.)
The circumference of a circle (the perimeter of a circle): The circumference of a circle is the perimeter -- the distance around the outer edge. Circumference = where r = the radius of the circle and pi = 3.141592...
The area of a circle: Area = where r = the radius of the circle and pi = 3.141592...
A chord of a circle: A chord of a circle is a line segment that connects one point on the edge of the circle with another point on the circle. (The diameter is a chord -- it's just the longest chord!)
An arc of a circle: An arc of a circle is a segment of the circumference of the circle.
The formula for the arc length of a circle: Arc length of a circle in radians: Arc Length = Arc length of a circle in degrees: Arc Length =
A sector of a circle: A sector of a circle is a pie shaped portion of the area of the circle. Technically, the piece of pie is between two segments coming out of the center of the circle.
The area of a sector of a circle: Sector area of a circle in radians: Sector Area = Sector area of a circle in degrees: Sector Area =
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# math
This is the sequence 1,3,6,10,15 the pattern is addin 1 more than last time but what is the name for this pattern
These are called the triangular numbers
The sequence is
1
3=1+2
6=1+2+3
10=1+2+3+4
15=1+2+3+4+5
You can also observe this pattern
x
_________
x
xx
__________
x
xx
xxx
__________
x
xx
xxx
xxxx
to see why they're called triangular numbers. I think the Pythagoreans (around 700 B.C.E.) were the ones who gave them this name. I do know the Pythagoreans tried to assign numbers to many different objects like this.
Triangular Numbers
The number of dots, circles, spheres, etc., that can be arranged in an equilateral or right triangular pattern is called a triangular number. The 10 bowling pins form a triangular number as do the 15 balls racked up on a pool table. Upon further inspection, it becomes immediately clear that the triangular numbers, T1, T2, T3, T4, etc., are simply the sum of the consecutive integers 1-2-3-4-.....n or Tn = n(n + 1)/2, namely, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66,78, 91, etc.
Triangular numbers are the sum of the balls in the triangle as defined by Tn = n(n + 1)/2.
Order.n...1........2...............3.....................4.............................5....................6.....7.....8.....9
.............O.......O...............O....................O............................O
....................O...O.........O...O...............O...O.......................O...O
..................................O...O...O.........O....O....O...............O....O....O
......................................................O....O...O....O.........O.....O...O....O
.................................................................................O....O....O....O....O
Total......1........3.................6....................10..........................15..................21...28...36...45...etc.
The sum of a series of triangular numbers from 1 through Tn is given by S = (n^3 + 3n^2 + 2n)/6.
After staring at several triangular and square polygonal number arrangements, one can quickly see that the 1st and 2nd triangular numbers actually form the 2nd square number 4. Similarly, the 2nd and 3rd triangulars numbers form the 3rd square number 9, and so on. By inspection, one can see that the nth square number, Sn, is equal to Tn + T(n - 1) = n^2. This can best be visualized from the following:
.........Tn - 1...3...6...10...15...21...28...36...45...55...66...78...91
.........T(n - 1)........1...3....6....10...15...21...28...36...45...55...66...78
.........Sn.........1...4...9....16...25...36...49...64...81..100.121.144.169
A number cannot be triangular if its digital root is 2, 4, 5, 7 or 8.
Some interesting characteristice of Triangular numbers:
The numbers 1 and 36 are both square and triangular. Some other triangular squares are 1225, 41,616, 1,413,721, 48,024,900 and 1,631,432,881. Triangular squares can be derived from the series 0, 1, 6, 35, 204, 1189............Un where Un = 6U(n - 1) - U(n - 2) where each term is six times the previous term, diminished by the one before that. The squares of these numbers are simultaneously square and triangular.
The difference between the squares of two consecutive rank triangular numbers is equal to the cube of the larger numbers rank.
Thus, (Tn)^2 - (T(n - 1))^2 = n^3. For example, T6^2 - T5^2 = 441 - 225 = 216 = 6^3.
The summation of varying sets of consecutive triangular numbers offers some strange results.
T1 + T2 + T3 = 1 + 3 + 6 = 10 = T4.
T5 + T6 + T7 + T8 = 15 + 21 + 28 + 36 = 100 = 45 + 55 = T9 + T10.
The pattern continues with the next 5 Tn's summing to the next 3 Tn's followed by the next6 Tn's summing to the next 4 Tn's, etc.
The sum of the first "n" cubes is equal to the nth triangular number. For instance:
n............1.....2.....3.....4.......5
Tn..........1.....3.....6....10.....15
n^3.........1 + 8 + 27 + 64 + 100 = 225 = 15^2
Every number can be expressed by the sum of three or less triangular numbers, not necessarily different.
1 = 1, 2 = 1 + 1, 3 = 3, 4 = 3 + 1, 5 = 3 + 1 + 1, 6 = 6, 7 = 6 + 1, 8 = 6 + 1 + 1, 9 = 6 + 3, 10 = 10, etc.
Alternate ways of finding triangular squares.
From Tn = n(n + 1)/2 and Sn = m^2, we get m^2 = n(n + 1)/2 or 4n^2 + 4n = 8m^2.
Adding one to both sides, we obtain 4n^2 + 4n + 1 = 8m^2 + 1.
Factoring, we find (2n + 1)^2 = 8m^2 + 1.
If we allow (2n + 1) to equal "x" and "y" to equal 2m, we come upon x^2 - 2y^2 = 1, the famous Pell Equation.
We now know that the positive integer solutions to the Pell equation, x^2 - 2y^2 = +1 lead to triangular squares. But how?
Without getting into the theoretical aspect of the subject, sufficeth to say that the Pell equaion is closely connected with early methods of approximating the square root of a number. The solutions to Pell's equation, i.e., (x,y), often written as (x/y) are approximations of the square root of D in x^2 - 2y^2 = +1. Numerous methods have evolved over the centuries for estimating the square root of a number.
Diophantus' method leads to the minimum solutions to x^2 - Dy^2 = +1, D a non square, by setting x = my + 1 which leads to y = 2m/(D - m^2).
From values of m = 1.......n, many rational solutions evolve.
Eventually, an integer solution will be reached.
For instance, the smallest solution to x^2 - 2y^2 = +1 derives from m = 1 resulting in x = 3 and y = 2 or sqrt(2) ~= 3/2..
Newton's method leads to the minimum solution sqrt(D) = sqrt(a^2 + r) = (a + D/a)/2 ("a" = the nearest square) = (3/2).
Heron/Archimedes/El Hassar/Aryabhatta obtained the minimum solution sqrt(D) = sqrt(a^2 +-r) = a +-r/2a = (x/y) = (3/2).
Other methods exist that produce values of x/y but end up being solutions to x^2 - Dy^2 = +/-C.
Having the minimal solutions of x1 and y1 for x^2 - Dy^2 = +1, others are derivable from the following:
(x + ysqrtD) = (x1 + y1sqrtD)^n, n = 1, 2, 3, etc.
Alternitive approach
Given x = p and y = q satifying x^2 - 2y^2 = +1, we can write (x + sqrtD)(x - sqrtD) = 1.
x = [(p + qsqrt(2))^n + (p - qsqrt(2))^n]/2
y = [(p + qsqrt(2))^n - (p - qsqrt(2))^n]/(2sqrt(2))
Having the minimum solution of x = 3 and y = 2, the next few solutions derive from n = 2 and 3 where x = 17, y = 12, x = 99 and y = 70 respectively.
Alternative approach
Subsequent solutions can also be obtained by means of the following:
x^2 - 2y^2 = +1 can be rewritten as x^2 - 2y^2 = (x + yqrt(2)(x - ysqrt(2)) = +1.
Using the minimum solution of x = 3 and y = 2, we can now write
.................(3 + sqrt(2))^2(3 - sqrt(2))^2 = 1^2 = 1
.................(17 + 12sqrt(2))(17 - 12sqrt(2)) = 1
.................289 - 2(144) = 17^2 - 2(12)^2 = 1 the next smallest solution.
The next smallest solution is derivable from
.................(3 + sqrt(2))^3(3 - sqrt(2))^3 = 1^2 = 1 which works out to
.................(99 + 70sqrt(2))(99 - 70sqrt(2)) = 1 or
.................99^2 - 2(70)^2 = 1.
Similarly, (3 + sqrt(2))^4(3 - sqrt(2))^4 = 1^2 = 1 leads to
.................(577 + 408sqrt(2))(577 - 408sqrt(2)) = 1 and
.................577^2 - 2(408)^2 = 1.
Regardless of the method, we ultimately end up with the starting list of triangular squares.
..x........y........n.......m........Tn = Sm^2
..3........2........1........1..............1
.17......12.......8........6..............36
.99......70......49......35...........1225
577....408....288.....204.........41,616 etc.
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How would you describe the pattern of this sequence? 2/3, 4/4, 6/5, 8/6, 10/7 The formula is 2n/n+2 where n is 1, 2, 3, 4, 5
1. ### Math(Geometry)
In my geometry book I am really stumped on these question.This is what it says to do, "Find a pattern for each sequence. Use the pattern to show the next two terms."And here are th questions I am stuck on with the #'s.
2. ### algebra
Describe the pattern in each sequence and determine the next term of the sequence. 11, 17, 23, 29, … The pattern in each sequence is that after each sequence, the number is added by 6. The next term of the sequence would be 35.
3. ### math
What is the pattern to this sequence? 1,3,4,7,11,18,29 I figured it was plus the last 2 numbers which will then equal the next number. But I got messed up and can't figure the next 3 numbers in this pattern, any help is helpful.
4. ### Math
1. What are the next two terms of the following sequence? -3, 1, 5, 9... 2. What are the next two terms of the following sequence? -2, 4, -8, 16... 3. What is the common difference of the following arithmetic sequence? 13, -7,
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Ch 1. Stress and Strain Multimedia Engineering Mechanics NormalStress Shear andBearing Stress NormalStrain Hooke'sLaw ThermalEffects IndeterminateStructures
Chapter 1. Stress/Strain 2. Torsion 3. Beam Shr/Moment 4. Beam Stresses 5. Beam Deflections 6. Beam-Advanced 7. Stress Analysis 8. Strain Analysis 9. Columns Appendix Basic Math Units Basic Equations Sections Material Properties Structural Shapes Beam Equations Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll
MECHANICS - EXAMPLE Example Plate with Applied Loads and Displacement A 0.5 cm thick rectangular plate is pulled in tension by two loads in the x and y directions as shown. The total deflection in the x and y direction is 0.021 cm and 0.009 cm, respectively. What is the Poisson's ratio of the material? The z direction has no load and the deflection is not known. Solution Stresses and Strains on Plate This problem involves loading from two directions, and thus requires at least the 2-D Hooke's Law. The 3-D Hooke's Law could be used, but since σz is zero, those equations will reduce to the 2-D equations. The equations are, εx = (σx - ν σy)/E εy = (σy - ν σx)/E The strains and stresses in the x and y direction need to be calculated. σx = Px/Ax = 5/[(0.05)(0.005)] = 20 MPa σy = Py/Ay = 9/[(0.10)(0.005)] = 18 MPa εx = 0.021/10 = 0.0021 cm/cm εy = 0.009/5 = 0.0018 cm/cm Substituting the stresses and strains into the 2-D Hooke's Law equations, gives 0.0021 E = 20 - ν 18 0.0018 E = 18 - ν 20 Solving for ν gives, ν = 0.1876
Practice Homework and Test problems now available in the 'Eng Mechanics' mobile app
Includes over 400 problems with complete detailed solutions.
Available now at the Google Play Store and Apple App Store.
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# Tag Info
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### Why is in Matlab exp(pi * sqrt(163)/3) - 640320 = -2.3283e-10
First, that expression is not an integer, but very close to an integer $640320$. In fact, as @Peter mentioned, this is related to some deep theory of modular $j$-functions and Heegner numbers, e.g. ...
• 15.2k
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### What unique value does Cramer's rule offer?
Cramer‘s rule is used in a lot of proofs. You can use it to get the $(i,j)$-th entry of the matrix $A^{-1}$ (see adjugate matrix), it be used to prove the that every smooth $C^1$ diffeomorphism is ...
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### Prove that $1 \leq \|A^{-1}\| _2\|A-B\|_2$
Let $v$ be a nonzero unit vector in $\ker(B)$. Then $$\|A^{-1}\|_2\|A-B\|_2 \ge\|I-A^{-1}B\|_2 =\sup_{\|u\|_2=1}\|(I-A^{-1}B)u\|_2 \ge\|(I-A^{-1}B)v\|_2 =\|v\|_2=1.$$
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### Why is this matrix always symmetric?
An initial simplification helps. Absorb $\frac{h^2}{8}$ into $B$ and forget about it. Now note that since $B$ is symmetric, our given matrix is symmetric if and only if $X:=B(I-A^{-1}B)^{-1}$ is ...
### Intersection of two ellipses at exactly 2 points
Assume that $A, B, k$ are given where $A, B$ are symmetric positive definite $2 \times 2$ matrices, and $k \gt 0$. The solution of $x^T A x = k$ is the ellipse $x = v_1 \cos t + v_2 \sin t$ ...
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### Intersection of two ellipses at exactly 2 points
We can assume WLOG (up to division by a constant) that the first ellipse $(E)$ has equation $$X^TAX=1$$ Let $X^TBX=\ell$ be the second variable ellipse that we will call $(E_{\ell})$. Result : the ...
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### Why is easier to get inverse of mass matrix?
The claim that "Inverting the mass matrix is significantly easier than solving a linear system involving the stiffness matrix" is more of a slogan than a precise theorem. This answer gives ...
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### Finding the bounds of the missing values of a symmetric positive semidefinite matrix
The determinant of your matrix is $-(x-1)^2$. Since a positive semidefinite matrix must have nonnegative determinant, the only possible value is $x=1$. You still need to check that the matrix is ...
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### LU factorization distribution over addition
The answer is almost certainly no. The key fact to notice is that your matrix $A = I + \alpha D$ is a rank $n$ perturbation of the matrix $\alpha D$. You cannot profit from applying the Sherman-...
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### Efficient algorithm of rank-one update of the Cholesky decomposition
Tensorflow uses the implementation recorded in Krause & Igel (2015). reference: Krause, O., & Igel, C. (2015, January). A more efficient rank-one covariance matrix update for evolution ...
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### Symmetric matrix with integer values is positive semidefinite
Again this is not true. $$\det\begin{pmatrix} 9&8&7&6\\ 8&9&0&0\\ 7&0&9&0\\ 6&0&0&9 \end{pmatrix}=-5508.$$ Also, if you replace the $0$s with $1$s the ...
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Transcript
```One-Sample t-test
What do these problems we have been working on have in
common?
In a population of American graduate students, individuals
earn on average \$7.25/hour on alcohol, with a standard
students from Brooklyn earn a different wage than grad
students in general. I gather together 9 graduate students
from this program and calculate the average amount they
earn an hour: \$11.00. Use an alpha level of .01.
How do I know I need to be using a z-test?
1. We are comparing a sample mean to a KNOWN population mean.
2. We KNOW the population s.
What to do if you do not know the Population
Standard Deviation (s)?
Use the best estimate of sx
You must correct for the
uncertainty of this estimate.
T-score for a single sample mean
Where sx_ =
s
What to do if you do not know the Population
Standard Deviation (s)?
_
x
_
x
Probability
Hypothesis testing with the t-statistic
Retain H0
Outcome âtâ
Reject H0
t-crit
One-tailed test
One-Sample t-test
z-test is used when we know both m and s
t-test is when we know m but not s
The Sampling Distribution of the t-test
This table lists the critical value of the t-statistic for the
Degrees of freedom and a level.
Given the level and the df you can find the critical value
of the t-statistic that divides the outcomes into reject or retain the null.
The average IQ score of Americans is 100. I believe that my Statistics
class (you guys!) have different IQs than the general population. I force
all 15 of you to take an IQ test, and I calculate a mean of 110 (s= 20). Use
an alpha level of .05 to determine if this class has a different IQ than the
population.
How do I know I need to be using a one-sample t-test?
Step 1: State the null and alternative hypotheses:
H0: My stats class does not have a different IQ than the average American
H1: My stats class has a different IQ than the average American
Step 2: Find the critical value.
Go to t-table! Must figure out of this is one- or two-tailed, and df.
The Sampling Distribution of the t-test
This table lists the critical value of the t-statistic for the
Degrees of freedom and a level.
Given the level and the df you can find the critical value
of the t-statistic that divides the outcomes into reject or retain the null.
The average IQ score of Americans is 100. I believe that my Statistics
class (you guys!) have different IQs than the general population. I force
all 15 of you to take an IQ test, and I calculate a mean of 110 (s= 20). Use
an alpha level of .05 to determine if this class has a different IQ than the
population.
How do I know I need to be using a one-sample t-test?
Step 1: State the null and alternative hypotheses:
H0: My stats class does not have a different IQ than the average American
H1: My stats class has a different IQ than the average American
Step 2: Find the critical value.
Go to t-table! Must figure out of this is one- or two-tailed, and df.
+/-2.145
The average IQ score of Americans is 100. I believe that my Statistics
class (you guys!) have different IQs than the general population. I force
all 15 of you to take an IQ test, and I calculate a mean of 110 (s= 20). Use
an alpha level of .05 to determine if this class has a different IQ than the
population.
Step 3: Calculate the obtained statistic:
tobt ï½
x ï mx
sx
=
- 100
110
________
= 10/5.17 = 1.93
20/sqrt(15)
Step 4: Make a decision.
I
I
-2.15
2.15
Step 4: Retain the null hypothesis.
The Sampling Distribution of the t-test
This table lists the critical value of the t-statistic for the
Degrees of freedom and a level.
Given the level and the df you can find the critical value
of the t-statistic that divides the outcomes into reject or retain the null.
```
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# Cylindrical coordinate system
A cylindrical coordinate system with origin O, polar axis A, and longitudinal axis L. The dot is the point with radial distance ρ = 4, angular coordinate φ = 130°, and height z = 4.
A cylindrical coordinate system is a three-dimensional coordinate system that specifies point positions by the distance from a chosen reference axis, the direction from the axis relative to a chosen reference direction, and the distance from a chosen reference plane perpendicular to the axis. The latter distance is given as a positive or negative number depending on which side of the reference plane faces the point.
The origin of the system is the point where all three coordinates can be given as zero. This is the intersection between the reference plane and the axis.
The axis is variously called the cylindrical or longitudinal axis, to differentiate it from the polar axis, which is the ray that lies in the reference plane, starting at the origin and pointing in the reference direction.
The distance from the axis may be called the radial distance or radius, while the angular coordinate is sometimes referred to as the angular position or as the azimuth.The radius and the azimuth are together called the polar coordinates, as they correspond to a two-dimensional polar coordinate system in the plane through the point, parallel to the reference plane. The third coordinate may be called the height or altitude (if the reference plane is considered horizontal), longitudinal position,[1] or axial position.[2]
Cylindrical coordinates are useful in connection with objects and phenomena that have some rotational symmetry about the longitudinal axis, such as water flow in a straight pipe with round cross-section, heat distribution in a metal cylinder, electromagnetic fields produced by an electric current in a long, straight wire, accretion disks in astronomy, and so on.
They are sometimes called "cylindrical polar coordinates"[3] and "polar cylindrical coordinates",[4] and are sometimes used to specify the position of stars in a galaxy ("galactocentric cylindrical polar coordinates").[5]
## Definition
The three coordinates (ρ, φ, z) of a point P are defined as:
• The axial distance or radial distance ρ is the Euclidean distance from the z-axis to the point P.
• The azimuth φ is the angle between the reference direction on the chosen plane and the line from the origin to the projection of P on the plane.
• The axial coordinate or height z is the signed distance from the chosen plane to the point P.
### Unique cylindrical coordinates
As in polar coordinates, the same point with cylindrical coordinates (ρ, φ, z) has infinitely many equivalent coordinates, namely (ρ, φ ± n×360°, z) and (−ρ, φ ± (2n + 1)×180°, z), where n is any integer. Moreover, if the radius ρ is zero, the azimuth is arbitrary.
In situations where someone wants a unique set of coordinates for each point, one may restrict the radius to be non-negative (ρ ≥ 0) and the azimuth φ to lie in a specific interval spanning 360°, such as [−180°,+180°] or [0,360°].
### Conventions
The notation for cylindrical coordinates is not uniform. The ISO standard 31-11 recommends (ρ, φ, z), where ρ is the radial coordinate, φ the azimuth, and z the height. However, the radius is also often denoted r or s, the azimuth by θ or t, and the third coordinate by h or (if the cylindrical axis is considered horizontal) x, or any context-specific letter.
The coordinate surfaces of the cylindrical coordinates (ρ, φ, z). The red cylinder shows the points with ρ = 2, the blue plane shows the points with z = 1, and the yellow half-plane shows the points with φ = −60°. The z-axis is vertical and the x-axis is highlighted in green. The three surfaces intersect at the point P with those coordinates (shown as a black sphere); the Cartesian coordinates of P are roughly (1.0, −1.732, 1.0).
Cylindrical coordinate surfaces. The three orthogonal components, ρ (green), φ (red), and z (blue), each increasing at a constant rate. The point is at the intersection between the three colored surfaces.
In concrete situations, and in many mathematical illustrations, a positive angular coordinate is measured counterclockwise as seen from any point with positive height.
## Coordinate system conversions
The cylindrical coordinate system is one of many three-dimensional coordinate systems. The following formulae may be used to convert between them.
### Cartesian coordinates
For the conversion between cylindrical and Cartesian coordinates, it is convenient to assume that the reference plane of the former is the Cartesian xy-plane (with equation z = 0), and the cylindrical axis is the Cartesian z-axis. Then the z-coordinate is the same in both systems, and the correspondence between cylindrical (ρ,φ,z) and Cartesian (x,y,z) are the same as for polar coordinates, namely
{\displaystyle {\begin{aligned}x&=\rho \cos \varphi \\y&=\rho \sin \varphi \\z&=z\end{aligned}}}
in one direction, and
{\displaystyle {\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}}}\\\varphi &={\begin{cases}0&{\mbox{if }}x=0{\mbox{ and }}y=0\\\arcsin \left({\frac {y}{\rho }}\right)&{\mbox{if }}x\geq 0\\\arctan \left({\frac {y}{x}}\right)&{\mbox{if }}x>0\\-\arcsin \left({\frac {y}{\rho }}\right)+\pi &{\mbox{if }}x<0\end{cases}}\end{aligned}}}
in the other. The arcsin function is the inverse of the sine function, and is assumed to return an angle in the range [−π/2,+π/2] = [−90°,+90°]. These formulas yield an azimuth φ in the range [−90°,+270°]. For other formulas, see the polar coordinate article.
Many modern programming languages provide a function that will compute the correct azimuth φ, in the range (−π, π), given x and y, without the need to perform a case analysis as above. For example, this function is called by atan2(y,x) in the C programming language, and atan(y,x) in Common Lisp.
### Spherical coordinates
Spherical coordinates (radius r, elevation or inclination θ, azimuth φ), may be converted into cylindrical coordinates by:
θ is elevation: θ is inclination: {\displaystyle {\begin{aligned}\rho &=r\cos \theta \\\varphi &=\varphi \\z&=r\sin \theta \end{aligned}}} {\displaystyle {\begin{aligned}\rho &=r\sin \theta \\\varphi &=\varphi \\z&=r\cos \theta \end{aligned}}}
Cylindrical coordinates may be converted into spherical coordinates by:
θ is elevation: θ is inclination: {\displaystyle {\begin{aligned}r&={\sqrt {\rho ^{2}+z^{2}}}\\\theta &=\arctan \left({\tfrac {z}{\rho }}\right)\\\varphi &=\varphi \end{aligned}}} {\displaystyle {\begin{aligned}r&={\sqrt {\rho ^{2}+z^{2}}}\\\theta &=\arctan \left({\tfrac {\rho }{z}}\right)\\\varphi &=\varphi \end{aligned}}}
## Line and volume elements
See multiple integral for details of volume integration in cylindrical coordinates, and Del in cylindrical and spherical coordinates for vector calculus formulae.
In many problems involving cylindrical polar coordinates, it is useful to know the line and volume elements; these are used in integration to solve problems involving paths and volumes.
The line element is
${\displaystyle \mathrm {d} \mathbf {r} =\mathrm {d} \rho \,{\boldsymbol {\hat {\rho }}}+\rho \,\mathrm {d} \varphi \,{\boldsymbol {\hat {\varphi }}}+\mathrm {d} z\,\mathbf {\hat {z}} .}$
The volume element is
${\displaystyle \mathrm {d} V=\rho \,\mathrm {d} \rho \,\mathrm {d} \varphi \,\mathrm {d} z.}$
The surface element in a surface of constant radius ρ (a vertical cylinder) is
${\displaystyle \mathrm {d} S_{\rho }=\rho \,\mathrm {d} \varphi \,\mathrm {d} z.}$
The surface element in a surface of constant azimuth φ (a vertical half-plane) is
${\displaystyle \mathrm {d} S_{\varphi }=\mathrm {d} \rho \,\mathrm {d} z.}$
The surface element in a surface of constant height z (a horizontal plane) is
${\displaystyle \mathrm {d} S_{z}=\rho \,\mathrm {d} \rho \,\mathrm {d} \varphi .}$
The del operator in this system leads to the following expressions for gradient, divergence, curl and Laplacian:
{\displaystyle {\begin{aligned}\nabla f&={\frac {\partial f}{\partial \rho }}{\boldsymbol {\hat {\rho }}}+{\frac {1}{\rho }}{\frac {\partial f}{\partial \varphi }}{\boldsymbol {\hat {\varphi }}}+{\frac {\partial f}{\partial z}}\mathbf {\hat {z}} \\[8px]\nabla \cdot {\boldsymbol {A}}&={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\left(\rho A_{\rho }\right)+{\frac {1}{\rho }}{\frac {\partial A_{\varphi }}{\partial \varphi }}+{\frac {\partial A_{z}}{\partial z}}\\[8px]\nabla \times {\boldsymbol {A}}&=\left({\frac {1}{\rho }}{\frac {\partial A_{z}}{\partial \varphi }}-{\frac {\partial A_{\varphi }}{\partial z}}\right){\boldsymbol {\hat {\rho }}}+\left({\frac {\partial A_{\rho }}{\partial z}}-{\frac {\partial A_{z}}{\partial \rho }}\right){\boldsymbol {\hat {\varphi }}}+{\frac {1}{\rho }}\left({\frac {\partial }{\partial \rho }}\left(\rho A_{\varphi }\right)-{\frac {\partial A_{\rho }}{\partial \varphi }}\right)\mathbf {\hat {z}} \\[8px]\nabla ^{2}f&={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\left(\rho {\frac {\partial f}{\partial \rho }}\right)+{\frac {1}{\rho ^{2}}}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}\end{aligned}}}
## Cylindrical harmonics
The solutions to the Laplace equation in a system with cylindrical symmetry are called cylindrical harmonics.
## References
1. ^ Krafft, C.; Volokitin, A. S. (1 January 2002). "Resonant electron beam interaction with several lower hybrid waves". Physics of Plasmas. 9 (6): 2786–2797. Bibcode:2002PhPl....9.2786K. doi:10.1063/1.1465420. ISSN 1089-7674. Retrieved 9 February 2013. ...in cylindrical coordinates (r,θ,z) ... and Z = vbzt is the longitudinal position...
2. ^ Groisman, Alexander; Steinberg, Victor (1997). "Solitary Vortex Pairs in Viscoelastic Couette Flow". Physical Review Letters. 78 (8): 1460–1463. Bibcode:1997PhRvL..78.1460G. doi:10.1103/PhysRevLett.78.1460. ...where r, θ, and z are cylindrical coordinates ... as a function of axial position...
3. ^ Szymanski, J. E. (1989). Basic Mathematics for Electronic Engineers: models and applications. Tutorial Guides in Electronic Engineering (no. 16). Taylor & Francis. p. 170. ISBN 978-0-278-00068-1.
4. ^ Nunn, Robert H. (1989). Intermediate Fluid Mechanics. Taylor & Francis. p. 3. ISBN 978-0-89116-647-4.
5. ^ Sparke, Linda Siobhan; Gallagher, John Sill (2007). Galaxies in the Universe: An Introduction (2nd ed.). Cambridge University Press. p. 37. ISBN 978-0-521-85593-8.
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Quiz-1-Math-20D-W09-Solutions
# Quiz-1-Math-20D-W09-Solutions - y = 2-e x 3 2 y y(0 = 0 and...
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Quiz #1 Math 20D January 16 SOLUTIONS 1. ( § 2.1, #30. 10 points) Find the value of y 0 for which the solution of the initial value problem y 0 - y = 1 + 3 sin t, y (0) = y 0 remains finite as t → ∞ . ANSWER: (1) p ( t ) = - 1 (2) μ ( t ) = e R p ( t ) dt = e - t (some students will be able to figure this out directly, that is OK) (3) Thus y ( t ) = 1 μ ( t ) Z g ( t ) μ ( t ) dt + C = e t Z e - t (1 + 3 sin t ) dt + C = e t - e - t + 3 Z e - t sin tdt + C (Give say 7 points if students get to here correctly.) (4) Compute R e - t sin tdt by integration by parts twice. In particular: Let u = sin t, dv = e - t dt, du = cos tdt v = - e - t . Then Z e - t sin tdt = - e - t sin t + Z e - t cos tdt. Now let u = cos t, dv = e - t dt, du = - sin tdt v = - e - t . Then Z e - t cos tdt = - e - t cos t - Z e - t sin tdt. Hence Z e - t sin tdt = - e - t sin t - e - t cos t - Z e - t sin tdt, so that Z e - t sin tdt = - 1 2 e - t (sin t + cos t ) . So y ( t ) = e t - e - t + 3 Z e - t sin tdt + C = - 1 - 3 2 (sin t + cos t ) + Ce t . Now y (0) = y 0 , so that y 0 = y (0) = - 1 - 3 2 + C = - 5 2 + C. Therefore y ( t ) = - 1 - 3 2 (sin t + cos t ) + y 0 + 5 2 e t .
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The only way this can be bound is if y 0 + 5 2 = 0 , that is y 0 = - 5 2 . 2. ( § 2.2, #24. 10 points) Solve the initial value problem
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Unformatted text preview: y = 2-e x 3 + 2 y , y (0) = 0 and determine where the solution attains its maximum value. ANSWER: This is a separable equation: (3 + 2 y ) dy = (2-e x ) dx, so (integrate) 3 y + y 2 = 2 x-e x + C. Since y (0) = 0 , we have 0 =-1 + C, so that C = 1 . We get 2 x-e x + 1 = 3 y + y 2 = ± y + 3 2 ¶ 2-9 4 , so that ± y + 3 2 ¶ 2 = 2 x-e x + 13 4 . We take the positive square root to get: y =-3 2 + r 2 x-e x + 13 4 . The x value where y attains its maximum is where 2 x-e x + 13 4 attains its maximum. Settin the derivative to be zero, we get 0 = d dx ± 2 x-e x + 13 4 ¶ = 2-e x . So it is where e x = 2 , that is, x = ln2 ....
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I will assign readings listed below, which will have associated exercises due in class the next day. Readings will come from my supplemental notes and from the textbook, which is the 3rd Edition of University Calculus: Early Transcendentals by Hass et al published by Addison Wesley (Pearson). I will also (hopefully) assign some videos of me working out examples, especially when I want to show you a different way of doing things from the textbook's.
1. Review:
• Date assigned: January 7 Monday.
• Date due: January 8 Tuesday.
• Reading from the textbook: Review at least Sections 5.4 and 5.5.
• Exercises due:
1. If F′(x) = f(x) for all x, then what is ∫ f(x) dx, that is the indefinite integral of f(x) with respect to x?
2. If ∫ f(x) dx = F(x) + C, then what is ∫bx=af(x) dx, that is the definite integral of f(x) with respect to x as x runs from a to b?
2. Work:
• Date assigned: January 8 Tuesday.
• Date due: January 9 Wednesday.
• Reading from the textbook: Pages 376–380 (§6.5).
• Exercises due:
1. Suppose that a variable force is applied on object moving from a to b along the x-axis, such that the amount of the force in the direction of the positive x-axis is a function F of the object's position along the x-axis. Write down an integral for the work done on the object by that force.
2. Suppose that a spring with a spring constant k is stretched or compressed a distance x from its equilibrium length. Write down a formula for the force of restoration on the spring.
3. Moments:
• Date assigned: January 9 Wednesday.
• Date due: January 10 Thursday.
• Reading from the textbook: Pages 382–389 (§6.6).
• Exercises due:
1. If you wish to find the total mass of the plate described in Exercise 6.6.15 on page 389 of the textbook, will you do an integral with respect to x or an integral with respect to y?
2. Give the formula for the centre of mass (, ȳ) in terms of the total mass M and the moments Mx and My.
4. Differential equations:
• Date assigned: January 10 Thursday.
• Date due: January 11 Friday.
• Reading from my notes (first set):
• Section 6 (the middle of page 4);
• Sections 8&9 (most of pages 5–7);
• Optional: Sections 10&11 (the rest of page 7 and page 8).
• Reading from the textbook: Pages 403–409 (§7.2).
• Exercises due: Fill in the blanks with vocabulary words:
1. An equation with differentials or derivatives in it is a(n) _____ equation.
2. A quantity is growing _____ if it is growing in such a way that its rate of growth is proportional to its size.
5. Integration by parts:
• Date assigned: January 14 Monday.
• Date due: January 15 Tuesday.
• Reading from my notes (first set): Section 7 (the bottom of page 4 and the top of page 5).
• Reading from the textbook: Pages 423–427 (§8.1).
• Exercises due: Use the formula ∫ u dv = uv − ∫ v du for integration by parts.
1. To find ∫ x ex dx, what should be u and what should be dv?
2. To find ∫ x ln x dx, what should be u and what should be dv?
6. Trigonometric integration:
• Date assigned: January 15 Tuesday.
• Date due: January 16 Wednesday.
• Reading from the textbook: Pages 429–434 (§8.2).
• Exercises due:
1. To find ∫ sin2x dx, what trigonometric identity would you use? That is, sin2x = _____?
2. For which (if any) of the following other integrals would you also need to use the identity from the previous exercise?
1. ∫ sin2x cos2x dx,
2. ∫ sin2x cos3x dx,
3. ∫ sin3x cos2x dx,
4. ∫ sin3x cos3x dx.
7. Trigonometric substitution:
• Date assigned: January 16 Wednesday.
• Date due: January 17 Thursday.
• Reading from the textbook: Pages 435–438 (§8.3).
• Exercise due: Give a trigonometric substitution that will help to find ∫ √(x2 + 25) dx. (This one might take a little more work than most of these daily homework exercises need, but keep in mind that you just need to state the trigonometric substitution; you do not have to work out the entire integral.)
8. Partial fractions:
• Date assigned: January 17 Thursday.
• Date due: January 18 Friday.
• Reading from the textbook: Pages 440–445 (§8.4).
• Exercise due: Suppose that a rational expression has the denominator (x2 + 3)(x + 3)2.
1. What are the denominators of its partial fractions?
2. For each of these, indicate the maximum degree of the numerator. (If you wish, you may do this by writing a general form for each numerator, such as A or Ax + B.)
9. Integration using computers and tables:
• Date assigned: January 22 Tuesday.
• Date due: January 23 Wednesday.
• Skim pages T1–T6 (the table of integrals).
• Optional: Look at these some or all of these computer algebra systems that will do integrals (as well as much more):
• Sage (free to download, free for most uses online, \$14.00 per month or more to make intensive computations online);
• Wolfram Mathematica (\$160.00 to download, \$80.00 per year online, free 15-day trial);
• Wolfram Alpha (free for some uses online, \$4.75 per month for advanced features, \$2.99 for a smartphone app with intermediate features).
• Exercise due: Which entry in the table of integrals in the back of the textbook (pages T1–T6) tells you how to integrate ∫ (x2 + 3)−1/2 dx without using hyperbolic functions (sinh, cosh, etc) or their inverses? (Hints: That table doesn't use negative exponents; it uses fractions instead. And it doesn't use fractional exponents; it uses roots instead.)
10. Numerical integration:
• Date assigned: January 23 Wednesday.
• Date due: January 24 Thursday.
• Reading from the textbook: Pages 452–459 (§8.6).
• Exercises due:
1. Which numerical method of integration approximates a function with a piecewise-linear continuous function?
2. Which numerical method of integration approximates a function with a piecewise-quadratic continuous function?
11. Improper integrals:
• Date assigned: January 24 Thursday.
• Date due: January 25 Friday.
• Reading from the textbook: Pages 462–467 (all of §8.7 except for Tests for Convergence and Divergence).
• Exercise due: Consider the integral ∫x=−∞ (x2 + |x|1/2)−1 dx, that is the integral of (x2 + |x|1/2)−1 dx as x runs from −∞ to ∞. List all of the reasons why this integral is improper.
12. Infinite sequences:
• Date assigned: January 25 Friday.
• Date due: January 28 Monday.
• Reading from the textbook: Pages 478–487 (§9.1).
• Reading from my notes (first set): the introduction to Chapter 6 (page 36 and the top of page 37).
• Exercises due: Exercises 2 and 4 from §9.1 (on page 487) in the textbook.
13. Infinite series:
• Date assigned: January 28 Monday.
• Date due: January 29 Tuesday.
• Page 490 and through the top of page 492 (§9.2: introduction);
• Example 6 on page 494 and through page 497 (all of the rest of §9.2 except for Geometric Series).
• Reading from my notes (first set): Sections 6.1&6.2 (the rest of page 37 and through the top of page 40).
• Exercises due: Fill in the vocabulary words:
1. limn→∞an, the limit of an as n goes to infinity, is the limit of a(n) ___.
2. Σn=0an, the sum of an as n runs from zero to infinity, is the sum of a(n) ___.
14. Evaluating special series:
• Date assigned: January 30 Wednesday.
• Date due: January 31 Thursday.
• Reading from my notes (second set): Section 6.3 (the rest of page 40 and the first half of page 41).
• Reading from the textbook: The rest of page 492 and through the end of Example 5 on page 494 (§9.2: Geometric Series).
• Exercises due: Finish these formulas and attach any conditions necessary for them to be true:
1. The sum of bn+1 − bn as n runs from i to infinity (where i is a natural number and b is an infinite sequence of real numbers): Σn=i (bn+1 − bn) = _____.
2. The sum of rn as n runs from i to infinity (where i is a natural number and r is a real number): Σn=irn = _____.
15. The Integral Test:
• Date assigned: January 31 Thursday.
• Date due: February 1 Friday.
• Reading from the textbook: Pages 499–504 (§9.3).
• Reading from my notes (second set): Section 6.4 through The p-Series Test (the rest of page 41 and through the top half of page 43).
• Exercises due:
1. Does the Integral Test apply to the function f(x) = sin2x)? Why or why not?
2. For which values of p does Σn=1 (1/np), the sum of 1/np as n runs from 1 to infinity, converge?
16. Comparison tests for integrals:
• Date assigned: February 1 Friday.
• Date due: February 4 Monday.
• Reading from the textbook: The rest of page 468 and through page 470 (§8.7: Tests for Convergence and Divergence).
• Exercises due: Suppose that you want to know whether ∫x=1 (x − 1)/x2 dx, the infinite integral of (x − 1)/x2, converges.
1. Knowing that the infinite integral of 1/x2 converges, can you use the Direct Comparison Test to decide?
2. Knowing that the infinite integral of 1/x2 converges, can you use the Limit Comparison Test to decide?
3. Knowing that the infinite integral of 1/x diverges, can you use the Direct Comparison Test to decide?
4. Knowing that the infinite integral of 1/x diverges, can you use the Limit Comparison Test to decide?
17. Comparison tests for series:
• Date assigned: February 4 Monday.
• Date due: February 5 Tuesday.
• Reading from my notes (second set): the rest of Section 6.4 through The Limit Comparison Test (the rest of page 43).
• Reading from the textbook: Pages 506–509 (§9.4).
• Exercises due: Suppose that you want to know whether Σn=1 (n − 1)/n2, the infinite series of (n − 1)/n2, converges.
1. Knowing that the infinite series of 1/n2 converges, can you use the Direct Comparison Test to decide?
2. Knowing that the infinite series of 1/n2 converges, can you use the Limit Comparison Test to decide?
3. Knowing that the infinite series of 1/n diverges, can you use the Direct Comparison Test to decide?
4. Knowing that the infinite series of 1/n diverges, can you use the Limit Comparison Test to decide?
18. Absolute convergence:
• Date assigned: February 6 Wednesday.
• Date due: February 7 Thursday.
• Reading from my notes (second set): The Absolute Convergence Test from Section 6.4 (the top of page 44).
• Pages 510&511 (§9.5: introduction);
• Page 519 and through the paragraph following Example 5 on page 520 (§9.6: Conditional Convergence; Rearranging Series).
• Exercises due:
1. If a series converges, is it necessarily true that its series of absolute values also converges?
2. If the series of absolute values converges, is it necessarily true that the original series converges?
19. The Ratio and Root Tests:
• Date assigned: February 7 Thursday.
• Date due: February 8 Friday.
• Reading from my notes (second set): the rest of Section 6.4 through The Root Test (the rest of the top half of page 44).
• Reading from the textbook: Pages 512–515 (§9.5: The Ratio Test; The Root Test).
• Exercises due:
1. Under what circumstances does the Ratio Test not tell you whether a series converges?
2. Under what circumstances does the Root Test not tell you whether a series converges?
3. If the Ratio Test doesn't tell you, is it possible that the Root Test will?
20. Alternating series:
• Date assigned: February 8 Friday.
• Date due: February 11 Monday.
• Reading from my notes (second set): the rest of Section 6.4 (the rest of page 44).
• Pages 516–518 (§9.6: introduction);
• The rest of page 520 (§9.6: Summary of Tests).
• Exercises due: Identify which of these series are alternating:
1. The sum of (−1)n/(2 + n);
2. The sum of (2 + (−1)n)/n;
3. The sum of (cos n)/n.
21. Power series:
• Date assigned: February 11 Monday.
• Date due: February 12 Tuesday.
• Reading from the textbook: Pages 522–530 (§9.7).
• Exercises due: Which of the following are power series (in the variable x)?
1. Σn=0n2(x − 3)n
2. Σn=5 (2x − 3)n
3. Σn=0 (√x − 3)n
4. 5 + 7x − 3x3
22. Taylor polynomials:
• Date assigned: February 13 Wednesday.
• Date due: Feburary 14 Thursday.
• Reading from my notes (second set): Chapter 7 through Section 7.1 (page 45 and through the first half of page 47).
• Pages 534–536 (§9.8: Taylor Polynomials), but don't worry yet about when they talk about infinite series and convergence;
• Page 537 and through Theorem 24 on page 539 (§9.9: introduction; the very beginning of Estimating the Remainder).
• Exercises due:
1. If a function f is to have a good approximation on an interval by a polynomial of degree at most k, then it's best if its derivative of what order is close to zero on that interval? (Its first derivative, its second derivative, its kth derivative, or what?)
2. Given a function f and a number a, if f is differentiable k times (at least) at a, then let Pk be the Taylor polynomial of order k generated by f at a and let Rk be the Taylor remainder of order k generated by f at a (so Pk and Rk are also functions). True or false: For a given real number x, the limit, as k → ∞, of Rk(x) is zero, if and only if the limit, as k → ∞, of Pk(x) (which is the sum of an infinite Taylor series) converges to f(x).
23. Taylor series:
• Date assigned: Feburary 14 Thursday.
• Date due: February 15 Friday.
• Reading from my notes (second set): The beginning of Section 7.2 (the rest of page 47).
• Reading from the textbook: Pages 532–534 (the rest of §9.8).
• Exercises due:
1. Fill in the blank: If f is a function whose derivatives of all orders exist everywhere, then the Maclaurin series generated by f is the Taylor series generated by f at ___.
2. True or false: Whenever a is a constant and f is a function whose derivatives of all orders exist at a, if the Taylor series generated by f at a converges anywhere, then it must converge to f there.
24. The Binomial Theorem:
• Date assigned: February 15 Friday.
• Date due: February 18 Monday.
• Reading from the textbook: Page 543 and through Example 2 on page 545 (§9.10: introduction; The Binomial Series for Powers and Roots).
• Reading from my notes (second set): The parts of Section 7.2 about the Binomial Theorem (the first line in the list on page 48 and the paragraph in the middle of page 48).
• Exercises due:
1. Using the Binomial Theorem, expand (x + 1)6.
2. Using the Binomial Theorem, write (1 + x2)−1 as an infinite series (assuming that x2 < 1 so that the series converges).
25. More common Taylor series:
• Date assigned: February 18 Monday.
• Date due: February 19 Tuesday.
• The rest of page 539 and through page 541 (the rest of §9.9);
• The rest of page 545 and through page 548 (the rest of §9.10).
• Reading from my notes (second set): The rest of Section 7.2 (the rest of page 48 and page 49).
• Exercises due:
1. Following Exercise 2 from the previous assignment, integrate your answer to get an infinite series for atan x = tan−1x.
2. Can any limit using L'Hôpital's Rule be done using Taylor polynomials or series instead? Explain why or why not.
26. Graphs in three dimensions:
• Date assigned: February 20 Wednesday.
• Date due: February 21 Thursday.
• Reading from the textbook: Pages 596–599 (§11.1).
• Reading from my notes (second set): the first half of page 1 (introduction, §1).
• Exercises due:
1. In a right-handed rectangular coordinate system using the variables x, y, and z, if you curl the fingers of your right hand from the direction of the positive x-axis to the direction of the positive y-axis and stick out your thumb, then in what direction approximately should your thumb point?
2. What is the name of a shape whose equation in a three-dimensional rectangular coordinate system is linear and contingent? (For example, 2x + 3y + 5z = 8.)
3. What is the equation in the rectangular (x, y, z)-coordinate system of a sphere whose radius is r and whose centre is (h, k, l)?
27. Vectors:
• Date assigned: February 21 Thursday.
• Date due: February 22 Friday.
• Reading from my notes (last set): the rest of page 1 and through the first half of page 5 (§§2–4).
• From page 601 to the very top of page 605 (§11.2: introduction; Component Form; Vector Algebra Operations);
• The bottom half of page 606 (§11.2: Midpoint of a Line Segment).
• Exercises due:
1. Give a formula for the vector from the point (x1, y1) to the point (x2, y2).
2. If u, v, and w are vectors, simplify the expression 2(u + 3v) − 6(v − 3w) − 18(w + u/9).
28. Length and angle:
• Date assigned: February 22 Friday.
• Date due: February 25 Monday.
• Reading from my notes (last set): the rest of page 5 and through the top of page 7 (§5).
• Pages 605&606 (§11.2: Unit Vectors);
• Page 607&608 (§11.2: Applications).
• Exercises due:
1. Give a formula for the magnitude (or norm, or length) of the vector ⟨a, b, c⟩.
2. If |u + v|2 = |u|2 + |v|2, then what is the angle between u and v? (Hint: Use the Pythagorean Theorem, the Law of Cosines, or the formula for angles in terms of lengths in my notes; any one of these could do the job! You might also want to draw a picture.)
29. The dot product:
• Date assigned: February 25 Monday.
• Date due: February 26 Tuesday.
• Reading from my notes (last set): the rest of page 7 and through the top half of page 10 (§§6–8).
• Reading from the textbook: Pages 610–615 (§11.3).
• Exercises due:
1. State a formula for the dot product u ⋅ v of two vectors using only their lengths |u| and |v|, the angle ∠(u, v) between them, and real-number operations.
2. State a formula for the vector projection of u onto v using only dot products and real-number operations (so in particular, no lengths or angles unless expressed using dot products).
30. The cross product:
• Date assigned: February 26 Tuesday.
• Date due: February 27 Wednesday.
• Reading from my notes (last set): the rest of page 10 and through the top of page 14 (§§9–12).
• Reading from the textbook: Pages 618–622 (§11.4).
• Exercises due:
1. State a formula for the magnitude |u × v| of the cross product of two vectors u and v, using only their lengths |u| and |v|, the angle ∠(u, v) between them, and real-number operations.
2. If u and v are vectors in 2 dimensions, then is u × v a scalar or a vector?
3. If u and v are vectors in 3 dimensions, then is u × v a scalar or a vector?
31. Parametrized curves:
• Date assigned: February 28 Thursday.
• Date due: March 1 Friday.
• Reading from the textbook: Pages 557–562 (§10.1).
• Reading from my notes (last set):
• Most of page 15 and the first paragraph of page 16 (about the first half of §14);
• Optional: the rest of page 16 and the top of page 17 (the rest of §14).
• Exercises due: Define a parametrized curve by (x, y) = (2t2, 3t3) for 0 ≤ t ≤ 2.
1. Which variable(s) is/are the parameter(s)?
2. What are the beginning/initial point and the ending/final/terminal point of the curve?
32. Geometry with vectors:
• Date assigned: March 1 Friday.
• Date due: March 4 Monday.
• Reading from the textbook: Pages 624–630 (§11.5).
• Reading from my notes (last set): the rest of page 14 and the top of page 15 (§13).
• Exercises due:
1. Give a parametrization for the line through the point (x0, y0, z0) and parallel to the vector ⟨a, b, c⟩.
2. Give an equation for the plane through the point (x0, y0, z0) and perpendicular to the vector ⟨a, b, c⟩.
33. Calculus with parametrized curves:
• Date assigned: March 4 Monday.
• Date due: March 5 Tuesday.
• Reading from the textbook: Page 564 and through the end of Example 3 on page 566 (§10.2: introduction; Tangents and Areas).
• Reading from my notes (last set): the rest of page 17 and page 18 (§15).
• Exercises due: If x and y are each functions of t:
1. Give a formula for the derivative of y with respect to x in terms of the derivatives of x and y with respect to t. (There is basically only one possible correct answer to this.)
2. Give a formula for the second derivative of y with respect to x in terms of derivatives of x and y with respect to t. (There is more than one possible correct answer to this, and you only need to give one of them, but make sure that all of the derivatives appearing are with respect to t as required!)
34. Arclength of parametrized curves:
• Date assigned: March 5 Tuesday.
• Date due: March 6 Wednesday.
• Reading from the textbook: Pages 566 to 572 (the rest of §10.2).
• Reading from my notes (last set): page 19 (§16).
• Exercises due:
1. If a curve is parametrized by x = f(t) and y = g(t) for a ≤ t ≤ b, then what integral gives the length of this curve?
2. How does changing the parametrization of a curve affect its arclength?
35. Polar coordinates:
• Date assigned: March 7 Thursday.
• Date due: March 8 Friday.
• Reading from the textbook: Pages 574–577 (§10.3).
• Exercises due: True or false:
1. If a point in the coordinate plane is given by polar coordinates (r, θ), then r ≥ 0 and 0 ≤ θ < 2π.
2. Every point in the coordinate plane can be given by polar coordinates (r, θ) such that r ≥ 0 and 0 ≤ θ < 2π.
36. Graphs in polar coordinates:
• Date assigned: March 8 Friday.
• Date due: March 11 Monday.
• Reading from the textbook: Pages 578–580 (§10.4).
• Exercises due: Suppose that a curve is parametrized in polar coordinates by r = f(θ) for some differentiable function f. (In the following answers, refer directly to only f, its derivatives, and θ.)
1. What is the slope of the curve at a given value of θ?
2. Under what circumstances is this slope undefined?
37. Area in polar coordinates:
• Date assigned: March 11 Monday.
• Date due: March 12 Tuesday.
• Reading from the textbook: Pages 581&582 and most of page 583 (§10.5: introduction; Area in the Plane).
• Exercises due:
1. Can the area of a region in the plane ever be negative?
2. What is the formula for the area of the region satisfying f(θ) ≤ r ≤ g(θ) and α ≤ θ ≤ β in polar coordinates? (Assume that α and β are real numbers with α ≤ β and β − α ≤ 2π, and that f and g are continuous functions defined at least on [α, β] with 0 ≤ f(θ) ≤ g(θ) whenever α ≤ θ ≤ β.)
38. Length in polar coordinates:
• Date assigned: March 12 Tuesday.
• Date due: March 13 Wednesday.
• Reading from the textbook: The rest of page 583 and page 584 (§10.5: Length of a Polar Curve).
• Exercises due:
1. Can the length of a curve ever be negative?
2. What is the formula for the length of the curve given by r = f(θ) and α ≤ θ ≤ β in polar coordinates? (Assume that α and β are real numbers with α ≤ β, and that f is a continuously differentiable function defined at least on [α, β] with (f(θ1), θ1) always defining a different point in polar coordinates than (f(θ2), θ2).)
That's it!
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This web page was written between 2003 and 2019 by Toby Bartels, last edited on 2019 January 18. Toby reserves no legal rights to it.
The permanent URI of this web page is `http://tobybartels.name/MATH-1700/2019WN/homework/`.
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RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3
# RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3
## RD Sharma Class 10 Solutions Constructions Exercise 11.3
Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
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Solution:
Steps of construction :
(i) Draw a circle with O centre and 6 cm radius.
(ii) Take a point P, 10 cm away from the centre O.
(iii) Join PO and bisect it at M.
(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.
(v) Join PT and PS.
Then PT and PS are the required tangents.
Question 2.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3 cm.
(ii) Draw a diameter and produce it to both sides.
(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.
(iv) Bisect PO at M and QO at N
(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.
(vi) Join PS, PT, QS’ and QT’.
Then PS, PT, QS’ and QT’ are the required tangents to the given circle.
Question 3.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. [CBSE 2013] Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.
(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.
(iii) Bisect AB at M.
(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.
(v) Join AS, AT, BS’and BT’.
Then AS, AT, BS’ and BT’ are the required tangent.
Question 4.
Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3.5 cm
(ii) Take a point P which is 6.2 cm from O.
(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.
(iv) Join PT and PS.
PT and PS are the required tangents to circle.
Question 5.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°. [CBSE 2013] Solution:
Steps of construction :
Angle at the centre 180° – 45° = 135°
(i) Draw a circle with centre O and radius 4.5 cm.
(ii) At O, draw an angle ∠TOS = 135°
(iii) At T and S draw perpendicular which meet each other at P.
PT and PS are the tangents which inclined each other 45°.
Question 6.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Solution:
Steps of Construction :
Draw a line segment BC = 8 cm
From B draw an angle of 90°
Draw an arc $$\breve { BA }$$ = 6cm cutting the angle at A.
Join AC.
ΔABC is the required A.
Draw ⊥ bisector of BC cutting BC at M.
Take M as centre and BM as radius, draw a circle.
Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification :
∠ABC = 90° (Given)
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also AE is a tangent to the circle.
Question 7.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length. [CBSE 2016] Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.
Steps of construction :
(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.
(ii) Taking any point P on outer circle. Join OP.
(iii) Bisect OP, let M’ be the mid-point of OP.
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.
(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.
(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.
Actual calculation:
In right angle ΔOMP, ∠PMO = 90°
∴ PM2 = OP2 – OM2
[by Pythagoras theorem i.e. (hypotenuse)2 = (base)2 + (perpendicular)2] ⇒ PM2 = (5)2 – (3)2 = 25 – 9 = 16
⇒ PM = 4 cm
Hence, the length of both tangents is 4 cm.
### RD Sharma Class 10th Solutions Chapter 11 Constructions Exercise 11.3 Q1
RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3 Q2
Q3
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Numerical Examples - Stereographic Projection
# Numerical Examples for Stereographic Projection #
## SPHERE #
### Forward Equations #
Given
Radius of sphere: $R=\;\;$ units Center: $\phi_1=\;$ ° $\lambda_0=\;$ ° Central scale factor: $k_0=\;$ Point: $\phi=\;$ ° $\lambda=\;$ °
Find $x, y, k$
Using equations (21-4), (21-2), and (21-3) in order,
\eqalign { k &= 2\times 1.0/[1+\sin40^\circ\sin30^\circ+\cos40^\circ\cos30^\circ\cos(-75^\circ-(-100^\circ))] \cr &= 1.0402304 }
\eqalign{ x &= 1.0\times1.0402304\cos30^\circ\sin(-75^\circ-(-100^\circ)) \cr &= 0.3807224\;\text{units} }
\eqalign{ y &= 1.0\times1.0402304[\cos40^\circ\sin30^\circ - \sin40^\circ\cos30^\circ\cos(-75^\circ-(-100^\circ))] \cr &= -0.1263802\;\text{units} }
Examples of other forward equations are omitted, since the above equations are general.
### Inverse Equations #
Inversing forward example:
Given $R, \phi_1, \lambda_0, k_0$ for forward example
Point: $x=\;$ units $y=\;$ units
Find $\phi, \lambda$
Using equations (21-18) and (21-19),
$$\rho = [0.3807224^2 + (-0.1263802)^2]^{1/2} = 0.4011502\;\text{units}$$
\eqalign{ c &= 2 \arctan[0.4011502/(2\times1.0\times1.0)] \cr &= 22.6832261^\circ }
Using equations (21-14) and (21-15),
\eqalign{ \phi =& \arcsin[\cos22.6832261^\circ\sin40^\circ + (-0.1263802)\sin22.6832261^\circ \cr & \cos40^\circ/0.4011502] \cr =& 29.9999991^\circ }
\eqalign{ \lambda =& -100^\circ + \arctan[-0.1503837\sin12.9082572^\circ/(0.2233906 \cr &\cos40^\circ\cos12.9082572^\circ - (-0.1651911)\sin40^\circ \sin12.9082572^\circ)] \cr =& -109.9999978^\circ }
## ELLIPSOID #
### Oblique Aspect #
#### Forward Equations #
Given:
Clarke 1866WGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00676866 $e=$ 0.0822719 Center: $\phi_1=$ ° $\lambda_0=$ ° Central scale factor: $k_0=$ Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$
From equation (3-1),
\eqalign{ \chi_1 =& 2\arctan\{ \tan(45^\circ + 40^\circ/2)[(1-0.0822719\sin40^\circ)/ \cr & (1+0.0822719\sin40^\circ)]^{0.0822719/2} \} -90^\circ \cr =& 2\arctan2.1351882-90^\circ \cr =& 39.8085922^\circ }
\eqalign{ \chi =& 2\arctan\{ \tan(45^\circ + 30^\circ/2)[(1-0.0822719\sin30^\circ)/ \cr & (1+0.0822719\sin30^\circ)]^{0.0822719/2} \} -90^\circ \cr =& 2\arctan1.7261956-90^\circ \cr =& 29.8318339^\circ }
From equation (14-15),
\eqalign{ m_1 &= \cos40^\circ/(1-0.0067687\sin^240^\circ)^{1/2} \cr &= 0.7671179 }
\eqalign{ m &= \cos30^\circ/(1-0.0067687\sin^230^\circ)^{1/2} \cr &= 0.8667591 }
From equation (21-27),
\eqalign{ A =& 2\times6378206.4\times0.9999\times0.7671179/\{ \cos39.8085922^\circ \cr & [1+\sin39.8085922^\circ\sin29.8318339^\circ + \cos39.8085922^\circ \cr & \cos29.8318339^\circ\cos(-90^\circ-(-100^\circ))]\} \cr =& 6450107.68\;\text{m} }
From equations (21-24), (21-25), and (21-26),
\eqalign{ x &= 6450107.68\cos29.8318339^\circ\sin(-90^\circ-(-100^\circ)) \cr &= 971630.79\;\text{m} }
\eqalign{ y =& 6450107.68[\cos39.8085922^\circ\sin29.8318339^\circ \cr & - \sin39.8085922^\circ\cos29.8318339^\circ\cos(-90^\circ-(-100^\circ))] \cr =& -1063049.26\;\text{m} }
\eqalign{ k &= 6378206.40\cos29.8318339^\circ/(6378206.40\times0.8667591) \cr &= 1.0121248 }
#### Inverse Equations #
Inversing forward example:
Given
$x=\;$m $y=\;$m
Find: $\phi, \lambda$
From equation (14-15),
\eqalign{ m_1 &= \cos40^\circ/(1-0.0067687\sin^240^\circ)^{1/2} \cr &= 0.7671179 }
From equation (3-11), as in the forward oblique example,
$$\chi_1 = 39.8085922^\circ$$
From equations (20-18) and (21-38),
\eqalign{ \rho &= [971630.79^2 + (-1063049.26)^2]^{1/2} \cr &= 1440187.53\text{ m} }
\eqalign{ c_e =& 2\arctan[1440187.57\cos39.8085922^\circ/(2\times6378206.40 \cr & \times0.9999)\times0.7671179] \cr =& 12.9018251^\circ }
From equation (21-37),
\eqalign{ \chi =& \arcsin[\cos12.9018251^\circ\sin0.6947910040690184+(-1063049.3)\sin12.9018251^\circ\cr & \cos39.8085922^\circ/1440187.57] \cr =& 29.8318335^\circ }
Using $\chi$ as the first trial $\phi$ in equation (3-4),
\eqalign{ \phi =& 2\arctan\{ \tan(45^\circ + 29.8318335^\circ/2)[(1-0.0822719\sin29.8318335^\circ)/ \cr & (1+0.0822719\sin29.8318335^\circ)]^{0.0822719/2} \} -90^\circ \cr =& 29.9991438^\circ }
Using this new trial value in the same equation for $\phi$, not for $\chi$,
\eqalign{ \phi =& 2\arctan\{ \tan(45^\circ + 29.8318335^\circ/2)[(1-0.0822719\sin29.9991438^\circ)/ \cr & (1+0.0822719\sin29.9991438^\circ)]^{0.0822719/2} \} -90^\circ \cr =& 29.9999953^\circ }
Repeating with $29.9999953^\circ$ in place of $29.9991438^\circ$ , the next trial $\phi$ is
$$\phi = 29.9999996^\circ$$
The next trial calculation produces the same $\phi$ to seven decimals. Therefore, this is $\phi$.
Using equation (21-36),
\eqalign{ \lambda =& -100^\circ+\arctan[971630.8\sin12.9018251^\circ \cr & (1440187.57\cos39.8085922^\circ\cos12.9018251^\circ \cr & -(-1063049.30)\sin39.8085922^\circ\sin12.9018251^\circ)] \cr =& -100^\circ+\arctan(216946.86/1230366.77) \cr =& -100^\circ+10.0000000^\circ \cr =& -90.0000000^\circ }
Instead of the iterative equation (3-4), series equation (3-5) may be used (omitting terms with $e^8$ here for simplicity):
\eqalign{ \phi =& 29.8318335^\circ\times \pi/180^\circ + (0.0067687/2 + 5\times0.0067687^2/24 \cr & 0.0067687^3/12)\sin(2\times29.8318335^\circ)+(7\times0.0067687^2/48 \cr & + 29\times0.0067687^3/240)\sin(4\times29.8318335^\circ) \cr & + (7\times0.0067687^3/120)\sin(6\times29.8318335^\circ) \cr =& 0.5235988\text{ radian} \cr =& 29.9999995^\circ }
### Polar Aspect With Known $k_0$ #
#### Forward Equations #
Given:
InternationalWGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00672267 $e=$ 0.0819919 Center: $\phi_1=$ -90° (South Pole)90° (North Pole) $\lambda_0=$ ° Central scale factor: $k_0=$ Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$
Since this is the south polar aspect, for calculations change signs of $x, y, \phi_1, \lambda_1$, and $\lambda_0$: $\lambda_0=100^\circ$ , $\phi=75^\circ$ , $\lambda=-150^\circ$
Using equations (15-9) and (21-33),
\eqalign{ t &= \tan(45^\circ-75^\circ/2)/[(1-0.0822719\sin 75^\circ)/(1+0.0822719\sin 75^\circ)]^{0.0822719/2} \cr &= 0.1325179 }
\eqalign{ \rho =& 2\times6378388.0\times0.994\times0.1325120/[(1+0.0819919)^{(1+0.0819919)} \cr & \times(1-0.0819919)^{(1-0.0819919)}] \cr =& 1674638.30\text{ m} }
Using equations (21-30) and (21-31),
\eqalign{ x &= 1674638.31\sin(-150^\circ-100^\circ) \cr &= 1573645.26\text{ m} }
\eqalign{ y &= -1674638.31\cos(-150^\circ-100^\circ) \cr &= 572760.03\text{ m} }
Changing signs of x and y for the south polar aspect,
$$x = -1573645.26\text{ m}$$
$$y = -572760.03\text{ m}$$
From equation (14-15),
\eqalign{ m &= \cos 75^\circ/(1-0.0067227\sin^275^\circ)^{1/2}\cr &= 0.2596346 }
From equation (21-32),
\eqalign{ k &= 1674638.31/(6378388.0\times0.2596346) \cr &= 1.0112244 }
#### Inverse Equations #
Inversing forward example:
Given
$x=\;$m $y=\;$m
Find: $\phi, \lambda$
Since this is the south polar aspect, for calculations change signs as stated in text: $\lambda_0=100^\circ$ , $x=1573645.3\text{ m}$ , $y=572760.0\text{ m}$ .
From equations (20-18) and (21-39),
\eqalign{ \rho &= (1573645.3^2 + 572760.0^2)^{1/2} \cr &= 1674638.33\text{ m} }
\eqalign{ t =& 1674638.33\times[(1+0.0819919)^{(1+0.0819919)} \cr & (1-0.0819919)^{(1-0.0819919)}]^{1/2}/(2\times6378388.0\times0.994) \cr =& 0.1325120 }
To iterate with equation (7-9), use as the first trial $\phi$,
\eqalign{ \phi &= 90^\circ - 2\arctan 0.1325120 \cr &= 74.9031986^\circ }
Substituting in (7-9),
\eqalign{ \phi =& 90^\circ - 2\arctan\{0.1325120\times[(1-0.0819919\sin74.9031986^\circ)/ \cr & (1+0.0819919\sin74.9031986^\circ)]^{0.0819919/2} \} \cr =& 74.9999558^\circ }
Using this second trial $\phi$ in the same equation instead of $74.9031986^\circ$ ,
$$\phi = 74.9999997^\circ$$
The third trial gives the same value to seven places.
From equation (20-16), using ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range,
\eqalign{ \lambda &= 100^\circ + \arctan[1573645.3/(-572760.0)] \cr &= -150.0000016^\circ }
The sign of $\phi$ and $\lambda$ must be reversed for the south polar aspect. Finally,
$$\phi = -74.9999997^\circ$$
$$\lambda = 150.0000016^\circ$$
### Polar Aspect With Known $\phi_c$ #
#### Forward Equations #
Given:
InternationalWGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00672267 $e=$ 0.0819919 Standard parallel: $\phi_c=$ ° $\lambda_0=$ ° Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$
Since this is the south polar aspect, for calculations change signs of $x, y, \phi_c, \lambda_1$, and $\lambda_0$: $\phi_c=71^\circ$ , $\lambda_0=100^\circ$ , $\phi=75^\circ$ , $\lambda=-150^\circ$
Using equation (15-9),
\eqalign{ t &= \tan(45^\circ-75^\circ/2)/[(1-0.0819919\sin 75^\circ)/(1+0.0819919\sin 75^\circ)]^{0.0819919/2} \cr &= 0.1325120 }
For $t_c$ substitute $71^\circ$ in place of $75^\circ$ in (15-9), and
\eqalign{ t_c &= \tan(45^\circ-71^\circ/2)/[(1-0.0819919\sin 71^\circ)/(1+0.0819919\sin 71^\circ)]^{0.0819919/2} \cr &= 0.1684118 }
From equations (14-15) and (21-34),
\eqalign{ m_c &= \cos 71^\circ/(1-0.0067227\sin^271^\circ)^{1/2}\cr &= 0.3265509 }
\eqalign{ \rho &= 6378388.0\times0.3265509\times0.1325120/0.1684118 \cr &= 1638869.54\text{ m} }
Equations (21-30), (21-31), and (21-32) are used as in the preceding south polar example,
\eqalign{ x &= 1638869.54\sin(-150^\circ-100^\circ) \cr &= 1540033.61\text{ m} }
\eqalign{ y &= -1638869.54\cos(-150^\circ-100^\circ) \cr &= 560526.39\text{ m} }
Changing signs of x and y for the south polar aspect,
$$x = -1540033.61\text{ m}$$
$$y = -560526.39\text{ m}$$
\eqalign{ m &= \cos 75^\circ/(1-0.0067227\sin^275^\circ)^{1/2}\cr &= 0.2596346 }
\eqalign{ k &= 1638869.54/(6378388.0\times0.2596346) \cr &= 0.9896256 }
#### Inverse Equations #
Inversing forward example:
Given
$x=\;$m $y=\;$m
Find: $\phi, \lambda$
Since this is the south polar aspect, for calculations change signs as stated in text: $\phi_c=71^\circ$ , $\lambda_0=100^\circ$ , $x=1540033.6\text{ m}$ , $y=560526.4\text{ m}$ .
From equations (15-9) and (14-15), as calculated in the corresponding forward example,
\eqalign{ t_c &= \tan(45^\circ-71^\circ/2)/[(1-0.0819919\sin 71^\circ)/(1+0.0819919\sin 71^\circ)]^{0.0819919/2} \cr &= 0.1684118 }
\eqalign{ m_c &= \cos 71^\circ/(1-0.0067227\sin^271^\circ)^{1/2}\cr &= 0.3265509 }
From equations (20-18) and (21-40),
\eqalign{ \rho &= [1540033.6^2 + 560526.4^2]^{1/2} \cr &= 1638869.53\text{ m} }
\eqalign{ t &= 1638869.53\times0.1684118/(6378388.0\times0.3265509) \cr &= 0.1325120 }
For the first trial $\phi$ in equation (7-9)
\eqalign{ \phi &= 90^\circ - 2\arctan 0.1325120 \cr &= 74.9031989^\circ }
Substituting in (7-9),
\eqalign{ \phi =& 90^\circ - 2\arctan\{0.1325120\times[(1-0.0819919\sin74.9031989^\circ)/ \cr & (1+0.0819919\sin74.9031989^\circ)]^{0.0819919/2} \} \cr =& 74.9999561^\circ }
Replacing $74.9031989^\circ$ with $74.9999561^\circ$ , the next trial $\phi$ is
$$\phi = 75.0000001^\circ$$
The next iteration results in the same $\phi$ to seven places.
From equation (20-16), using ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range,
\eqalign{ \lambda &= 100^\circ + \arctan[1540033.6/(-560526.4)] \cr &= -149.9999997^\circ }
The sign of $\phi$ and $\lambda$ must be reversed for the south polar aspect. Finally,
$$\phi = -75.0000001^\circ$$
$$\lambda = 149.9999997^\circ$$
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$t=-13$
Using properties of equality, the solution to the given equation, $14=t+27$, is \begin{array}{l}\require{cancel} 14-27=t+27-27 \\\\ -13=t \\\\ t=-13 .\end{array}
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# Search by Topic
#### Resources tagged with Factors and multiples similar to Amazing Alphabet Maze:
Filter by: Content type:
Stage:
Challenge level:
### There are 105 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Little Squares
##### Stage: 1 Challenge Level:
Look at the squares in this problem. What does the next square look like? I draw a square with 81 little squares inside it. How long and how wide is my square?
### Heads and Feet
##### Stage: 1 Challenge Level:
On a farm there were some hens and sheep. Altogether there were 8 heads and 22 feet. How many hens were there?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Six in a Circle
##### Stage: 1 Challenge Level:
If there is a ring of six chairs and thirty children must either sit on a chair or stand behind one, how many children will be behind each chair?
### Constant Counting
##### Stage: 1 Challenge Level:
You can make a calculator count for you by any number you choose. You can count by ones to reach 24. You can count by twos to reach 24. What else can you count by to reach 24?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Doubling Fives
##### Stage: 1 Challenge Level:
This activity focuses on doubling multiples of five.
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### What's Left?
##### Stage: 1 Challenge Level:
Use this grid to shade the numbers in the way described. Which numbers do you have left? Do you know what they are called?
### Multiples
##### Stage: 1 Challenge Level:
This package will help introduce children to, and encourage a deep exploration of, multiples.
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Multiples Grid
##### Stage: 2 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Scoring with Dice
##### Stage: 2 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Add the Weights
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Gran, How Old Are You?
##### Stage: 2 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Domino Pick
##### Stage: 1 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Nineteen Hexagons
##### Stage: 1 Challenge Level:
In this maze of hexagons, you start in the centre at 0. The next hexagon must be a multiple of 2 and the next a multiple of 5. What are the possible paths you could take?
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### The Moons of Vuvv
##### Stage: 2 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Lots of Lollies
##### Stage: 1 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Fitted
##### Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Factor Lines
##### Stage: 2 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### In the Money
##### Stage: 2 Challenge Level:
There are a number of coins on a table. One quarter of the coins show heads. If I turn over 2 coins, then one third show heads. How many coins are there altogether?
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### Two Primes Make One Square
##### Stage: 2 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### What Is Ziffle?
##### Stage: 2 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Tom's Number
##### Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Ip Dip
##### Stage: 1 and 2 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
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http://www.statisticshowto.com/ancillary-statistic/
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# Ancillary Statistic: Simple Definition and Example
The term “ancillary statistic” is one of those terms that mean something slightly different depending on where you read about it.
Most (but not all) authors agree that an ancillary statistic is a distribution constant statistic that can be combined with a maximum likelihood estimator to create a minimally sufficient statistic. The ancillary statistic in this sense is defined as one part of a sufficient statistic that has a parameter free marginal distribution.
Some authors will describe an ancillary statistic as simply a statistic whose distribution doesn’t depend on the model parameters. In this context, an ancillary statistic is basically a summary of the data. Standing alone, it doesn’t give any information about the parameter P. For example, an ancillary statistic could be an estimator for a random sample size, but it isn’t an estimator for any specific sample size (Kardaun, 2006).
## As a Complement to Sufficient Statistics
Given the above definitions, it should be easy to see why ancillaries are sometimes referred to as the complement of a sufficient statistic. While a sufficient statistic contains all of the parameter’s information, an ancillary contains no information about the model parameter. Basu’s Theroem (as cited in Gosh, 2011) summarizes this idea:
If U is a complete statistic and a sufficient statistic for a parameter(θ), and if V is an ancillary statistic for θ, then U and V are independent.
Ancillary statistics are used in compound distributions and unconditional inference.
## Specific Types of Ancillary Statistic
• First order ancillary: an ancillary is first order if the statistic’s expected value is independent of the population.
• Trivial ancillary: Defined as the constant statistic V(X) ≡ c ∈ ℝ (Shao, 2008).
If an ancillary contains no information about population parameters, why use it at all? Ning-Zhing & Jian (2008) state two reasons why you might choose ancillary over sufficiency:
1. Invariance to the parameter. Invariant statistics are not easily changed by transformations, like simple data shifts.
2. Independence to sufficient statistics.
## References
Ghosh M. (2011) Basu’s Theorem. In: DasGupta A. (eds) Selected Works of Debabrata Basu. Selected Works in Probability and Statistics. Springer, New York, NY
Kardaun (2006). Classical Methods of Statistics: With Applications in Fusion-Oriented Plasma Physics. Springer Science & Business Media.
Ning-Zhing, S. & Jian, T. (2008). Statistical Hypothesis Testing: Theory and Methods. World Scientific.
Shao, J. Mathematical Statistics. Springer Science & Business Media.
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https://math.stackexchange.com/questions/4510020/rouches-theorem-application-and-find-mulitplicity
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# Rouche's Theorem application and find mulitplicity
Determine the number of zeros, with multiplicity, of the polynomial $$f(z) = 1+4z^3 +z^{10} +2z^{12}$$ inside the annulus $$\{z \in \mathbb{C}:1<|z|<2\}$$
First we consider the unit disk $$|z|<1$$.
Let $$g(z)=-z^{10}$$
Then $$|f(z)+g(z)|= |1+4z^3+2z^{12}|= 7 < 8 = |f(z)|$$
So by Rouche's theorem, $$f$$ and $$g$$ have the same number of zeros. Since $$g$$ has 10, $$f$$ has 10.
Now consider the disk $$|z|<2$$.
Let $$g(z)=-2z^{12}$$.
Then $$|f(z)+g(z)|= |1+2^4+2^{10}| < 2^{13}$$.
So So by Rouche's theorem, $$f$$ and $$g$$ have the same number of zeros. Since $$g$$ has 12, $$f$$ has 12.
Thus, in $$\{z \in \mathbb{C}:1<|z|<2\}$$, we get 2 zeros.
Now for the multiplicity, usually we want to check the derivative.
So we have $$f(z) = 1+4z^3 +z^{10} +2z^{12}$$ and $$f'(z)= 12z^2+10z^9+24z^{11}$$. But from here, I'm not sure how to find the zeros.
• $|f(z)| = 8$ $(|z|=1)$ is wrong. In fact $f(-1)=0$.
– Gerd
Aug 11, 2022 at 7:52
• For $|z|=1$ the term $4z^3$ is dominating. So you should take $g(z)=4z^3.$ For $f'$ the term $24z^{11}$ is dominating for $|z|=2$ and for $|z|=1.$ Aug 11, 2022 at 13:22
• @RyszardSzwarc How to show that $4z^3$ is dominating? Note that the inequality of Rouche's theorem on the boundary is strict. Moreover, Rouche's theorem includes the multiplicity of the roots and therefore, I think that there's no need to check $f'(z)$. Aug 11, 2022 at 14:23
• @on1921379 My bad. I have missed one term. Concerning multiplicity you cannot rely on Rouche theorem without studying $f'.$ For example $f(z)= 2z^n-1$ has simple roots but $2z^n$ doesn't, in the open unit ball. Aug 11, 2022 at 16:27
• @RyszardSzwarc You are right, Thanks. Aug 11, 2022 at 16:57
As was proved by the asker the function $$f(z)=1+4z^3+z^{10}+2z^{12}$$ has $$12$$ roots in the open unit ball $$|z|<2$$ as the term $$2z^{12}$$ is dominating on the circle $$|z|=2.$$
For $$0<\delta<1$$ consider the circle $$|z|=1-\delta.$$ When $$\delta\to 0^+$$ we get $$4|z|^3= 4-12\delta+o(\delta^2)$$ $$1+|z|^{10}+2|z|^{12}= 1+(1-10\delta)+2(1-12\delta)+o(\delta^2)=4-34\delta+o(\delta^2)$$ Therefore for $$\delta>0$$ small enough the term $$4z^3$$ is dominating on the circle $$|z|=1-\delta.$$ Hence the function $$f(z)$$ has $$3$$ roots in the open ball $$|z|<1-\delta$$ for every $$\delta>0$$ small enough. Hence $$f(z)$$ has $$3$$ roots in $$|z|<1.$$
It remains to determine the number of roots on the circle $$|z|=1.$$ Assume $$z_0$$ satisfies $$f(z_0)=0$$ and $$|z_0|=1.$$ Then $$-4z_0^3=1+z_0^{10}+2z_0^{12}$$ hence $$4=|1+z_0^{10}+2z_0^{12}|$$ The equality implies $$z_0^{10}=z_0^{12}=1,$$ i.e. $$z_0=\pm 1.$$ We have $$f(-1)=0$$ and $$f(1)\neq 0.$$
Summarizing the function $$f(z)$$ has $$12-3-1=8$$ roots in the region $$1<|z|<2.$$
Concerning multiplicity, for every root $$z_0$$ with multiplicity greater than $$1,$$ there holds $$f'(z_0)=0.$$ We have $$f'(z)= 12z^2+10z^9+24z^{11}$$ The term $$24z^{11}$$ is dominating on the circle $$|z|=1$$ as well as Therefore all $$11$$ roots of $$f'(z)$$ are located in the open ball $$|z|<1.$$ Thus all the roots in the region $$1\le |z|<2$$ are single.
Remark It turns out that also $$3$$ roots of $$f(z)$$ located in $$|z|<1$$ are single as well. It follows from the fact that (according to Wolphram Alpha GCD algorithm or this) the polynomials $$f(z)$$ and $$f'(z)$$ are relatively prime, i.e. they do not have a common divisor. That means they do not have a common root.
• Very nice explanation.+1 Aug 13, 2022 at 1:09
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https://www.sageintelligence.com/tips-and-tricks/excel-tips-tricks/2010/07/pmt-function-loans/
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# PMT Function Loans
Question: I’m looking at taking out a loan from the bank. I know how much for, and what the interest rate would be, but how do I figure out what the payments would be per month?
Answer: The PMT function can help
Process (Excel 2003 and 2007):
When it comes to finance, personal or business, knowing what you’re paying or how much you’re spending is paramount. Knowing how you can save money by adjusting your payments is a very handy thing, and we can use the PMT function to not only work this out, but see what we would have to pay before we enter into a loan agreement.
The PMT function has the following syntax:
=PMT(rate,nper,pv,[fv],[type])
Rate: this is the interest rate to be paid per period
Nper: this is the number of periods in the life of the loan (i.e. the number of payments to be made)
Pv: the present value of the loan amount
[Fv]: the future value (optional)
[Type]: shows whether payment is to be made on the first day or the last day of the month
In this example, we will use the following information to see how much we would need to pay back in monthly instalments on a \$100,000 loan over seven years.
1. Select cell B9, type in =C5/12 and press Enter
In order to work out the amount per period, we need to divide the yearly interest rate (12%) by the number of periods in the year. In this case, we’re making payments on a monthly basis, hence dividing the rate by 12.
2. Select cell C9, type in =C4*12 and press Enter.
The loan is over 7 years, with payments made monthly. Therefore we need to work out the total number of periods, hence multiplying the loan years by 12.
3. Select cell D9, type in =C3 and press Enter.
We could directly reference cell C3 when we create our formula, but for consistency in the example layout we will reference it to our demonstration line.
4. Select cell E9, type =pmt(B9,C9,D9) and press Enter
The formula references the interest rate for the period (rate = B9), the number of periods (nper = C9) and the present value of the loan (pv = D9). As [fv] and [type] are in square brackets, they’re optional values and don’t need to be entered. When you press enter, you will see the monthly amount to be repaid:
The amount is represented as a negative figure because this is money that you will be paying out. If you would like to see it as a positive number, simply multiply the result by -1.
5. Select cell F9, type in =E9*C9 and press Enter
You can now see how much will be repaid over the life of the loan. However, you can start playing with the numbers to see what effect it will have on the overall amount.
6. Select cell C4 (the loan years amount), type in 5 and press Enter
Changing the number of years to 5 changes the number of total periods from 84 to 60 (seen in cell C9). This means an increase in the monthly payment amount from \$1,765.27, however it decreases the total amount repaid from \$148,282.96 to \$133,466.69
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https://googlebestbuy.com/salary-one-quarter-of-the-oldest-employees-were/
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# Salary one-quarter of the oldest employees were
Salary Analysis
Our goal is to analyze
salaries in a company based on a survey had happened among its employees. 378
employees took part in our survey. In the survey asked employees about the
following things:
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·
Age (in years)
·
Gender
·
Job seniority (in years)
·
Department (office,
production, sales)
·
Salary
We want to get some
information about distribution above variables and check which variables have
the greatest impact on salaries.
Gender
Let’s see gender
frequency distribution looks like in this company:
Man
240
Woman
138
More than half of the
company’s employees are men – our company is involved in the production of a
certain product, so employs many employees which working physically – hence a
large number of men in company.
Department
As it was mentioned at
the beginning, our company have three departments – office, production and
sales. Let’s see how many people work in each of them.
Office
24
Production
274
Sales
80
As we expected – the
vast majority of employees are production employees – around 70 percent of all
employees.
Age
The next variable that
was asked in the survey was the employee’s age. Let’s count the basic
statistics for this variable:
Statistic
Value
Number
378
Central tendency
Mean
40,1
Median
40
Mode
31
Measures of position
Minimum
23
Maximum
62
First quartile
33
Third quartile
47
Measures of variability
Range
39
Inter-quartile Range
7
Variance
77,97
Standard Deviation
8,83
Coefficient of variation
0,22
Quartile coefficient of dispersion
0,18
Skewness
0,27
Kurtosis
-0,65
The average age of the
employee was 40.1 years. Half of the employees were 40 years old or less and
the largest number of employees participating in the study was 31 years old.
The youngest was 23 years old, the oldest – 63 years old. One-fourth of the youngest
employees were aged 33 and under, one-quarter of the oldest employees were aged
40 and over.
Observed values of the variable vary
around 8, 83 years of the average value, about 22%.
Skewness it is close to zero, so the age distribution can be considered as
symmetrical.
Negative kurtosis speaks of a larger dispersion of results compared to the
standard normal distribution.
Let’s look how it looks
box-plot of age:
As it was mentioned
earlier, the distribution of age can be considered symmetrical in relation to
the average.
Let’s look how it looks histogram of age:
Histogram shows a slight
right-sided asymmetry in the age distribution.
Job seniority
Let us now analyze the
employees’ seniority:
Statistic
Value
Number
378
Central tendency
Mean
11,2
Median
11
Mode
9
Measures of position
Minimum
0
Maximum
36
First
quartile
6
Third
quartile
15
Measures of variability
Range
36
Inter-quartile
Range
5
Variance
50,52
Standard
Deviation
7,11
Coefficient
of variation
0,63
Quartile
coefficient of dispersion
0,43
Skewness
0,73
Kurtosis
0,48
The average job
seniority of the employee was 11.2 years. Half of the employees work in the
company for 11 years and more, while the most employees work with them have
been working in the company for 9 years. Minimum of job seniority is 0, the
longest working employee has been working in the company for 36 years. Quarter
of employees work in the company for 6 years and shorter, quarter of employees
work in the company for 15 years and longer.
Observed values of the variable vary
around 7,11 years of the average value, about 63%, therefore, the variability
of seniority is high.
The skewness is positive so the distribution of seniority is positive skew- the
mass of the distribution is concentrated on the left of the figure.
Positive kurtosis says that the distribution is leptokurtic – the distribution
is more concentrated than the standard normal distribution.
Box-plot:
Histogram:
As we can see – the
internship distribution at the company is right-angled, so the majority of
employees are inexperienced.
Salary
Let us now analyze the
employees’ salaries:
Statistic
Value
Number
378
Central tendency
Mean
3634,1
Median
3300
Mode
3400
Measures of position
Minimum
1700
Maximum
19500
First
quartile
2700
Third
quartile
3950
Measures of variability
Range
17800
Inter-quartile
Range
625
Variance
3146299,10
Standard
Deviation
1773,78
Coefficient
of variation
0,49
Quartile
coefficient of dispersion
0,19
Skewness
4,07
Kurtosis
25,60
The average salary in
this company was 3634,1 PLN. Half of employees earn 3300 and less, the most
people earn 3400. The person who earning the least earns 1700, the most –
19500. Quarter of employees earn 2700 or less, one fourth of the earners earn
the most earn 3950 and more.
Observed values of the variable vary
around 1773,78 PLN of the average value, about 49%, therefore, so salaries are
moderately varied.
The skewness is strong
positive so the distribution of seniority is positive skew- employees with
below-average earnings prevail.
Strong positive kurtosis
says that the distribution is leptokurtic – the distribution is significantly
more concentrated than the standard normal distribution.
Box-plot:
Histogram:
On the charts you can
see what we wrote above – strong right-sided asymmetry of the distribution.
Correlation and regression
The only quantitative
variables in the study were age, seniority and salaries. We want to check
whether we are able to estimate salaries with the help of age or seniority. For
this purpose, let’s count the Pearson correlation coefficient between salaries
and age and seniority:
Pearson correlation coefficient with
Earnings
Age
Job seniority
0,405
0,413
We see that both age and
internship can explain the variability of earnings equally. So let’s build a
linear regression model where the dependent variable will be salaries (Y) and
an independent seniority (X).
As we can see the relationship
between salaries and seniority, we can describe the formula:
So, if seniority increases by one year,
the salary will increase by 103,19 PLN.
New person at work will get averagely 2473,3 PLN.
And for example – a person with 10 years of work experience should earn an
average 3505,2 PLN.
Seniority can explain the variability of
salaries in 17% – this is a weak dependency.
Estimation
We will try to perform
point and interval estimation for the salaries of employees.
Average
·
Point estimation
? = 3634,1 ± 91,2
·
Interval estimation –
level of confidence = 0,95
P(3455,3
< µ < 3812,9) = 0.95 Therefore, at 95% the average salaries in this company are in the range from 4716,6 to 5074,3. Standard deviation · Point estimation ? = 1773,78 · Interval estimation - level of confidence = 0,95 P(1655,71< ? < 1910,13) = 0.95 Therefore, at 95% standard deviation of salaries in this company are in the range from 1655,71 to 1910,13. Hypothesis Let's check whether women and men earn the same amount in this company: ?1 - woman salaries ?2 -man salaries Woman Salaries Man Salaries Mean 3712,68 3588,96 Standard deviation 1832,00 1741,70 number of observations 138 240 Z 0,644 p-value 0,520 Therefore, at the significance level of 0.05, we can say that there is no reason to argue that men and women earn different amounts.
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## Algebra: A Combined Approach (4th Edition)
$7\sqrt {7}-4\sqrt[3] x$
$3\sqrt {7}-\sqrt[3] {x}+4\sqrt {7}-3\sqrt[3] x$ $(3+4)\sqrt {7}-(1+3)\sqrt[3] x$ Combine like radicals. $7\sqrt {7}-4\sqrt[3] x$
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Homework Help Question & Answers
# Consider the following time series data. Week 1 2 3 4 5 6 Value 20 13...
Consider the following time series data. Week 1 2 3 4 5 6 Value 20 13 15 11 17 14 Using the naïve method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy: a. Mean absolute error (MAE) b. Mean squared error (MSE) c. Mean absolute percentage error (MAPE) Round your answers to two decimal places. MAE = MSE = MAPE = Using the average of all the historical data as a forecast for the next period, compute the same three values. Round your answers to two decimal places. MAE = MSE = MAPE = Which method appears to provide the more accurate forecasts for the historical data? Naïve method (most recent value) Average of all the historical data pt be graded, but may be reviewed and considered by your instructor.
#### Homework Answers
ReportAnswer #1
Question: Consider the following time series data.
Answer:
a. Using Naive method as the forecast for the next week, compute the following measures of forecat accuracy:
Naive Method: The naive forecast method is the simplest forecast method of all. The last period's actual value is taken as the current forecast value.
Overall calculation:
Excel formulas:
b. Using the average of all the historical data as a forecast for the next period, compute the same three values.
Overall calculation:
Excel formulas:
c. Which method appears to provide the more accurate for the historical data?
Average of all the historical data, provides the better and accurate forecast in terms of MAE, MSE and MAPE when compared to the values of Naive method and its MAE, MSE and MAPE. Since, the error values are much lowest and more closest to the actual data.
Formulas for your reference:
Mean Absolute Error (MAE) = |Error| / n
Here: |Error| = |Value - Forecast|
Mean Squared Error (MSE) = Error^2 / n
Here: Error^2 = (Value - Forecast)^2
Mean Absolute Ppercentage Error (MAPE) = Error% / n
Here: Error% = (|Error| / Value) x 100
Know the answer?
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http://www.riddlesandanswers.com/v/235862/you-have-12-pills-and-they-all-got-the-same-weight-except-for-one-which-hasnt-got-the-same-weight/
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# 12 PILLS RIDDLE
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## 12 Pills Riddle
You have 12 pills and they all got the same weight, except for one, which hasn't got the same weight. You don't know if it is heavier or easier. You have one scale to weight the pills. You now have to find out, which pill is the right one (the one with a different weight), but you can use the scale only three times. How do you know, which one is the right one?
Hint:
E = easier in "1", H = heavier in "1".
1: Weight 4:4. If they balance go to "2", if they don't balance, go to "3".
2: Balance 1:1 of the pills you didn't weight yet. Then weight one you didn't weight and one you did weight. If they balanced in the first weighing, and balanced in the second weighing, the last pill is the right one. If they balanced in the first weighing and didn't balance in the second, the one you didn't use before is the right pill. If they didn't balance at all, it's the pill you weighed twice. If they didn't balance in the first weighing, but balanced in the second, it is the first pill.
3: Weight EHH : EHH. If they balance, weight one you already weighed, with an unweighed and go to "4". If they don't balance go to "5".
4: If they balance, the one you didn't weight at all is the right pill. If they don't balance, the one you only weighed once is the right one.
5: Give away every pill that was once easier AND once heavier. You should only have EHH left. Weight H:H. If they balance, E is the right one. If the don't balance, the one which was only heavier the whole time, is the right pill.
Did you answer this riddle correctly?
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https://www.jiskha.com/questions/1595871/i-have-two-piece-of-ribbon-one-is-17-5cm-longer-than-the-other-and-1-3-of-one-ribbon-is
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# maths
i have two piece of ribbon one is 17.5cm longer than the other and 1/3 of one ribbon is the same as 3/4 of the other what is the size of the smaller ribbon
1. 👍 2
2. 👎 1
3. 👁 797
1. s = smaller , r = longer
s + 17.5 = r
r / 3 = 3 s / 4 ... 4 r = 9 s
solve the system
1. 👍 0
2. 👎 0
2. that doesn't answer the question though does it ?
1. 👍 1
2. 👎 0
3. X=x
Y=(x+17.5)
1/3Y=3/4x
1/3(x+17.5)=3/4x
x+17.5=3/4x*3
x+17.5=9x/4
17.5=(9x/4)-x
17.5=(9x-4x)/4
17.5*4=5x
70=5x
14=x
So, shortest piece of ribbon is 14cm
1. 👍 1
2. 👎 0
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Simplify 9/12 to the easiest form. Online simplify fractions calculator to mitigate 9/12 to the lowest terms quickly and also easily.
You are watching: What is 9/12 in simplest form
Click come see full answer Correspondingly, what is the simplest type of 9 12?
The components of 12 space 1, 2, 3, 4, 6, and also 12. Therefore the greatest common factor of 9 and also 12 is 3. For this reason you division both political parties by 3. Therefore the simplest form that 912 is 34 .
Similarly, what is the shortest term because that 4 10? 410 in lowest terms is 25 .
In this regard, what is the shortest term that 3 9?
How to reduce (simplify) to shortest terms ordinary (common) math fraction 3/9? as a ideal fraction. (numerator smaller sized than denominator): 3/9 = 1/3 together a decimal number: 3/9 ≈ 0.33. Together a percentage: 3/9 ≈ 33.33%
What is the portion of 9 12?
The number 2 the was supplied to divide the two numbers that comprise the fraction is called a common factor or a divisor of the numerator and the denominator the the fraction. 9/12 = (9 ÷ 3)/(12 ÷ 3) = 3/4 Feb 23 20:16 UTC (GMT)
360/1,430 = (360 ÷ 10)/(1,430 ÷ 10) = 36/143 Feb 23 20:16 UTC (GMT)
39 Related question Answers Found
### What is 9 end 12 as a decimal?
0.75 is a decimal
and 75/100 or 75% is the percent for 9/12.
75%
### What is the simplest kind of 10 12?
What is 10/12 Simplified? - 5/6 is the simplified fraction for 10/12. Simplify 10/12 to the simplest form.
### What is the simplest type of 10 25?
Simplify 10/25 to the simplest form.
10/25 streamlined
Answer: 10/25 = 2/5
### What is 10 the end of 25 together a fraction?
Convert fraction
(ratio) 10 / 25 Answer: 40%
### What is the simplest type of 4 12?
Simplify 4/12 to the simplest form. Virtual simplify fractions calculator to mitigate 4/12 come the lowest state quickly and easily.
4/12 streamlined
Answer: 4/12 = 1/3
### What is the simplest type of 10 15?
Simplify 10/15
to the simplest form. Virtual simplify fractions calculator to mitigate 10/15 to the lowest state quickly and easily.
10/15 simplified
Answer: 10/15 = 2/3
### How carry out you simplify 9 6?
Reduce 9/6 to shortest terms
uncover the GCD (or HCF) that numerator and denominator. GCD that 9 and also 6 is 3. 9 ÷ 36 ÷ 3. Diminished fraction: 32. Therefore, 9/6 streamlined is 3/2.
### What is 3 9 together a fraction?
Fraction come decimal switch table
portion Decimal
3/9 0.33333333
4/9 0.44444444
5/9 0.55555556
6/9 0.66666667
### What is the simplest type of 9 3?
Latest diminished (simplified) fractions
9/3 = (9 ÷ 3)/(3 ÷ 3) = 3 Feb 23 20:49 UTC (GMT)
34/9 currently reduced (simplified) come lowest terms 34 > 9 => improper fraction Rewrite: 34 ÷ 9 = 3 and also remainder = 7 => 34/9 = (3 × 9 + 7)/9 = 3 + 7/9 = = 3 7/9, mixed number (mixed fraction) Feb 23 20:49 UTC (GMT)
### Can you simplify 3 9?
Simplify 3/9 come the easiest form. Virtual simplify fractions calculator to reduce 3/9 to the lowest state quickly and also easily.
3/9 streamlined
Answer: 3/9 = 1/3
### Can you leveling 2 9?
What is 2/9 simplified
? here we will certainly simplify 2/9 come its simplest kind and convert it come a mixed number if necessary. In the fraction 2/9, 2 is the numerator and 9 is the denominator.
### What is the simplest type of 4 8?
Yes, 1 and also 2 are prime numbers. This fraction, 1/2, is the simplest kind of 4/8.
### What is the simplest kind of 12 8?
Simplify 12/8 come the simplest form.
12/8 simplified
Answer: 12/8 = 3/2 = 1 1 2
### What is the simplest kind of 3 12?
Simplify 3/12
to the simplest form.
3/12 simplified
Answer: 3/12 = 1/4
### What is the simplest type of 16 20?
Simplify 16/20
to the simplest form.
See more: Here’S Exactly What Does It Mean To Dream About Your Ex Dying?
16/20 simplified
Answer: 16/20 = 4/5
### How execute you resolve lowest terms?
This way that there is no number, other than 1, that deserve to be divided evenly into both the numerator and the denominator. To mitigate a portion to lowest terms
, division the numerator and also denominator by your Greatest typical Factor (GCF).
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(a) We need to protect from the rain a cake that is in the shape of an equilateral triangle of side 2.1. All we have are identical tiles in the shape of an equilateral triangle of side 1. Find the smallest number of tiles needed. 6 tiles are needed. Since each edge is more than 2 units long, there must be three tiles on each edge. Allowing one triangle at each corner to be on two edges, there must still be 6 tiles touching the edges. And 6 tiles are sufficient; put one tile exactly in each corner, and the remaining part is a truncated edge-1.2 triangle. 3 tiles are sufficient for covering any triangle up to edge 1.5, by putting one in each corner. 4 tiles cover up to edge 2, and in general, n^2 tiles cover up to edge n. By putting these patterns in the middle and one triangle at each corner, 6 tiles can cover up to a triangle of edge 2.25 and 7 can cover up to a triangle of edge 2.5. (b) Suppose the cake is in the shape of an equilateral triangle of side 3.1. Will 11 tiles be enough to protect it from the rain? Yes, cover a size 2 triangle in one corner with 4 tiles and a single tile in each other corner. In the two edge gaps of length 0.1, place triangles so that they just touch the other triangles, with a triangle of size 0.1 hanging off. The part left over is a hexagon of sides 1.1, 0.1, 1, 0.2, 1, 0.1. This is a truncated triangle of edge 1.3, which can be covered with 3 tiles. Indeed, this works if the last triangle is up to size 1.5, which corresponds with an overall triangle size of 3 and 1/6. With 12 tiles, the last triangle can be up to edge 2, and so the overall triangle can be up to edge 3 and 1/3. What about 8 tiles? If you put 4 in one corner, and one in each of the other two corners, the space in the middle can be covered with two outward-pointing triangles up to a certain limit. At the limit, two projecting parts of these triangles (of size X) do not themselves overlap, and at the same time the corner triangles overlap the 4-triangle group by X. So an edge = 2 + 2X = 3 - X, so X = 1/3 and the edge is 2 and 2/3. tiles largest triangle coverable 1 1 3 3/2 4 2 6 9/4 7 5/2 8 8/3 9 3 11 19/6 12 10/3 I don't see any real pattern here except at the whole number triangle size values. Joe Devincentis
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Solve the given differential equation
• chwala
In summary, the constant may be manipulated and it does not matter where one places the constant in these kind of problems.
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
separation of variables
I am on differential equations today...refreshing.
Ok, this is a pretty easier area to me...just wanted to clarify that the constant may be manipulated i.e dependant on approach. Consider,
Ok I have,
##\dfrac{dy}{6y^2}= x dx##
on integration,
##-\dfrac{1}{6y} + k = \dfrac{x^2}{2}##
##k= \dfrac{x^2}{2} + \dfrac{1}{6y}##
using ##y(1)=0.04## we shall get,
##k=\dfrac{28}{6}##
##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##
...
aaargh looks like i will get the same results...cheers
So, you don't have a problem, right?
Mark44 said:
So, you don't have a problem, right?
Correct, I do not have a problem.
Where one places the constant doesn't really matter in these kind of problems.
chwala said:
Where one places the constant doesn't really matter in these kind of problems.
It does matter in certain cases.
If I worked the problem like this ...
##\frac {dy}{y^2} = 6x~dx##
##\frac{-1}y = 3x^2## (Omitting the constant for now)
##y = \frac{-1}{3x^2} + C## (Adding the constant now)
The above gives the wrong value for C when you substitute in the initial condition.
I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
The constants will be different, but when you solve for y in either equation, the solutions will be the same.
chwala said:
##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##
You should write the solution in the form y = f(x), as was shown in the solution.
chwala
Mark44 said:
It does matter in certain cases.
If I worked the problem like this ...
##\frac {dy}{y^2} = 6x~dx##
##\frac{-1}y = 3x^2## (Omitting the constant for now)
##y = \frac{-1}{3x^2} + C## (Adding the constant now)
The above gives the wrong value for C when you substitute in the initial condition.
I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
The constants will be different, but when you solve for y in either equation, the solutions will be the same.
You should write the solution in the form y = f(x), as was shown in the solution.
@Mark44 I realized that it would just give the same solution, thus left it hanging.
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## Discriminant Analysis in R
Discrimination tests are more important in sensory analysis. The main idea behind sensory discrimination analysis is to identify any significant difference or not.
Here are the details of different types of discrimination methods and p-value calculations based on different protocols/methods.
This article will discuss different types of methods and discriminant analysis in r.
## Places to visit in and around Thrissur
If you want to enjoy all the glories of the cultural beauty of the state of Kerala, Thrissur is one of the important places to include in your travel list. Thrissur district is blessed...
## paired t test tabled value vs p-value
When do you use paired t-test and how to apply the same in a practical situation? In this article, we will talk about paired t-test analysis calculation based on a mathematical formula and using...
## Null Hypothesis
Why we need a null hypothesis test?. Hypothesis testing is an important stage in statistics. The test evaluates two mutually exclusive statements about a population to determine which statement is better supported by the...
## t-test in R-How to Perform T-tests in R
Student’s t-test is the deviation of the estimated mean from its population mean expressed in terms of standard error. In this article talking about how to perform a t-test in R, its assumptions, and...
## Places to visit in Ernakulam District
Places to visit near Ernakulam district, Are you planning to visit Kerala?
Each of the many places you visit is of historical and colonial significance.
If you are planning to visit places in Kerala, good planning is essential and this information will help you a lot.
Let’s take a look at the best places to visit in the Ernakulam district.
## Proportion test in R
How to do a proportion test in R and what are the conditions that need to meet for the proportion test? The sampling method for each population is simple random sampling. The samples are...
## One Sample Analysis in R
Hypothesis: One sample analysis in R-In statistics, we can define the corresponding null hypothesis (H0) as follow: 1. H0:m=μ, 2. H0:m≤μ 3. H0:m≥μ The corresponding alternative hypotheses (Ha) are as follow: 1. Ha:m≠μ (different) 2. Ha:m>μ (greater) 3. Ha:m<μ (less) Outlier Detection: Out.fun<-function{abs(x-mean(x,na.rm=TRUE))>3*sd(x,na.rm=TRUE)} ddply(data,.(sample, variable),transform,outlier.team=out.fun(value)) column heading...
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# Search results
1. ### Position as function of time - satellite in elliptical orbit (spherical coordinates)
Speaking in terms of fuel, what is a more efficient way of putting something into orbit? Should you go for low orbit and increasing speed to maintain the orbit? Or spending the fuel to get further out, then accelerate to the slower speed needed to orbit at that range with the same object...
2. ### Math to the Moon
I have a question for someone who may know a bit more about the calculations involved in this process. Lets say for instance you are standing on earth and somehow fly off the ground and go straight to the moon. Throughout the flight you will be experiencing different forces of gravity pulling...
3. ### Getting something to the moon.
So we could say again being as far away as possible before launch point, say we are searching for a constant speed of 2 g's to over come and ultimately reverse the the force of gravity basically like were free falling away from earth to the moon. Over distance we can increasingly reduce the for...
4. ### Getting something to the moon.
So you mean to say it is better to get it overwith instead of the slow approach or you mean it is easier to get there by first getting into orbit?
5. ### Getting something to the moon.
Lets say though we are not looking for orbital velocity but just a point a to point b path to the moon. I know that the traditional approach is to go into orbit, then basically "take off" from there as you can use your already stored up speed etc etc and in the end is probably a better idea, but...
6. ### Getting something to the moon.
But in this scenario, lets say we take the Balloon and this platform holding the rocket to around 25 miles up in the air. By balloon this would take about 45 minutes to an hour give or take some depending on the ballon and load it carries. Either way, at this point, the rocket itself has pulled...
7. ### Getting something to the moon.
This is all theoretical of course. I would like to pt together a small presentation on what it would take to get an object to the moon. For the sake of argument, we will say the object weighs 10 pounds. The shape can be determined by the most efficient shape given the method of propulsion etc...
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# How would you show $\mathbb{Q}=\langle \frac{1}{n!}: n\in \mathbb{N} \rangle$?
The containment of one side is obvious, but I can't see how to show $\mathbb{Q} \subseteq\langle \frac{1}{n!}: n\in \mathbb{N} \rangle$, and would prefer an answer that shows the set containment via algebraic techniques.
I am mean how to show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)?
That is, I am considering the rationals as an additive group and $\langle \cdot \rangle$ means the subgroup generated by the set.
• Is this a question about rings, groups, $\Bbb Z$-modules, $\Bbb Q$ vector spaces? – user2520938 Jan 4 '17 at 10:46
• If $n>b$ then $n!=b\times m$ so $\frac ab=\frac {am}{n!}$. – lulu Jan 4 '17 at 10:49
• Note: I am assuming that you meant "show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)" . If you meant something else, you should clarify. – lulu Jan 4 '17 at 10:50
• @lulu Yes your interpretation is correct and that is the question I was asking. – Mark Jan 4 '17 at 10:52
• Say you want $\frac{a}{b},\ a,b > 0$, then $\frac{a}{b}= a(b-1)! \cdot \frac{1}{b!}$. – Zubzub Jan 4 '17 at 11:02
It follows from the Fundamental Theorem of Arithmetic that $\Bbb Q$ is generated, as a group, by the set $$\left\{\frac1{p^k}\text{ such that p is prime and k>0}\right\}$$ (one can make this set of generators smaller, but that is irrelevant here). Then it is enough to show that each $1/p^k$ can be obtained from the $1/n!$, but then it is enough to notice that $$\frac1{p^k}=(p^k-1)!\frac1{(p^k)!}.$$
HINT: The generated subgroup also contains any finite sum of the same element $k$ times since $$\frac{k}{n!} = \underbrace{\frac{1}{n!} + \ldots \frac{1}{n!}}_{k \text{ times}}.$$
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1. ## co-ordinate
$\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
and BF: 4x+3y-1=0,
$CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
2. Originally Posted by mp3qz
$\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
and BF: 4x+3y-1=0,
$CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
What is the meaning of this statement?
3. Originally Posted by mp3qz
$\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
and BF: 4x+3y-1=0,
$CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
We are given the equation for line BF.
We are given the equation for line AD.
Line AC is perpendicular to BF.
A little math
$\angle CAD$ = 8deg 7min 48sec
Which is half the angle CAB.
Thus we know the equation of line AB.
It is required that Line CH be perpendicular to AB.
The equation for Line CH: y = $\frac {-3x}{4} - \, 1.75$
Since H is not restricted to being on the line AB, the point C can occur anywhere on the line.
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1. imaginary numbers
need to solve 6-6i/-9i
came up with the answer 2/3 + 2/3i
What am I doing wrong. It's showing up as incorrect.
Thank you.
2. Re: imaginary numbers
You're missing a negative sign
$\frac{6-6i}{-9i}=\frac{2-2i}{-3i}=\frac{2-2i}{-3i}\cdot \frac{i}{i}=\frac{2i-2i^2}{-3i^2}=\frac{2i+2}{3}$
3. Re: imaginary numbers
Hello, needhelpnow!
Simplify: $\frac{6-6i}{-9i}$
Came up with the answer 2/3 + 2/3i
What am I doing wrong? It's showing up as incorrect.
Your answer, $\tfrac{2}{3} + \tfrac{2}{3}i$, is correct.
. . it might be read as: $\tfrac{2}{3} + \tfrac{2}{3i}$
Try entering it as. 2/3 + 2i/3. or .2/3 + (2/3)i.
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# Distribution of $\max\{x+y_1,x+y_2, …, x+y_n\}$ for i.i.d. uniform random variables [closed]
Let $x$, $y$ and $z$ be independent random variables, uniformly distributed over the same interval. What is the cumulative distribution of $\max\{x+y,x+z\}$?
And more generally, let $x, y_1, y_2, ... y_n$, all be independent random variables, uniformly distributed over the same interval. What is the cumulative distribution of $\max\{x+y_1,x+y_2, ..., x+y_n\}$?
I wonder if the fact that $x$ being included in $w_i=x+y_i$ makes $w_i$ and $w_j$ interdependent.
## closed as off-topic by Did, kingW3, Shailesh, projectilemotion, Davide GiraudoMar 9 '17 at 17:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, kingW3, Shailesh, projectilemotion, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
• Thoughts? Personal input? – Did Mar 9 '17 at 6:57
Suppose $X$, $Y$ and $Z$ are iid Unif$(0,1)$,
then $T := \max(X+Y, X+Z) = X + \max(Y, Z)$. CDF of $W := \max(Y, Z)$ is $F_W(w) = \Pr(W \leq w) = \Pr(Y \leq w, Z \leq w) = \Pr(Y \leq w)\Pr(Z \leq w) =\begin{cases} 0 & \text{if } w\leq 0 \\ w^2 & \text{if } 0 < w< 1 \\ 1 & \text{if } w\geq 1 \end{cases}$ Therefore, density of $W$ is $f_W(w) = \begin{cases} 2 w & \text{if } 0 < w< 1 \\ 0 & \text{otherwise } \end{cases}$.
Since $X$, $Y$ and $Z$ are independent and $W = \max(Y, Z)$, so $X$ and $W$ are also independent. To find the distribution of $T = X+W$, we can use convolution:
$f_T(t) = \int\limits_{-\infty}^{\infty}f_X(x)f_W(t-x)dx = \begin{cases} \int\limits_{0}^{t}2(t-x)dx = t^2 & \text{if } 0 < t\leq 1 \\ \int\limits_{t-1}^{1}2(t-x)dx = 2t -t^2 & \text{if } 1 < t\leq 2 \\ 0 & \text{otherwise} \end{cases}$
CDF of $T$ is
$F_T(t) = \begin{cases} 0 & \text{ if } t\leq 0 \\ \frac{t^3}{3} & \text{if } 0 < t\leq 1 \\ t^2 - \frac{t^3}{3} - \frac{1}{3} & \text{if } 1 < t\leq 2 \\ 1 & \text{if } t > 2\end{cases}$
• Can you explain how you got the bounds for the integral in the convolution formula? – Parseval May 16 '18 at 13:10
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# Excel (Only) 15 questions in Time Value of Money Problem
Subject Finance
Topic Excel (Only) 15 questions in Time Value of Money Problem
Paper details
Time Value of Money Problems Please answer the Time Value of Money Problems in Excel that is attached. Please submit all answers in a single Excel file. Please be sure to show that you understand the concepts. Please show all calculations, step by step, performed by using Excel, and all answers should be thoroughly explained Attachments: 1. Instructions 2. Time Value of Money Problems 3. Copy of Time Value of Money Examples 4. Copy of Stock Valuation 5. Copy of PPP Present Value if I can attach it 6. Copy of PPP Risk and Rates of Return if I can attach it 7. Copy of PPP WACC if I can attach it GUIDANCE This is time value of money tools that are essential for financial valuation of bonds, stocks. annuities, obligations, and all assets and investment opportunities Additional Info: Time Value of Money. For an illustrated short lecture, Risk and Rates of Return. A Power Point slide presentation that fully explains and illustrates this important subject is in the attached file. Also attached in the Week 4 Content area are PowerPoint presentations on: Present Value (also known as Discounted Cash Flow, or Time Value of Money). This presentation will explain present value, and show you how to use the Excel financial functions, which will quickly and easily calculate rates of interest, growth rates, and present and future values of cash flows. For your information, the rate of interest is the same thing as the rate of growth, because the rate of interest is the rate at which money grows. To find the rate of interest (or growth) I recommend using the Excel RATE function for a quick and easy solution. Another tip: the “rule of 72” is a handy rule of thumb. It says that any number that is growing at the rate i will double in approximately n periods, and i times n = 72 (approximately). For example: At a 6% rate of interest, \$100 will double in 72/6 = 12 periods (because 6 x 12 = 72). In 8 periods, at what rate of interest will \$200 double? Answer: at 72/8 = 9% per period (because 8 x 9 = 72). Reminder 1. Show all calculations. 2. Use Excel in order to accomplish item 1 above. 3. Describe and justify each assumption that is made, 4. Explain, support and justify every statement you make. 5. Support your conclusions with evidence (such as facts and statistics from reliable sources, quotes from the readings, or a well-reasoned logical argument). Examples from your own experience are also helpful. Please bear in mind that analyses and conclusions are only as good as the data and assumptions that they are based on; So ALWAYS support all of your numbers or calculations with reasoning and explanations, and be sure to provide sources for all of your data. HELPFUL EXCEL GUIDELINES 1) all calculations, step by step, and a financial calculator cannot show us all your calculations, step by step; 2) Here are some very important and helpful guidelines for using Excel a) Whenever your answer is a calculated number in a cell of an Excel spreadsheet please do not just type that number into the cell. b) For example, if your calculation involves the numbers in cells C2, A3 and A5, and you want to divide the number in C2 by the number in A3 less the number in A5, c) Use a formula like: =C2/(A3-A5) to show your calculation. Just type =C2/(A3-A5) into the cell. 3. You can also add a comment to any cell in Excel. If you don’t know how to do this in Excel, look up “comment” in Excel help. 4. Excel financial functions are very useful, quick and easy to use. (To access a tutorial on Excel financial functions click here: http://www.tvmcalcs.com/calculators/excel_tvm_functions/excel_tvm_functions_page1:) 5. More information on Excel financial functions is below: a) To access functions in Excel, click on INSERT, then on FUNCTION. b) Select the type of function you want to use, such as FINANCIAL, STATISTICAL, MATHEMATICAL, etc. c) Within the type of function select the exact function you want. For example, from FINANCIAL functions, select Future Value (FV) 6. Excel functions have the following form – 1) FV(rate, nper, pmt, pv) 2) where FV = Future Value, and this function’s variables are: 3) rate = rate of return, 4) nper = # of periods, 5) pmt = amount of payment each period, and 6) pv = present value. F. Use only Excel functions in this assignments wherever possible because: Here are examples of useful Excel functions: Description Function Name Variable 1 Variable 2 Variable 3 Variable 4 Future Value FV rate nper pmt pv Present Value PV rate nper pmt fv Payment per Period PMT rate nper pv fv Rate of Return RATE nper pmt pv fv Number of Periods NPER rate pmt pv fv Net Present Value NPV rate value 1 value 2 value n Internal Rate of Return IRR values 1,2,3 etc guess (% IRR)
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