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Online Tution » Important Question » Kinematics Equations
Kinematics Equations and It’s Derivation,Variables,Dynamic Equation
Kinematic Equations
Kinematics Equations: Kinematics Equations are very important in the field of physics. These equations can be used to describe the motion of any object in terms of its position, velocity, and acceleration. Generally, this equation is most used in the field of engineering, and that is why students want to get more and more information about Kinematics Equations.
But in standard books, it has been explained in such a standard way that a student may face a lot of difficulty in understanding it. Therefore, in today’s article, we have provided you with all the information regarding the use and construction of Kinematics Equations with explanations in the simplest languages. If you also want to get all the information regarding Kinematics Equations, then stay with us till the end of today’s article.
What are Kinematics equations?
Kinematics Equations are a set of mathematical equations used to describe the motion of an object in terms of its position, velocity, and acceleration. Under this, the force is not taken into account, which is the root cause of the motion. These equations link together five kinematic variables, including displacement (Δx), initial velocity (v0), final velocity (v), time interval (t), and constant acceleration (a). Using these equations, we can calculate one or more of these four.
If the other four are known that allows us to determine different aspects of an object’s motion. Kinematics equations are widely used in the field of physics to describe the motion of objects in various contexts such as free fall, projectile motion, and circular motion.
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How do Kinematics equations relate to the motion of an object?
Kinematics Equations are used to describe the motion of an object in terms of its position, velocity, and acceleration. These equations provide a mathematical framework for computing unknown aspects of an object’s motion based on unknown variables.
For examplewe know the initial velocity, final velocity, and time interval of an object, then we can use Kinematics Equations to calculate the displacement traveled by the object during that time.
Similarly, if we know the displacement, initial velocity, and time interval of an object, then we can calculate its final velocity. Kinematics Equations enable us to provide a systematic way of computing the unknown variables of motion of an object and to describe different types of motion such as free fall projectile motion and circular motion.
Name the five variables that link together Kinematics Equations
• Displacement (Δx): the change in position of the object from its initial position to its final position.
• Initial velocity (v0): The velocity of the object at the beginning of its motion.
• Final velocity (v): the velocity of the object at the end of its motion.
• Time interval (t): the duration of the motion of the object.
• Constant acceleration (a): the rate at which the object’s velocity changes over time, which is assumed to be constant in Kinematics equations.
Check: CBSE Class 12 All Subject Answer Keys 2023
Difference between Kinematics and Dynamics equations
Kinematics and Dynamics are two important branches of Physics, which deals with the motion of objects. But they have different focuses and use different sets of equations. Here we have listed below 10 differences between Kinematics and Dynamics Equations-
• Focus: Kinematics focuses on describing the motion of an object in terms of position, velocity, and acceleration, without considering the forces that can cause motion. Whereas Dynamics focuses on the relationship between the motion of an object and the forces that cause it
• Equations: Kinematics Equations describe motion in terms of displacement, initial velocity, final velocity, time interval, and constant acceleration. Whereas Dynamics Equations describe motion in terms of force, mass, and acceleration.
• Derivation: Kinematics Equations are derived from the analysis of motion without considering the forces that cause it. Whereas Dynamic Equations are derived from Newton’s laws of motion which tell us how the motion of an object changes in response to external forces.
• Applications: Kinematics Equations are used to solve problems related to the motion of an object. For example, determining the distance covered or the time taken to cover a certain distance. Dynamic equations are used to solve problems related to forces acting on an object. For example, calculating the force required to move an object.
• Variables: Kinematics Equations relate 5 variables together, whereas Dynamics Equations relate three variables together.
• Types of Motion: Kinematics Equations can describe different types of motion, such as linear motion, circular motion, and projectile motion. But dynamic equations are only used to describe the motion of an object in response to external forces.
• Forces: Kinematics Equations do not take into account the forces acting on an object, while Dynamic Equations describe the forces acting on an object, which can affect its motion.
• Units: The units of measurement used for Kinematics Equations are different from those used for Dynamic Equations.
• Predictive Ability: Kinematics Equations can predict the future motion of an object given its initial position, while Dynamic Equations can predict the future motion of an object given its initial position and forces acting on it.
• Real-Life Examples: Kinematics Equations are often used in real-life examples such as determining the speed of a car, while Dynamics Equations are used in engineering applications.
Derivation of Kinematics equations
The derivation of Kinematics Equations involves analyzing the motion of an object under constant acceleration. This is done using basic principles of calculus, and the definitions of velocity and acceleration. Here is a step-by-step derivation of the Kinematics equations:
Assuming constant acceleration (a), we can use the definition of acceleration to write:
a = dv/dt
where v is the velocity of the object and t is time.
We can integrate both sides of this equation concerning time to get:
∫a dt = ∫dv
Integrating the left side gives:
at + C1 = v
where C1 is the constant of integration.
Using the definition of velocity, we can write:
v = dx/dt
where x is the displacement of the object.
Substituting v with at + C1, we get:
dx/dt = at + C1
We can integrate both sides of this equation concerning time to get:
∫dx = ∫at + C1 dt
Integrating the left side gives:
x = C2 + at^2/2 + C1t
where C2 is the constant of integration.
Now, we can use the initial conditions of the object’s motion to solve for the constants C1 and C2. Let x0 be the initial position of the object and v0 be its initial velocity at time t = 0. Then:
x0 = C2
v0 = C1
Substituting these values into the equation for x, we get:
x = x0 + v0t + (1/2)at^2
This equation relates the displacement of an object to its initial position, initial velocity, acceleration, and time interval. We can also use the equations we derived earlier to express the final velocity of the object in terms of its initial velocity, acceleration, and time interval:
v = v0 + at
These equations, together with the equations for acceleration and velocity, form the set of Kinematics equations.
Four basic kinematics equations
These are the four basic kinematics equations:
• First basic kinematics equation:
v=v0+at
• Second basic kinematics equation:
x=(v+v02)t
• Third basic kinematics equation:
x=v0t+12at2
• Fourth basic kinematics equation:
v2=vo2+2ax
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Significance of Kinematics equations in the field of physics
Kinematics equations are essential in the field of physics because they provide a mathematical framework for describing the motion of objects. They are particularly useful in analyzing the motion of objects under constant acceleration, which is a common scenario in many physical systems.
Kinematics equations allow us to calculate various aspects of the motion of an object, such as its velocity, displacement, and acceleration at certain initial conditions, and are important in the study of mechanics, which is the branch of physics that deals with the motion of objects and the forces that produce them.
What are some real-life examples of Kinematics equations?
Kinematics Equations can be applied to many real-life introductory topics, such as the motion of a car on a highway, the motion of a ball thrown in the air, and the motion of a satellite with respect to earth.
These can be used to base engineering applications such as designing rockets, and projectiles, or calculating trajectories. Within sports, Kinematics Equations can be used to analyze the motion of athletes, such as during a race. Sprinter’s speed, or the speed of a diver during a dive.
What is the role of time interval in Kinematics equations?
The role of time interval in Kinematics Equations is very important because it determines the period of motion of an object. The time interval is usually denoted by the variable “t” and is measured in seconds. Knowing the time interval and other initial conditions such as initial position, velocity, and acceleration of objects, we can use Kinematics Equations to calculate various aspects of motion.
For example, the time interval is important in determining the rate of change of these variables of initial position, velocity, and displacement time, which can be described by the acceleration of the object.
How is constant acceleration related to Kinematics equations?
Constant acceleration is a common phenomenon in many physical systems and is related to the Kinematics Equations in that they are specifically designed to describe the motion of objects under constant acceleration. Constant acceleration means that the description of the object is uniform and does not change with time.
This allows us to use Kinematics Equations to calculate various aspects of an object’s motion such as its final position and displacement. Examples of constant acceleration include free-fall motion, projectile motion, and circular motion.
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FAQs
What are some common misconceptions about Kinematics equations?
A common misconception about Kinematics Equations is that they can be used to describe any type of motion, which is not true. Kinematics Equations apply only to objects that move with constant acceleration or at rest. Another misconception is that Kinematics Equations can be used to determine the cause of motion which is not the case. Kinematics Equations describe the motion of an object only in terms of position, velocity, and acceleration.
What are some limitations of using Kinematics equations to describe motion?
There are some limitations to the Kinematics Equations that can be used to describe the motion of objects with constant acceleration or constant velocity, and also do not take into account other factors that may affect the motion of an object, such as air. Pollution or external forces additionally require motion in a straight line and do not apply to objects moving along a curved path.
How can Kinematics equations be used to calculate the acceleration of an object?
Kinematics equations can be used to calculate the acceleration of an object by using the formula a = (v - v0)/t, where "a" is the acceleration, "v" is the final velocity, "v0" is the initial velocity, and "t" is the time interval. This formula assumes that the object is moving with constant acceleration.
What is the difference between speed and velocity in Kinematics equations?
Speed and velocity are related under Kinematics, but their concepts are also different depending on the equations. Speed refers to the speed at which an object moves and is usually measured in meters per second or kilometers per hour. On the other hand, refers to the rate at which an object is moving in a specific direction. It is measured in m/s or Km/h. The direction of movement opposite to the speed is taken into account and depending on the direction a positive or negative result can be obtained.
How are initial velocity and final velocity related in Kinematics equations?
Initial velocity and final velocity are related in Kinematics equations through the equation v = v0 + at, where "v" is the final velocity, "v0" is the initial velocity, "a" is the acceleration, and "t" is the time interval. This equation allows us to calculate the final velocity of an object based on its initial velocity, acceleration, and time interval.
What is the meaning of displacement in Kinematics equations?
In kinematics equations, displacement refers to the change in the position of an object over a given time interval, it is usually denoted by the symbol delta x and is measured in meters. The displacement can be calculated using this equation
Δx = v0t + 1/2at^2, where "v0" is the initial velocity, "t" is the time interval, and "a" is the acceleration. This equation assumes that the object is moving with constant acceleration.
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## #math - Sat 5 May 2007 between 00:11 and 00:41
### NY Lost Funds
ihope What's this about scribd? noodle_snacks heyanyone here ihope noodle_snacks: yep.By the way, on IRC, the general protocol is to simply ask a question, then wait at least, say, ten minutes to get an answer. noodle_snacks hehemy badanyway, no idea if i am in the right place, but: ihope So you're fine, but it's very annoying to see that someone's asked a question and then left when you could have answered it. noodle_snacks I have the equation x^4 - 3x^3 + 7x^2 - 8x + 6 = 0, given that 1+i is a factor, from that i know that 1-i is also a factorbut now i am getting confused as to the method to take out those two factors so i can use the quadratic formula to solve for the last two rootsany pointers? ihope noodle_snacks: you mean x-1-i and x-1+i are factors? noodle_snacks yeahsorry ihope Do you know synthetic division? noodle_snacks doesn't ring a bell noshould i look it up? ihope That's not a bad idea.The general way to remove a root from a polynomial is to divide by the corresponding factor: (x^4 - 3x^3 + 7x^2 - 8x + 6)/(x - (1+i)) is the polynomial with the root 1+i removed. Steve|Office Synthetic division was one of those things I learned in junior high school and then never remembered or used since. echelon-- lol. ihope If you want to remove both roots at once, divide by (x - (1+i))(x - (1-i)), whatever that is. noodle_snacks thanksis it better to simplify (x - (1+i))(x - (1+i))first? ihope x^2 - 2x + 2.If you use synthetic division, though, you'll have to do it one at a time.Steve|Office: what do you use when you want to solve a polynomial? Or do you not solve them? Steve|Office mbot, usually. noodle_snacks with synthetic division, it seems likely you end up with a remainder, do you just add that to the x^0 term? ihope noodle_snacks: if the thing you use to divide with is a root, you don't end up with a remainder.If you do end up with a remainder, it wasn't a root. noodle_snacks oh ok, must have been the example i was looking at rod-cal is not i an imaginary number? ihope rod-cal: i is an imaginary number, yes.I am not an imaginary number, but i is. :-) rod-cal Is not the science of imaginary numbers called 'complex numbers'? noodle_snacks thats what it says in my text book ihope rod-cal: a complex number is anything that's either a real number or an imaginary number.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > israg Structured version Visualization version GIF version
Theorem israg 25392
Description: Property for 3 points A, B, C to form a right angle. Definition 8.1 of [Schwabhauser] p. 57. (Contributed by Thierry Arnoux, 25-Aug-2019.)
Hypotheses
Ref Expression
israg.p 𝑃 = (Base‘𝐺)
israg.d = (dist‘𝐺)
israg.i 𝐼 = (Itv‘𝐺)
israg.l 𝐿 = (LineG‘𝐺)
israg.s 𝑆 = (pInvG‘𝐺)
israg.g (𝜑𝐺 ∈ TarskiG)
israg.a (𝜑𝐴𝑃)
israg.b (𝜑𝐵𝑃)
israg.c (𝜑𝐶𝑃)
Assertion
Ref Expression
israg (𝜑 → (⟨“𝐴𝐵𝐶”⟩ ∈ (∟G‘𝐺) ↔ (𝐴 𝐶) = (𝐴 ((𝑆𝐵)‘𝐶))))
Proof of Theorem israg
Dummy variables 𝑔 𝑤 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 israg.a . . . 4 (𝜑𝐴𝑃)
2 israg.b . . . 4 (𝜑𝐵𝑃)
3 israg.c . . . 4 (𝜑𝐶𝑃)
41, 2, 3s3cld 13467 . . 3 (𝜑 → ⟨“𝐴𝐵𝐶”⟩ ∈ Word 𝑃)
5 fveq2 6103 . . . . . 6 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (#‘𝑤) = (#‘⟨“𝐴𝐵𝐶”⟩))
65eqeq1d 2612 . . . . 5 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → ((#‘𝑤) = 3 ↔ (#‘⟨“𝐴𝐵𝐶”⟩) = 3))
7 fveq1 6102 . . . . . . 7 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (𝑤‘0) = (⟨“𝐴𝐵𝐶”⟩‘0))
8 fveq1 6102 . . . . . . 7 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (𝑤‘2) = (⟨“𝐴𝐵𝐶”⟩‘2))
97, 8oveq12d 6567 . . . . . 6 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → ((𝑤‘0) (𝑤‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)))
10 fveq1 6102 . . . . . . . . 9 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (𝑤‘1) = (⟨“𝐴𝐵𝐶”⟩‘1))
1110fveq2d 6107 . . . . . . . 8 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (𝑆‘(𝑤‘1)) = (𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1)))
1211, 8fveq12d 6109 . . . . . . 7 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → ((𝑆‘(𝑤‘1))‘(𝑤‘2)) = ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2)))
137, 12oveq12d 6567 . . . . . 6 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))
149, 13eqeq12d 2625 . . . . 5 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))) ↔ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2)))))
156, 14anbi12d 743 . . . 4 (𝑤 = ⟨“𝐴𝐵𝐶”⟩ → (((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2)))) ↔ ((#‘⟨“𝐴𝐵𝐶”⟩) = 3 ∧ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))))
1615elrab3 3332 . . 3 (⟨“𝐴𝐵𝐶”⟩ ∈ Word 𝑃 → (⟨“𝐴𝐵𝐶”⟩ ∈ {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))} ↔ ((#‘⟨“𝐴𝐵𝐶”⟩) = 3 ∧ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))))
174, 16syl 17 . 2 (𝜑 → (⟨“𝐴𝐵𝐶”⟩ ∈ {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))} ↔ ((#‘⟨“𝐴𝐵𝐶”⟩) = 3 ∧ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))))
18 df-rag 25389 . . . . 5 ∟G = (𝑔 ∈ V ↦ {𝑤 ∈ Word (Base‘𝑔) ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2))))})
1918a1i 11 . . . 4 (𝜑 → ∟G = (𝑔 ∈ V ↦ {𝑤 ∈ Word (Base‘𝑔) ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2))))}))
20 simpr 476 . . . . . . . 8 ((𝜑𝑔 = 𝐺) → 𝑔 = 𝐺)
2120fveq2d 6107 . . . . . . 7 ((𝜑𝑔 = 𝐺) → (Base‘𝑔) = (Base‘𝐺))
22 israg.p . . . . . . 7 𝑃 = (Base‘𝐺)
2321, 22syl6eqr 2662 . . . . . 6 ((𝜑𝑔 = 𝐺) → (Base‘𝑔) = 𝑃)
24 wrdeq 13182 . . . . . 6 ((Base‘𝑔) = 𝑃 → Word (Base‘𝑔) = Word 𝑃)
2523, 24syl 17 . . . . 5 ((𝜑𝑔 = 𝐺) → Word (Base‘𝑔) = Word 𝑃)
2620fveq2d 6107 . . . . . . . . 9 ((𝜑𝑔 = 𝐺) → (dist‘𝑔) = (dist‘𝐺))
27 israg.d . . . . . . . . 9 = (dist‘𝐺)
2826, 27syl6eqr 2662 . . . . . . . 8 ((𝜑𝑔 = 𝐺) → (dist‘𝑔) = )
2928oveqd 6566 . . . . . . 7 ((𝜑𝑔 = 𝐺) → ((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0) (𝑤‘2)))
30 eqidd 2611 . . . . . . . 8 ((𝜑𝑔 = 𝐺) → (𝑤‘0) = (𝑤‘0))
3120fveq2d 6107 . . . . . . . . . . 11 ((𝜑𝑔 = 𝐺) → (pInvG‘𝑔) = (pInvG‘𝐺))
32 israg.s . . . . . . . . . . 11 𝑆 = (pInvG‘𝐺)
3331, 32syl6eqr 2662 . . . . . . . . . 10 ((𝜑𝑔 = 𝐺) → (pInvG‘𝑔) = 𝑆)
3433fveq1d 6105 . . . . . . . . 9 ((𝜑𝑔 = 𝐺) → ((pInvG‘𝑔)‘(𝑤‘1)) = (𝑆‘(𝑤‘1)))
3534fveq1d 6105 . . . . . . . 8 ((𝜑𝑔 = 𝐺) → (((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2)) = ((𝑆‘(𝑤‘1))‘(𝑤‘2)))
3628, 30, 35oveq123d 6570 . . . . . . 7 ((𝜑𝑔 = 𝐺) → ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2))) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))
3729, 36eqeq12d 2625 . . . . . 6 ((𝜑𝑔 = 𝐺) → (((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2))) ↔ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2)))))
3837anbi2d 736 . . . . 5 ((𝜑𝑔 = 𝐺) → (((#‘𝑤) = 3 ∧ ((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2)))) ↔ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))))
3925, 38rabeqbidv 3168 . . . 4 ((𝜑𝑔 = 𝐺) → {𝑤 ∈ Word (Base‘𝑔) ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0)(dist‘𝑔)(𝑤‘2)) = ((𝑤‘0)(dist‘𝑔)(((pInvG‘𝑔)‘(𝑤‘1))‘(𝑤‘2))))} = {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))})
40 israg.g . . . . 5 (𝜑𝐺 ∈ TarskiG)
4140elexd 3187 . . . 4 (𝜑𝐺 ∈ V)
42 fvex 6113 . . . . . . . 8 (Base‘𝐺) ∈ V
4322, 42eqeltri 2684 . . . . . . 7 𝑃 ∈ V
44 wrdexg 13170 . . . . . . 7 (𝑃 ∈ V → Word 𝑃 ∈ V)
4543, 44ax-mp 5 . . . . . 6 Word 𝑃 ∈ V
4645rabex 4740 . . . . 5 {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))} ∈ V
4746a1i 11 . . . 4 (𝜑 → {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))} ∈ V)
4819, 39, 41, 47fvmptd 6197 . . 3 (𝜑 → (∟G‘𝐺) = {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))})
4948eleq2d 2673 . 2 (𝜑 → (⟨“𝐴𝐵𝐶”⟩ ∈ (∟G‘𝐺) ↔ ⟨“𝐴𝐵𝐶”⟩ ∈ {𝑤 ∈ Word 𝑃 ∣ ((#‘𝑤) = 3 ∧ ((𝑤‘0) (𝑤‘2)) = ((𝑤‘0) ((𝑆‘(𝑤‘1))‘(𝑤‘2))))}))
50 s3fv0 13486 . . . . . . 7 (𝐴𝑃 → (⟨“𝐴𝐵𝐶”⟩‘0) = 𝐴)
511, 50syl 17 . . . . . 6 (𝜑 → (⟨“𝐴𝐵𝐶”⟩‘0) = 𝐴)
5251eqcomd 2616 . . . . 5 (𝜑𝐴 = (⟨“𝐴𝐵𝐶”⟩‘0))
53 s3fv2 13488 . . . . . . 7 (𝐶𝑃 → (⟨“𝐴𝐵𝐶”⟩‘2) = 𝐶)
543, 53syl 17 . . . . . 6 (𝜑 → (⟨“𝐴𝐵𝐶”⟩‘2) = 𝐶)
5554eqcomd 2616 . . . . 5 (𝜑𝐶 = (⟨“𝐴𝐵𝐶”⟩‘2))
5652, 55oveq12d 6567 . . . 4 (𝜑 → (𝐴 𝐶) = ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)))
57 s3fv1 13487 . . . . . . . . 9 (𝐵𝑃 → (⟨“𝐴𝐵𝐶”⟩‘1) = 𝐵)
582, 57syl 17 . . . . . . . 8 (𝜑 → (⟨“𝐴𝐵𝐶”⟩‘1) = 𝐵)
5958eqcomd 2616 . . . . . . 7 (𝜑𝐵 = (⟨“𝐴𝐵𝐶”⟩‘1))
6059fveq2d 6107 . . . . . 6 (𝜑 → (𝑆𝐵) = (𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1)))
6160, 55fveq12d 6109 . . . . 5 (𝜑 → ((𝑆𝐵)‘𝐶) = ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2)))
6252, 61oveq12d 6567 . . . 4 (𝜑 → (𝐴 ((𝑆𝐵)‘𝐶)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))
6356, 62eqeq12d 2625 . . 3 (𝜑 → ((𝐴 𝐶) = (𝐴 ((𝑆𝐵)‘𝐶)) ↔ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2)))))
64 s3len 13489 . . . . 5 (#‘⟨“𝐴𝐵𝐶”⟩) = 3
6564a1i 11 . . . 4 (𝜑 → (#‘⟨“𝐴𝐵𝐶”⟩) = 3)
6665biantrurd 528 . . 3 (𝜑 → (((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))) ↔ ((#‘⟨“𝐴𝐵𝐶”⟩) = 3 ∧ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))))
6763, 66bitrd 267 . 2 (𝜑 → ((𝐴 𝐶) = (𝐴 ((𝑆𝐵)‘𝐶)) ↔ ((#‘⟨“𝐴𝐵𝐶”⟩) = 3 ∧ ((⟨“𝐴𝐵𝐶”⟩‘0) (⟨“𝐴𝐵𝐶”⟩‘2)) = ((⟨“𝐴𝐵𝐶”⟩‘0) ((𝑆‘(⟨“𝐴𝐵𝐶”⟩‘1))‘(⟨“𝐴𝐵𝐶”⟩‘2))))))
6817, 49, 673bitr4d 299 1 (𝜑 → (⟨“𝐴𝐵𝐶”⟩ ∈ (∟G‘𝐺) ↔ (𝐴 𝐶) = (𝐴 ((𝑆𝐵)‘𝐶))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∈ wcel 1977 {crab 2900 Vcvv 3173 ↦ cmpt 4643 ‘cfv 5804 (class class class)co 6549 0cc0 9815 1c1 9816 2c2 10947 3c3 10948 #chash 12979 Word cword 13146 ⟨“cs3 13438 Basecbs 15695 distcds 15777 TarskiGcstrkg 25129 Itvcitv 25135 LineGclng 25136 pInvGcmir 25347 ∟Gcrag 25388 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-3 10957 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-fzo 12335 df-hash 12980 df-word 13154 df-concat 13156 df-s1 13157 df-s2 13444 df-s3 13445 df-rag 25389 This theorem is referenced by: ragcom 25393 ragcol 25394 ragmir 25395 mirrag 25396 ragtrivb 25397 ragflat2 25398 ragflat 25399 ragcgr 25402 footex 25413 colperpexlem1 25422 colperpexlem3 25424 mideulem2 25426 opphllem 25427 lmiisolem 25488 hypcgrlem1 25491 hypcgrlem2 25492 trgcopyeulem 25497
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0
# How you will get 210 by mulitiplying 2 consecutive numbers?
Updated: 9/18/2023
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10y ago
14 x 15 = 210
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Related questions
### Can you find the 2 consecutive numbers with a product of 702 and 210?
702 = 26 x 27; 210 = 14 x 15.
### How do you work out the 2 consecutive numbers that make 702 and 210?
You cannot.If you have any two consecutive numbers, one of them must be odd and the other even. So their sum must be odd and therefore cannot be 702 nor 210
### How do you calculate the least common multiple of 2 consecutive numbers that is greater than 200 and a multiple of 7?
Consecutive numbers can't both be multiples of 7. The LCM of consecutive numbers is their product. 14 and 15 are consecutive numbers whose LCM is a multiple of 7 that is greater than 200.
4,5
### What are consecutive prime numbers?
2 and 3 are consecutive prime numbers.
### What are all the pairs of prime numbers that are consecutive numbers?
The numbers 2 and 3 are consecutive prime numbers. Are there other pairs of prime numbers which are consecutive numbers?
### How are 2 and 3 consecutive prime numbers?
2 and 3 are consecutive numbers and they are both prime.
### How many pairs of prime numbers are there which are consecutive odd integers?
There are an infinite number of prime numbers which are consecutive odd integers. Choose any natural number n. Take all primes up to any number n, take their product, and add 1 and subtract 1 from it. These 2 numbers are consecutive odd integers. eg 2*3*5*7 = 210 209 and 211 are primes which are consecutive odd integers.
### How do you get consecutive prime numbers?
The only consecutive prime numbers are 2 and 3.
### If there is 2 consecutive numbers and the sum is 110 what are the two consecutive numbers?
The numbers are 54 and 56.
No
### What are prime consecutive numbers?
2 and 3 are the only example of consecutive prime numbers.
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# Numerical sequences: corrected high school math exercises in PDF.
A series of math exercises for the final year of high school on numerical sequences.
This sheet involves the following concepts:
1. definition of a sequence;
2. sum of the terms of a sequence;
3. convergence of a numerical sequence;
4. asymptotic behavior of a sequence;
5. study suite and functions;
6. recurring suites.
Exercise #1:
u is the sequence defined by and, for any natural number n, .
With the spreadsheet, the first values of and were obtained below.
1. Conjecture an expression for as a function of n.
2. Validate this conjecture by reasoning by recurrence.
Exercise #2:
V is the sequence defined by and for any natural number n, .
Prove by recurrence that for any natural number n, .
Exercise #3:
Show by recurrence that, for any natural number n, .
Exercise #4:
In this figure:
• the triangles are rectangles.
Prove by recurrence that for any natural number n, .
Exercise #5:
Study, justifying, the limit in infinity of each of the following numerical series:
Exercise #6:
u is the geometric sequence of reason 0.8 and first term .
1. For any non-zero natural number n, express as a function of n.
2. Study the limit of the sequence .
Exercise #7:
Consider the sequence defined by and for all ,
.
1)Let f be the function defined on by .
a) Study the variations of f on .
b) Deduce that if , then f ‘ (x) .
2)Prove by recurrence that, for any natural number n, .
3)Determine the direction of variation of the sequence .
Exercise #8:
The sequence is defined by and for all ,
.
1)Using a calculator or spreadsheet, determine the first ten
terms of the sequence .
2)a)What conjecture can be made about the expression of as a function of n?
b) Prove this conjecture by recurrence.
Exercise #9:
Show by recurrence that, for any natural number n not zero,
.
Exercise #10:
Determine the limit of defined on using general theorems.
.
.
.
Exercise #11:
Let be the sequence defined by and, for all ,
.
Let be the sequence defined for any natural number n by :
.
1) Show that the sequence is geometric of reason .
Specify the first term.
2) Determine the expression of as a function of n and deduce that,
for any natural number n :
.
3) Determine the limit of the sequence .
Exercise #12:
Investigate whether the following sequences, defined on , are bounded.
.
.
.
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## AKA Minimum Standards: What Schools Legally need to teach
### 4th Grade Common Core Math
Common Core State Mandated Initiative from the US Department of Education. The information below includes the learning expectations for fourth grade math using the common core. This is the national standard for common core math your school may have more expectations but these are the minimum expectations from the federal government.
In Grade 4, instructional time should focus on three critical areas: (1) developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends; (2) developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers; (3) understanding that geometric figures can be analyzed and classified based on their properties, such as having parallel sides, perpendicular sides, particular angle measures, and symmetry.
1. Students generalize their understanding of place value to 1,000,000, understanding the relative sizes of numbers in each place. They apply their understanding of models for multiplication (equal-sized groups, arrays, area models), place value, and properties of operations, in particular the distributive property, as they develop, discuss, and use efficient, accurate, and generalizable methods to compute products of multi-digit whole numbers. Depending on the numbers and the context, they select and accurately apply appropriate methods to estimate or mentally calculate products. They develop fluency with efficient procedures for multiplying whole numbers; understand and explain why the procedures work based on place value and properties of operations; and use them to solve problems. Students apply their understanding of models for division, place value, properties of operations, and the relationship of division to multiplication as they develop, discuss, and use efficient, accurate, and generalizable procedures to find quotients involving multi-digit dividends. They select and accurately apply appropriate methods to estimate and mentally calculate quotients, and interpret remainders based upon the context.
2. Students develop understanding of fraction equivalence and operations with fractions. They recognize that two different fractions can be equal (e.g., 15/9 = 5/3), and they develop methods for generating and recognizing equivalent fractions. Students extend previous understandings about how fractions are built from unit fractions, composing fractions from unit fractions, decomposing fractions into unit fractions, and using the meaning of fractions and the meaning of multiplication to multiply a fraction by a whole number.
3. Students describe, analyze, compare, and classify two-dimensional shapes. Through building, drawing, and analyzing two-dimensional shapes, students deepen their understanding of properties of two-dimensional objects and the use of them to solve problems involving symmetry.
#### Operations and Algebraic Thinking
• Use the four operations with whole numbers to solve problems.
• Gain familiarity with factors and multiples.
• Generate and analyze patterns.
#### Number and Operations—Fractions
• Extend understanding of fraction equivalence and ordering.
• Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers.
• Understand decimal notation for fractions, and compare decimal fractions.
#### Measurement and Data
• Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit.
• Represent and interpret data.
• Geometric measurement: understand concepts of angle and measure angles.
#### Geometry
• Draw and identify lines and angles, and classify shapes by properties of their lines and angles.
#### Mathematical Practices
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.
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Homework Help
# What is the area of a triangle formed by the lines x - y = 8 and the x and y axis.
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What is the area of a triangle formed by the lines x - y = 8 and the x and y axis.
Posted by xataexa on March 24, 2012 at 3:17 PM via web and tagged with area, math, triangle
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The area of the triangle formed by the x and y axes and the line x - y = 8 has to be determined. The x and y axes are perpendicular to each other and the line x - y = 8 is the equation of the hypotenuse.
To determine the area the x and y intercepts need to be determined.
x - y = 8 has an x intercept of 8 and a y intercept of -8.
This gives the area of the triangle as (1/2)*8*8 = 32
The area of the triangle formed by the line x - y = 8 and the x and y axes is 32 square units.
Posted by justaguide on March 24, 2012 at 3:25 PM (Answer #1)
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1. ## Explain?
The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?
The distance a stone falls when dropped off a cliff is proportional to the square of the time it falls. If the stone falls 64.4 ft in 2 secs, how far would it have fallen in 3 secs?
No entiendo. I don't understand. Thanks for any help!
2. 2)
if stone fall 64.4 ft in 2 sec x ft would be in three seconds
64.4ft : 2 = x : 3
from this we got
2x=3*64.4
now
x=193.2/2
x=96.6ft
3. Originally Posted by brianfisher1208
The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?
The distance a stone falls when dropped off a cliff is proportional to the square of the time it falls. If the stone falls 64.4 ft in 2 secs, how far would it have fallen in 3 secs?
No entiendo. I don't understand. Thanks for any help!
For 2
The distance is not "linearly" proportional to time,
it's proportional to the square of the time elapsed since release,
as gravity accelerates the stone to the cliff base or water.
$\displaystyle distance\ fallen=kt^2$
$\displaystyle 64.4=k(2^2)=4k$
$\displaystyle k=\frac{64.4}{4}=16.1$
After 3 seconds
$\displaystyle d=16.1(3^2)\ feet$
4. 1) if car traveling 170 mph complete lap for 1.5 or(90 sec) for x speed (mph) he would travel the lap 1.45(87 sec)
170mph : 90 s = x mph : 87 s
90x = 170 * 87
90x = 14790mph*s
x = 14790 mph*s/90s
x = 164.33 mph...
here it is ....this should be fine
5. Archie Meade that was nice...didnt readed well...anyway i cant speak so well english and dont know good math terms ... but i checked now ..by the way thanks...
i dont know if this second answer is correct
6. Originally Posted by brianfisher1208
The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?
Inverse relationship... $\displaystyle r=\frac{k}{t}$
$\displaystyle 170=\frac{k}{\left(\frac{1.5}{60}\right)}$ by converting minutes to hours.
$\displaystyle k=\frac{170(1.5)}{60}=4.25$
$\displaystyle r=\frac{4.25}{\left(\frac{1.45}{60}\right)}=\frac{ 60(4.25)}{1.45}$ mph
$\displaystyle \frac{distance}{time}=average\ speed$
$\displaystyle (av\ speed)time=distance$
distance = lap, so it's the same both times
$\displaystyle 170(1.5)=x(1.45)$
$\displaystyle x=170\frac{1.5}{1.45}$
If the racecar takes a shorter time, it's average speed must have increased.
7. Originally Posted by icefirez
1) if car traveling 170 mph complete lap for 1.5 or(90 sec) for x speed (mph) he would travel the lap 1.45(87 sec)
170mph : 90 s = x mph : 87 s
90x = 170 * 87
90x = 14790mph*s
x = 14790 mph*s/90s
x = 164.33 mph...
here it is ....this should be fine
Hi icefirez,
you are almost there!
but you left out an important piece at the start....
$\displaystyle av.speed=\frac{d}{t}$
$\displaystyle d=(av.speed)t$
Distance is the same for both laps, so
$\displaystyle 170(90)=x(87)$
$\displaystyle x=\frac{170(90)}{87}$
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# 11 and 12 Grade Math
11 and 12 grade math practice the topics are divided into three parts. Part one deals with elementary Algebra, part two provides a basic course in trigonometry and part three considers elements of two dimensional Co-ordinate Geometry including solid geometry and mensuration.
Each topic that are covered in 11 and 12 grade math, concepts is enlightened with a summarization which includes important theorems, results and formula are discussed in each topic with numerous types of solved examples. Sufficient number of problems have been inserted in grade 11 and 12 practice math task worksheets beginning with easier followed gradually by harder ones.
It is expected that students should be acquainted with the basic 11 and 12 grade math concepts relating to each topic and should be able to apply those to simple elementary problems, preferably numerical.
Algebra:
In 11 and 12 grade math these are the topics which are covered in Algebra.
● Variation: Direct, inverse and joint variation, theorem of joint variation. Application to simple examples of time and worktime and distance, mensuration, physical laws, economics.
● Arithmetical Progression:
Definition of A. P., common difference, term, summation of terms. Sum of n natural numbers. Sum of the and cubes of first natural numbers, A. M.
● Geometric Progression: Definition of G. P., Common ratio, general term, summation of n terms, G. M.
● SurdsRational numbers. To show that √2 is not rational. Idea of irrational numbers, surds, quadratic surds, mixed surds, conjugate surds, properties of surds, if a + √b = 0 then a = 0, b = 0 ; if a + √b = c + √d , then a = c, b = d. Rationalization of surds. Square root of quadratic surds.
● Laws of Indices: Proofs for fundamental laws of indices for positive integers, statement for fractional, zero and negative indices : simple applications.
Logarithms: Definition, base, index, general properties of logarithms, common logarithm, characteristic and mantissa, antilogarithm, use of logarithmic tables.
Complex Numbers: Complex numbers, significance of the imaginary unit i, addition, multiplication and division, properties of complex numbers ; if a + ib = 0, then a= 0, b= 0 ; if a + ib = c + id, then a = c, b = d. Argand diagram. Modulus. Argument, complex conjugate. Square root of complex numbers, cube roots of unity and their properties.
● Theory of Quadratic Equations: Quadratic equations with real roots. Statement of fundamental theorem of algebra. Roots (two and only two roots), relation between roots and coefficients of a quadratic equation. Nature of roots, common roots. Nature of the quadratic expression ax$$^{2}$$ + bx + c — its sign and magnitude.
● Permutations: Definition. Theorem on permutations of n different things taken r at a time, things not all different, permutation with repetitions (circular permutation excluded).
● Combinations: Definition : Theorem on combination of n different things taken r at a time, things not all different. Basic identities. Division into two groups (circular combination excluded).
● Binomial Theorem for Positive Integral Index: Statement of the theorem, proof by method of induction. General term, number of terms, middle term, equidistant terms. Simple properties of binomial coefficients.
● Infinite Series: The power series Σxn. Binomial series (1 + x)n (n ≠ positive integer), exponential and logarithmic series with ranges of validity (statement only). Simple applications.
Trigonometry:
In 11 and 12 grade math these are the topics which are covered in Trigonometry.
Revision exercises of the topics covered in the syllabus of Secondary Mathematics.
The relation s = rθ.
● The Negative and Associated angles: - θ, 90° ± θ, 180° ± θ, 270° ± θ, 360° ± θ.
● Trigonometrical Ratios of Compound Angles: Geometrical methods (for Sine and Cosine only). Product formulae, sum & difference formulae.
● Multiple and Sub-multiple Angles: Simple problems.
Identities (conditional) of Trigonometrical Ratios (Sum of angles π or π/2)
General Solutions of Trigonometrical Equations.
Trigonometrical Inverses (specific mention of principal branch).
● Graphs of Trigonometrical Functions: y = sin mx, y = cos mx and y = tan mx, where m is an integer with stated values.
● Properties of Triangles: Basic relations between sides, angles, circus-radius and in-radius. Area of triangles in different forms. Simple and direct applications.
Plane Analytical Geometry, Mensuration & Solid Geometry:
In 11 and 12 grade math these are the topics which are covered in Plane Analytical Geometry, Mensuration & Solid Geometry.
Rectangular Cartesian Co-ordinates: Directed line and directed line segment, co-ordinate system on a directed line and rectangular Cartesian co-ordinate system in a plane.
Polar Co-ordinates: Notion of directed angles and polar co-ordinate system. (Radius vector o be taken as positive.)
Transformation from Cartesian to Polar Co-ordinates and vice-versa.
Distance between Two Points: Division of a line segment in a given ratio. Area of a triangle (all in terms of rectangular Cartesian co-ordinates). Application to geometrical properties. Verification of Apollonius’ Theorem.
Locus: Concept of locus by simple illustration. Equation of locus in term of rectangular Cartesian co-ordinates.
Equations of Straight Lines (in rectangular Cartesian co-ordinates only): Notion of inclination and slope of a line. Slope in terms of co-ordinates of two points on it. Equations of co-ordinate axes, equations of lines parallel to co-ordinate axes, slope-intercept form, point-slope form, equation of the line through two given points, intercept form, symmetric form, normal form. Every first degree equation represents a straight line.
Angle between Two Lines: Conditions of perpendicularity and parallelism of two lines. Equation of a line parallel to a given line. Equation of a line perpendicular to a given line, conditions that two lines may be identical.
Distance of a Point from a Given Line: Notion of a signed distance of a point from a line, position of a point with respect to a line, sides of a line. Equations of bisectors of angles between two lines, equation of bisector of an angle that contains the origin.
● Equations of Circles: Standard equation. Equation of a circle given center and radius. General equation of the form x2 + y2 + 2gx + 2fy + c = 0 represents a circle. Reduction to standard form (parallel. transformation assumed). Equation of a circle if end points of a diameter be given (all in terms of rectangular Cartesian co-ordinates). Parametric equation of a circle. Outside and inside points of a circle. Intersection of a line with a circle. Equation of a chord with respect to the middle point.
● Conic Section: Idea of conic sections as sections of cone. Focus— Directrix definitions of a conic section, eccentricity, classification according to the value of eccentricity.
● Parabola: Standard equation. Reduction of a parabola of the form x = ay2 + by + c or y = ax2 + bx + c to the standard form y2 = 4ax or x2 = 4ay respectively, elementary properties. Parametric equation.
● Ellipse and Hyperbola: Standard equations only. Conjugate hyperbola. Elementary properties. Parametric equation.
To investigate whether a point is inside, on or outside a conic. Intersection of a straight line with a conic, equation of chord of a conic with respect to the middle point.
● Diameters of Conic: Definition, equation of a diameter. Equation of a conjugate diameter: elementary properties of conjugate diameter (statement only).
Solid Geometry: Incidence relations between points and planes, lines and planes, coplanarity, skew lines, parallel planes. Intersecting planes—Two intersecting planes cut one another in a straight line and in no point outside it, perpendicular to a plane, projection of a line segment on a line and on a plane. Dihedral angle.
Corollary: Three straight lines intersecting pair wise or two parallel lines and its transversal lie in the same plane.
Theorems: Theorem 1: If a straight line is perpendicular to each of two intersecting straight lines at their point of intersection, it is also perpendicular to the plane in which they lie. (Apollonius’ Theorem may be used.)
Theorem 2: All straight lines drawn perpendicular to a given straight line at a given point are co-planar.
Theorem 3: If two straight lines are parallel and if one of them is perpendicular to a plane, then the other is also perpendicular to the same plane and its converse.
Theorem 3: Theorem of Three Perpendiculars
Mensuration:
Surface areas and volumes of prism and pyramid
`
Formula
Mathematical Induction
Variation
Surds
Complex Numbers
Arithmetic Progression
Geometric Progression
Logarithm
Trigonometry
Measurement of Angles
Trigonometric Functions
Converting Product into Sum/Difference and Vice Versa
Inverse Trigonometric Functions
Trigonometrical Table
Co-ordinate Geometry
Locus
The Straight Line
The Circle
The Parabola
The Ellipse
The Hyperbola
Solid Geometry
Mensuration
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# AP Physics C: Mechanics
## Presentation on theme: "AP Physics C: Mechanics"— Presentation transcript:
AP Physics C: Mechanics
Work & Energy
The ability of a physical system to cause change
What is Energy? The ability of a physical system to cause change What is its definition? Energy can come in many forms and is transferrable. It is a conserved quantity in the universe. What are its properties? Chemical, electrical, nuclear, thermodynamic, light, mechanical, kinetic, potential, elastic, etc. What forms can it be in?
Work (Energy) is a scalar quantity.
What is Work? The amount of energy transferred by a force acting through a displacement What is its definition? Work (Energy) is a scalar quantity. What are its properties? How can it be calculated? W=fscosθ
Work The area under an Force-displacement graph
Work Work is an energy transfer.
If energy is added to the system then W is positive. If energy is transferred from the system then W is negative.
Work Work is a scalar. The NET work on an object is the sum of the work done by each force acting on it.
Work Units of work 1 erg=10-7J System Unit Alternate name SI Nm
Joule (J) cgs dyne*cm erg English foot*pound - 1 erg=10-7J
Work-Energy Theorem The work done by a net force in displacing a particle equals the change in kinetic energy of the particle. Stress on the word particle. This is a drastic idealization.
Mechanical Potential Energy
Gravitational Elastic
Conservation of Energy
This is always true, but mechanical energy is only conserved when only conservative forces are present. Work done by a conservative force is path independent Forces such as friction and air drag are non-conservative. They result in energy dissipated as heat.
What is it?
Power The time rate at which work is being done.
How do we calculate it? If a net force acts on a particle and does work W over an interval of time Δt, then the average power is:
Power The time rate of energy transfer. Instantaneous Power
Power Watt are the units? 1hp=550ft*lbs/sec=746W
Note: A kilowatt-hour is often used by power companies. What is it a measure of?
Potential Energy and Conservative Forces
Potential energy is only associated with conservative forces. If an object or field does work on some external object, energy is transferred from the object or field to the external object: The energy transferred as work decreases the potential energy of the system from which it came.
Potential Energy and Conservative Forces
The conservative force equals the negative (partial) derivative of the potential energy with respect to x. The negative sign means that the final potential energy of the system is lower than the initial as the conservative force does work.
Potential Energy and Conservative Forces
Examples:
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Wizard
Joined: Oct 14, 2009
• Posts: 22916
July 4th, 2012 at 1:15:34 PM permalink
You are 40% of the way across a train bridge when you hear a train whistle behind you. You must choose to turn around or keep going forward to get off the bridge before the train runs you over. It turns out that regardless of which way you choose you will get off the bridge just in the knick of time. You can run 10 miles per hour. How fast is the train going?
As usual, please put solutions in
John Galt drives a Prius
.
It's not whether you win or lose; it's whether or not you had a good bet.
Nareed
Joined: Nov 11, 2009
• Posts: 11413
July 4th, 2012 at 1:28:08 PM permalink
Quote: Wizard
You are 40% of the way across a train bridge when you hear a train whistle behind you. You must choose to turn around or keep going forward to get off the bridge before the train runs you over.
If the whistle, along with the attached train, is behind you, wouldn't it run you over sooner if you turned around?
Random guess:
40 mph
Donald Trump is a fucking criminal
Doc
Joined: Feb 27, 2010
• Posts: 7076
July 4th, 2012 at 2:18:03 PM permalink
Quote: Nareed
If the whistle, along with the attached train, is behind you, wouldn't it run you over sooner if you turned around?
Yes, if it were going to run over you both ways, that would happen sooner if you turned back. But depending on speeds and where you are on the bridge, there may be reversing answers on which direction is better. In this case, they are equally marginal.
X = length of bridge
t1=time until train reaches bridge
t2=time until train has crossed bridge
S=train speed
t1=(0.4X/10)
t2=(0.6X/10)
t2-t1=X/S
0.6X/10 - 0.4X/10 = 0.2X/10 = X/S
S =10/0.2 = 50 mph
Nareed
Joined: Nov 11, 2009
• Posts: 11413
July 4th, 2012 at 2:23:02 PM permalink
Quote: Doc
Yes, if it were going to run over you both ways, that would happen sooner if you turned back. But depending on speeds and where you are on the bridge, there may be reversing answers on which direction is better. In this case, they are equally marginal.
Ok.
BTW, your spoiler label shows the result.
Donald Trump is a fucking criminal
Ibeatyouraces
Joined: Jan 12, 2010
• Posts: 11933
July 4th, 2012 at 2:31:23 PM permalink
deleted
DUHHIIIIIIIII HEARD THAT!
Doc
Joined: Feb 27, 2010
• Posts: 7076
July 4th, 2012 at 2:31:41 PM permalink
Quote: Nareed
Ok.
BTW, your spoiler label shows the result.
I know. I thought we were supposed to hide the solution, i.e. the how to figure it out, not the answer. If I misunderstood, I apologize.
Wizard
Joined: Oct 14, 2009
• Posts: 22916
July 4th, 2012 at 5:08:32 PM permalink
Quote: Doc
I know. I thought we were supposed to hide the solution, i.e. the how to figure it out, not the answer. If I misunderstood, I apologize.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4740
July 4th, 2012 at 5:15:32 PM permalink
Let L be the distance across the bridge you are when you hear the whistle. Since it is 40% of the way across the bridge, either you must turn around and run distance L, or keep going and run distance 1.5 L.
Let T be the distance the train is from the start of the bridge, and X be its velocity.
It takes you L/10 hours (if T is in miles) for you to reach the start of the bridge; it takes the train L/X hours (if L is in hours and X is in MPH) to get there. Therefore, L/10 = T/X.
It takes you 1.5 L/10 hours for you to reach the end of the bridge; it takes the train (T + 2.5L)/X hours. Therefore, 1.5 L/10 = (T + 2.5L)/X.
L/10 = T/X, so 1.5 L/10 = 1.5 T/X
1.5 T/X = (T + 2.5L)/X
1.5 T = T + 2.5 L
T = 5L
Substituting for T in L/10 = T/X, we get L/10 = 5L/X, so X = 50.
The train's velocity is 50 MPH.
ahiromu
Joined: Jan 15, 2010
• Posts: 2102
July 4th, 2012 at 5:28:29 PM permalink
L: Length of the bridge
You can get .4*L on the bridge before the train reaches said bridge. That means if you continue walking in the same direction as the train, you'll reach .8*L. In other words, in the time it takes you to run .2*L the train covers the entire length L... so the train is five times faster than you or 50 mph.
Did someone just watch Stand By Me?
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"
P90
Joined: Jan 8, 2011
• Posts: 1703
July 4th, 2012 at 5:52:12 PM permalink
Let bridge span be L, with L1 distance to the train.
Let train velocity be v.
Don't click the spoiler until reading the rest of the post.
At the moment you hear the whistle (t1), the train has traveled L1*v/c, c=340
At the moment you are being run over as you are exiting the bridge (t2), the train has traveled L1*v/340+v*.6L/4.4704
At the moment you run into the train as you backpedal, the train has traveled L1*v/340+v*.4L/4.4704
These are separated by the bridge: L1*v/340+v*.4L/4.4704+L=L1*v/340+v*.6L/4.4704
.4Lv/4.4704+L-.6Lv/4.4704=0
-.2Lv/4.4704=-L
Lv=L*4.4704/.2
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# How many atoms are in 3 moles of CO2?
Contents
## How many atoms are in a mole of CO2?
Let’s consider the carbon dioxide molecule. We know it has the formula CO2, and this tells us that: 1 mole carbon dioxide contains 6.02 x 1023 molecules. 1 mole of carbon atoms per mole of carbon dioxide = 6.02 x 1023 carbon atoms.
## How many CO2 are in 3 moles?
There are 1.806 X 1024 molecules in 3 moles of CO2.
## How many atoms are in 3 moles?
No. of atoms = 3 × 6.022 × 10^23 = 18.066 × 10^23 atoms.
## Does CO2 have 3 atoms?
Carbon dioxide is composed of one carbon atom covalently bonded to two oxygen atoms. It is a gas (at standard temperature and pressure) that is exhaled by animals and utilized by plants during photosynthesis. Carbon dioxide, CO2, is a chemical compound composed of two oxygen atoms and one carbon atom.
## How many atoms are in 2 moles of CO2?
Explanation: In one mole of carbon, there are 6.02×1023 atoms. So in two moles, there will be twice that: 1.204×1024 .
## How much atoms are in a mole?
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro’s Number (6.0221421 x 1023).
THIS IS IMPORTANT: Does Face Wash remove pimples?
## How many CO2 molecules are in 0.50 mole of CO2?
3.011×1023 individual carbon dioxide molecules….how any oxygen atoms in this molar quantity?
## What are moles of atoms?
A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
## How many atoms are in 3.0 moles of Sn?
There are 3×NA tin atoms in 3.0 mol of tin metal, where NA is Avogadro’s number, 6.022×1023 mol−1 .
## How do you find the number of atoms in 3 moles of carbon atoms?
Explanation:
1. 1 Mole is 6.02⋅1023 molecules.
2. Therefore 3 Moles of Carbon Dioxide CO2 =3(6.02⋅1023) or 18.06⋅1023 molecules of CO2.
3. There is one atom of Carbon and two atoms of Oxygen in each molecule of Carbon Dioxide CO2.
4. (3 atoms) 18.06⋅1023 molecules =54.18⋅1023 atoms.
5. 5.418⋅1024 atoms.
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# Difference between revisions of "1988 AIME Problems/Problem 12"
## Problem
Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$, $b$, $c$, and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$.
## Solution
Call the cevians AD, BE, and CF. Using area ratios ($\triangle PBC$ and $\triangle ABC$ have the same base), we have:
$\frac {d}{a + d} = \frac {[PBC]}{[ABC]}$
Similarily, $\frac {d}{b + d} = \frac {[PCA]}{[ABC]}$ and $\frac {d}{c + d} = \frac {[PAB]}{[ABC]}$.
Then, $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1$
The identity $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1$ is a form of Ceva's Theorem.
Plugging in $d = 3$, we get
$$\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1$$ $$3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)$$ $$3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27$$ $$9(a + b + c) + 54 = abc=\boxed{441}$$
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# An Interactive Higher-Order Thinking Tool
Look in any science or economics textbook and you will find graphs. The graphs tell a story which we want our students to understand. Typically, the graphs are mathematical models of systems that are dynamic and influenced by other variables not plotted. So, here come mathematical equations with four or five variables and we expect understanding. For beginning students, presenting these equations is not always the best way to provide an introduction to the investigation of a system. How about offering them a dynamic, graphical visualization of the model and hold off on the mathematical computations at this point?
Well, if you have a computer, you probably have Excel, and hence, the ability to produce an interactive graph of your model or a simulation of the system described by a target equation. Now you can pose questions and ask what is going to happen. Which variable has the greatest effect on the system? How does the graph change if we increase the value of a variable? An active learning environment has just been created in your classroom. Additionally, out-of-class investigative projects with further simulations can be assigned. This article provides the basis for an explanation of the development of interactive Excel spreadsheets. A basic knowledge of Excel to handle data and produce graphs is assumed. See Fryer (2002) or Sinex and Gage (2001) for help. No programming is required!
Constructing a Simple Interactive Excel Spreadsheet
Since we cannot just put an equation or function into Excel and plot it, we need to get the function in through calculation of data, and then plot the data. If the data is modified, the plot will adjust. Let's use as an example the standard form of the quadratic equation: y = ax2 + bx + c. How is the shape and position of the curve influenced by changing the coefficients a, b, and c? The coefficients are the variables we want to be able to adjust in value. Here is a diagram of what we are going to develop.
This is simply an adjustable computational spreadsheet, where the graph responds to changes in a, b or c. Our formula is the quadratic equation given above. In Excel, to lock-in the use of a single cell so it will not change when you drag a formula, you add the \$ (dollar sign) to the cell notation (ex. \$A\$1). The graphs must be done as xy scatter plots, as this is the only plot where the x-axis is a variable. The curves on these plots should be the smoothed ones, as Excel uses a cubic spline to fit the data. A regression line (trendline) can be displayed too, as it will adjust to changes in the data as the regression equation is recomputed. This allows monitoring (display in a cell) of parameters, such as slope. Keep in mind that graphs will rescale automatically to the range of the data. If you want a particular x-y scale, deselect auto-scale and set it. A tutorial in Excel on how to build this spreadsheet and add further interactivity is included (Excel tutorial steps 1-4, click on the tabs near the bottom of your screen).
To prevent students from "fouling up" the spreadsheets, I recommend turning on the protection feature. Go to Tools from the menu, select Protection, and then select Protect Sheet. Use of a password is optional. Unprotect is accomplished the same way. However, before this is done, go to each cell where data is going to change and deselect the protection for the cell (deselect "locked"). This is done by right clicking on the cell, select Format Cells… and select the Protection tab. If you want to hide your formulas from view, then select hidden. See the blue box in step 4 of the Excel tutorial.
Do you want the students to see the data as it changes? If not, you can click on the column and select hide. It will not be viewable or printed. You can also place the graph on top of the columns too.
Fostering Higher-Order Thinking
Once you have the interactive Excel spreadsheet, a series of questions needs to be designed that explore the effects of changing the variables. It's important to ask students to predict what will happen if a specific change is made. Then let them investigate it. College students are often afraid to make mistakes; however, predictions are just educated guesses. As students start predicting they elicit prior knowledge and set up the opportunity to enhance conceptual understanding. Use the interactive spreadsheet to explore and discover relationships and not as a confirmation tool. Bring the mathematics in later or, better yet, derive the equation from graphical exploration and analysis. The effects of multiple variables can easily be investigated. In science, developing a simple model by experimental data collection is easily followed and extended by examining the effect of other variables. Simulations can show the sensitivity of a model to certain variables. Models can range from simple to very complex mathematical ones as discussed in Liengme (2002).
The use of a reference line, a set model with specific known conditions, is a good tool to evaluate comparative graphs (Excel tutorial steps 5-6). The analysis and interpretation follows as variables are adjusted and the graphs of the model at different conditions appear for comparison to the set conditions. The use of such an interactive simulation allows topics that are difficult to illustrate with static graphs to be more understandable when viewed in a dynamic fashion.
Build an activity around the simulation that guides your students to discover behavior and relationships initially. To help guide students to use a simulation wisely, it is advisable to suggest that only one variable at a time be explored and that the full range of a variable be studied in an organized fashion – low to mid to high range. Here are a set of questions to investigate the behavior of the quadratic equation that can be used with the included tutorial.
• How does changing the "a" coefficient influence the shape of the curve?
• What is the difference between two curves for a > 0 and a < 0?
• What does changing the "b" coefficient do to the curve?
• What does changing the "c" coefficient do to the curve?
• If "a" approaches zero, what happens to the curve?
The table below provides several examples using interactive Excel spreadsheets in three disciplines. Second-semester general chemistry students use the Beer's Law Simulator after a laboratory where they have collected data to derive Beer's Law. The simulator allows them to develop a through understanding of variables that influence a Beer's Law measurement, a very powerful analytical tool in chemistry and biology. Understanding when the model may breakdown and not hold true is an important aspect to consider also. Assessment questions on possible errors during colorimetric analysis are now easy to compose and use.
DisciplineSimulationMathematicsInteractive Curves - Exploring functions in college algebraChemistryBeer's Law Simulator (pdf activity)EconomicsTrade
Adding "Bells and Whistles" to the Simulation
Excel has the ability to add more interactivity. Two very useful items for adjusting variables are the scroll bar and spinner, which are available from the forms toolbar, not the ones from the control toolbar (Bradley, 2002). The scroll bar and spinner, which can be linked to a cell, allow values to be changed by clicking and, on the scroll bar, dragging. They have some limitations, such as only positive values and whole numbers can be entered; however, this can be overcome by writing formulas (see Excel tutorial steps 6-7).
The comment box, which can be added to any cell, is a great way to add hints or short explanations of calculations or data. Just right click on any cell, and select Insert Comment. A cell with a comment box is recognized by a little red triangle in the upper right corner (see Excel tutorial step 6). When the cursor is placed on this cell or you mouse over, the comment box appears.
Using Excel 2002 (part of Office XP), it is possible to get a movable data point on a graph. To do this you set up a computation and fit the data points with a regression line. The regression line responds to input variable changes. Next add a new data point in the columns, where an x-value is supplied by a separate cell input and the y-value is computed using the regression results. This is done as a new series and categories (x-values) in first column option and you can add marker lines (Liengme, 2002). This point will then respond to changes in x, calculate y, and replot on the graph. This is a great way to introduce the idea of interpolation and explore behavior especially of non-linear models (see Excel tutorial steps 8-9).
The data validation option, (under Data on the menu bar and then select Validation…), allows you to set limits on input parameter boxes, show a limit message, and/or a warning message as well. On selecting the cell, the limits appear as a message similar to the comment boxes. This is very useful for novices exploring models for the first time (see Excel tutorial step 10).
Some Final Thoughts
Exploring and discovering with interactive Excel spreadsheets, aka simulations, is a great way to get students involved in active learning. All of this is accomplished from a purely computational venue with no macros or use of Visual Basic for Applications (VBA). Adding macros/VBA will allow more advanced simulations. For examples in economics, see Between the Sheets and in physical sciences, see Interactive Excel.
My freshman chemistry students are constantly involved using simple computational simulations. More advanced students might get involved developing or "dissecting" to understand the basis of the computation. Excel offers educators at all levels an inexpensive and powerful capability for dynamic graphical visualization to enhance learner-centered instruction.
Email: Scott Sinex
References
Helen Bradley (July, 2002) Excel's Best-Kept Secret: What? You're Not Using the Forms Toolbar?, WinPlanet, (accessed June 2003).
Fryer, W. (2002) Excel Spreadsheets: An Excel Shortcourse for Teachers, Technology and Learning Magazine, (accessed June 2003).
S.A. Sinex and B.A. Gage (2001) Using Excel for Handling, Graphing, and Analyzing Scientific Data: A Resource for Science and Mathematics Students, (accessed June 2003).
B.V. Liengme (2002) A Guide to Microsoft Excel 2002 for Scientists and Engineers, 3rd edition (opens in new tab), Butterworth-Heinemann, Oxford, England.
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# How to solve $V(n) = 2 \cdot V(n-1) + 2 \cdot n$
How to solve $$V(n) = 2 \cdot V(n-1) + 2\cdot n$$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?
• Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
– Did
Jan 20, 2019 at 11:03
• @user623855 2n ≠ 2^n.
– Did
Jan 20, 2019 at 11:04
Try to expand and generalize the case (by induction): $$V(n) = 2\times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = \sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $$V(1) = 2$$ in this case.
Let $$V(m)=f(m)+am+b$$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2\{f(m-1)+a(n-1)+b\}$$ $$2n=f(n)-2f(n-1)-an+2a-b$$
Set $$2a-b=0, -an=2$$ so that $$f(n)=2f(n-1)=2^rf(n-r),0\le r\le n$$
$$\implies a=-2, b=2(-2)=?$$
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# What is the slope of r=tantheta-theta at theta=pi/8?
Derivative with Polar Coordinates is (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)
$\frac{\setminus \partial r}{\setminus \partial \theta} = \frac{1}{\cos} ^ 2 \left(\theta\right) + 1$
$r \left(\theta = \frac{\pi}{8}\right) = 0.0 22$
$\frac{\setminus \partial x}{\setminus \partial y} = \frac{\tan \frac{\theta}{\cos} \left(\theta\right) + \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{1}{\cos} \left(\theta\right) + \cos \left(\theta\right) - r \sin \left(\theta\right)}$
$\frac{\setminus \partial x}{\setminus \partial y} \left(\theta = \frac{\pi}{8}\right) = 1.57$
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# Search by Topic
#### Resources tagged with Similarity similar to Von Koch Curve:
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### There are 30 results
Broad Topics > Transformations and their Properties > Similarity
### Von Koch Curve
##### Stage: 5 Challenge Level:
Make a poster using equilateral triangles with sides 27, 9, 3 and 1 units assembled as stage 3 of the Von Koch fractal. Investigate areas & lengths when you repeat a process infinitely often.
### Look Before You Leap
##### Stage: 4 Challenge Level:
Can you spot a cunning way to work out the missing length?
### Sierpinski Triangle
##### Stage: 5 Challenge Level:
What is the total area of the triangles remaining in the nth stage of constructing a Sierpinski Triangle? Work out the dimension of this fractal.
### Squareflake
##### Stage: 5 Challenge Level:
A finite area inside and infinite skin! You can paint the interior of this fractal with a small tin of paint but you could never get enough paint to paint the edge.
### Figure of Eight
##### Stage: 4 Challenge Level:
On a nine-point pegboard a band is stretched over 4 pegs in a "figure of 8" arrangement. How many different "figure of 8" arrangements can be made ?
### Strange Rectangle
##### Stage: 5 Challenge Level:
ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles.
### Squirty
##### Stage: 4 Challenge Level:
Using a ruler, pencil and compasses only, it is possible to construct a square inside any triangle so that all four vertices touch the sides of the triangle.
### From All Corners
##### Stage: 4 Challenge Level:
Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
### Golden Triangle
##### Stage: 5 Challenge Level:
Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.
##### Stage: 4 Challenge Level:
Find the area of the shaded region created by the two overlapping triangles in terms of a and b?
##### Stage: 4 Challenge Level:
Two ladders are propped up against facing walls. The end of the first ladder is 10 metres above the foot of the first wall. The end of the second ladder is 5 metres above the foot of the second. . . .
### Napkin
##### Stage: 4 Challenge Level:
A napkin is folded so that a corner coincides with the midpoint of an opposite edge . Investigate the three triangles formed .
### Pentabuild
##### Stage: 5 Challenge Level:
Explain how to construct a regular pentagon accurately using a straight edge and compass.
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### Two Triangles in a Square
##### Stage: 4 Challenge Level:
Given that ABCD is a square, M is the mid point of AD and CP is perpendicular to MB with P on MB, prove DP = DC.
### Chord
##### Stage: 5 Challenge Level:
Equal touching circles have centres on a line. From a point of this line on a circle, a tangent is drawn to the farthest circle. Find the lengths of chords where the line cuts the other circles.
### Nicely Similar
##### Stage: 4 Challenge Level:
If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle?
### All Tied Up
##### Stage: 4 Challenge Level:
A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be?
### Fit for Photocopying
##### Stage: 4 Challenge Level:
Explore the relationships between different paper sizes.
### Slippage
##### Stage: 4 Challenge Level:
A ladder 3m long rests against a wall with one end a short distance from its base. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m high. What is the maximum distance. . . .
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
### The Eyeball Theorem
##### Stage: 4 and 5 Challenge Level:
Two tangents are drawn to the other circle from the centres of a pair of circles. What can you say about the chords cut off by these tangents. Be patient - this problem may be slow to load.
### Chords
##### Stage: 4 Challenge Level:
Two intersecting circles have a common chord AB. The point C moves on the circumference of the circle C1. The straight lines CA and CB meet the circle C2 at E and F respectively. As the point C. . . .
### Folding Fractions
##### Stage: 4 Challenge Level:
What fractions can you divide the diagonal of a square into by simple folding?
### Sitting Pretty
##### Stage: 4 Challenge Level:
A circle of radius r touches two sides of a right angled triangle, sides x and y, and has its centre on the hypotenuse. Can you prove the formula linking x, y and r?
### Folding Squares
##### Stage: 4 Challenge Level:
The diagonal of a square intersects the line joining one of the unused corners to the midpoint of the opposite side. What do you notice about the line segments produced?
### Flower
##### Stage: 5 Challenge Level:
Six circles around a central circle make a flower. Watch the flower as you change the radii in this circle packing. Prove that with the given ratios of the radii the petals touch and fit perfectly.
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### Bus Stop
##### Stage: 4 Challenge Level:
Two buses leave at the same time from two towns Shipton and Veston on the same long road, travelling towards each other. At each mile along the road are milestones. The buses' speeds are constant. . . .
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Electromagnetic Induction Student's Results.html
# Electromagnetic Induction Student's Results.html - Exercise...
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Exercise 1: A wire moving through a constant, uniform magnetic field 1. When the wire moves from left to right through the field, in what direction do electrons in the wire move? Why? Explain, using the right hand rule. Well, the RHR tells us that the magnetic force is pointing up, the moving charge is moving left and the magnetic field line is tilted left. We find this using our three fingers (thumb, index finger and middle finger). 2. What happens to the buildup of charges at each end of the wire when the wire stops moving? Why? We can answer the question by the fact that when charges are sitting still or when it stops moving, they are unaffected by magnetic fields so it just stays neutral. 3. Is there any correlation between the speed of the wire and the amount of charge buildup at the ends of the wire? Explain, referring to an equation stated above. Well the same magnitude of force is is emitted when you move the force at the same speed. Only the direction of the force will change if you start moving the wire in the opposite direction, but the wire does seem to build more charges the faster you move it opposed to slowly moving it. Exercise 2: An induced emf 4. In Exercise 1, you saw how the wire’s velocity affected the amount of charge that accumulated. How does the wire’s velocity affect the emf in this exercise? What do the two exercises have in common? When changing the wires velocity between positive, it emits a positive voltage and when changing it to negative, it gives a negative voltage. So goes for magnetic field strength when increasing it moves up and when decreasing the V moves down. 5. What is the relationship, if any, between the velocity of the wire and the induced emf (is it linear, inverse, squared)? It seems to have an inverse relationship as it is changes directions dependent on positive or negative.
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• Fall '08
• Sherman
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# Bits & Pieces Adding Up to Solving π
by R. M. Stainforth
## (A)Squares and square roots
(1) In arithmetic a “square” is simply the result obtained by multiplying any number by itself.
25 is the square of 5 ( = 5 x 5 ) : 4 is the square of 2 ( = 2 x 2 )
and so on.
A “square root” is the opposite. It is the number which was multiplied by itself to obtain a square. In the examples above,
5 is the square root of 25 : 2 is the square root of 4
Likewise 10 is the square root of 100, 25 of 625, and so on. To save a lot of writing, mathematicians use a lot of ‘symbols’ to express mathematical ideas. The symbol for a square root is like this:
You can see for yourself that to write:
is a lot shorter than: “the square root of 100 is 10”
(2) In the examples above the square roots are whole numbers. The corresponding squares are sometimes called “perfect squares” because they have whole-number roots.
But every number has a square root and you can find it by a fairly simple calculation. But in most cases the roots are fractions which are called “irrational” because the formula for calculating them results in long decimals which could go on and on without ever coming to an end. For practical purposes we only use the first few numbers of the decimal. For instance, * You can check that with a calculator, but you will also find that to use only two places of decimals (1.73) or, at most, three places (1.732) is accurate enough for all ordinary use.
Similarly, *
Some calculators will give you the square root of any number if you press the symbol, or you can buy tables of them.
The calculations below can be done on a calculator to check on and
1.4142 x 1.4142 = 1.9999 * 1.73 x 1.73 = 2.9929 1.732 x 1.732 = 2.9998 * 1.73205 x 1.73205 = 2.999997202 *
For normal practical purposes, 1.99…=2, 2.99…=3, and so on.
(3) “Squares” in geometry.
A square, as you know, is a rectangle whose sides are all the same length. If you divide each side into equal segments, then join all the opposite points, you get a family of small, equal squares. For instance:
4 equal segments on each side, 16 small squares
So you see the connection. Four ‘squared’ becomes 16, and similarly for any number you choose. That explains the use of the geometrical word ‘square’ in arithmetic.
(4) Algebra and squares.
Don’t be frightened of the word algebra. Actually it’s from Arabic, like numerous other English words beginning with “al-”, which simply means “the”: alcohol, alchemy, alembic, and you can find others in a dictionary.
Algebra is really arithmetic, but it deals with numbers in a general sense. It uses a symbol, very often the letter x, which can stand for any number at all until you limit it in some way.
As a very simple example, let’s say that
We know immediately that x is the square root of z
so
Let’s now say that
then
x must be 5: there is no alternative (unless we go into negative numbers, but that’s a whole new field).
Try another simple case:
you are told that
(the unknown number y is twice the unknown x)
and at the same time
(the unknown x and y total 6 when added together)
Fiddle around in your head and you’ll quickly see (I hope!) that the only solution is that and .
Note that there are any number of solutions (answers) when you only have one of the equations, but only a single solution that satisfies both of them.
I hope this is enough to demonstrate to you that algebra is a shorthand way of expressing mathematical ideas and relationships. You just dealt with a pair of what are called ‘simultaneous equations’ and in algebra all we needed to write was:
Now try to express the concept in English without using symbols. The shortest I can find is:
“Find two numbers: they add up to 6 and one is twice the other.”
This requires 50 letters/numbers whereas the algebraic way needed only 10 letters/numbers/symbols—and this is a very simple case.
## (B)Descartes and Cartesian coordinates
(1) René Descartes was a French philosopher and mathematician born 400 years ago.
His name is preserved in the phrase ‘Cartesian coordinates.’ They are a way of converting mathematical problems, equations, formulæ, etc., into drawings which are much easier to understand than a jumble of numbers and symbols.
You can begin with graph paper, of which the simplest type shows a grid of small equal squares. Usually (but not necessarily) the bottom line and the left-hand margin are thicker than the lines of the grid, and they are marked off with equal divisions. Usually these divisions are marked with successive numbers, from left to right across the bottom and upwards on the left-hand margin.
Now, suppose you want an easy way to find half of any number you think of, without making an arithmetical calculation. You begin with a diagram like ‘Figure 1.’ Mark off some points (as done in red) where you know the answer. For instance, start off at 10 on the bottom row then move up to the level of half-of-ten, = 5, on the side line and mark the point where your pencil is. Do the same for 4/2, 8/4, 24/12 points (as in red) or any others you like of the same family.
Now you can’t help but notice that all these known points lie on a straight line.
Next you realize that if you choose any number you like, find its place on the bottom line, move straight up to the sloping line, and read across to the side line, that gives you half of your first number.
Just as an example, I chose 12 1/2 (see purple lines) and the diagram (‘graph’) shows that half of that is 6 1/4.
(2) Simply to save space and a lot of writing, we usually call the bottom line of a graph the “X-axis” and use it for plotting the numbers we know or choose to start with. They are often called “x.” We call the left-hand margin the “Y-axis” and use it for plotting the numbers derived from the “x” numbers. Quite often we call these derived numbers “y.”
In Figure 1 the Y-number for any point on the red line corresponds to half the X-number immediately below.
You can see that it needed two lines of writing to convey that idea, but in algebra we shorten it to simply
Let’s try another graph. See Figure 2. This time we plot the squares (remember them?) of the numbers on the bottom line (the X-axis). In algebra we say,
We agreed that Figure 2 would be the graph of the equation . The method of construction is just the same as on Figure 1, which plotted but this time you can see that the result is not a straight line. It is a curve (called a parabola). You can use it to get the square root or the square of a number. For example, you can check that 4 1/2 squared = 20, (there is some approximation, partly depending on how accurately the curve is plotted):
(3) A general point about graphs of algebraic equations. You might reasonably wonder why (or if) the red lines plotted on Figure 1 and Figure 2 stop where the X-axis and Y-axis meet. (This is called the zero-point.) The answer is NO. In the complete picture the red line would continue to the left BUT when it passes the zero-point it gets into the field of negative numbers, and that is a complication that we don’t need to discuss today.
You will meet the whole picture in the next section, when we get to the circle, but we’ll just take it for granted without comment.
(4) The circle in Cartesian coordinates
See Figure 3. It’s easy, with a compass, to draw a circle and then it seems natural (symmetrical) to make its centre the zero-point and to put the X-axis and the Y-axis through it, as shown. We’ll call its radius “r”. But how do we find an equation to match the circle???
The secret is that long ago a Greek mathematician named Pythagoras made a discovery in geometry. It was about triangles in which one of the three angles is a right-angle (90˚). The apparent shape can vary quite a lot as sketched below:
BUT
if we keep the long side the same length (which we’ll call “r”, as in the circle above), the combined area of the two squares drawn on its short sides is equal to the area of the square drawn on the long side—no matter whether it’s a “fat” or a “skinny” triangle.
That’s quite a mouthful to understand straight away, but if you look at the examples in Figure 4 you can see for yourself that it’s true.
Now look at Figure 3 again. You could draw any number of triangles like the two shown, with one of the small angles at the zero point and the other on the circle itself. In each one the longest side would be the same, = radius of the circle (= r), and the angle opposite the long side would be a right-angle.
So, treating this drawing as an algebraic graph and applying Pythagoras’ Theorem, you can see that the equation for the circle is , where r is the symbol we have chosen for the radius of the circle.
## (C)pi—at last!
Pi, always written as “π” is a letter in the Greek alphabet corresponding to our P and it is the accepted symbol for the ratio of the area of a circle to the area of a square which exactly contains the circle.
The area of the circle is πr2, by definition. The area of the square, as you can see, is 4r2 and the area of the “inside square is 2r2 (half as big). So obviously π is smaller than 4 but bigger than 2. Our problem is to find an exact figure for it.
Areas in general
If your diagram, a piece of land, the floor space in a building, or any other space, is bounded by straight lines, its area is quite easy to calculate. You can divide it into triangles and rectangles and the solution is straightforward.
But when the boundaries are curved the straightforward methods are no use.
We have to go to a branch of mathematics called calculus. It was invented by the famous Sir Isaac Newton about 300 years ago. His basic idea was to think of the diagram (graph) as if made up of a lot of parallel strips, all of the same width, each consisting of a narrow rectangle plus a small triangle caused by the slope of the curved graph. Depending on the algebraic equation, he calculated the area of the rectangular strips. Then he visualized the strips being made narrower and narrower (but more and more of them), so that the little triangles all became smaller and smaller until their total area could be dismissed as ZERO. I’m sorry I can’t be more precise, but that was his basic idea… and it worked!
As regards a circle, we decided that its formula is .
The radius can be as long or as short as we like, so we are going to say that it is one—one foot, one inch, one yard, one meter, any unit you like. In algebraic terminology, r=1 .*
So the equation for the circle becomes
which can be rearranged as
Now take the square root of each side and it becomes
By the square-root calculation mentioned earlier, we can find the square root of . It is a long series of fractions which could be calculated one after another without ever coming to an end, but they get progressively smaller until there’s no practical value in continuing. You have to accept my word for it, the answer is:
Now, back to areas.
Isaac Newton used special symbols for his calculus. One was a funny tall ‘S’ which we call the integral of a mathematical formula. Loosely speaking, if you have an equation stating that y = some formula with an x in it, then the integral of y, written , tells you the area under the curve on your graph.
In the present instance , so we want the integral of . To find this, we use the series of fractions above and apply a basic formula which says (you just have to believe me!)
The result (remembering that x = 1) is a new series of fractions, namely:
To make this series usable, we convert all the common fractions to decimals and add them up:
.16666… .02500… .00893… .00434… .00248… + .00158… .20899… and then we subtract them from the ‘1’ at the beginning: 1.00000… - .20899… = 0.79101…
and this number (0.79101…) is the numerical value of . This is the area of only a quarter-circle (the one on Figure 3 with the purple triangles in it). So the answer we are looking for is
This is supposed to be π but you know from your books that the true value of π is 3.14159…, so our answer is too high. The questions that arise are:
(a) What went wrong? (b) Can we fix it?
(a) The fractions in the series we used got progressively smaller, but even so we stopped calculating after only a few terms. If we had continued, the remaining fractions, even though smaller and smaller, would have added a little bit to the addition-sum above. If they only added 0.006 (a quite tiny amount), our first number would become 0.78501…, and 4 times that = 3.14064…, which is very close to the true value of π.
(b) To cure our difficulty, we need to find a series of fractions which get smaller, one after the other, much more quickly than in the first series we used. The secret is to find a comparable series, but one which uses “1/2” instead of “1” for the top line of the fractions
BECAUSE
whereas 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 always = 1
—no matter how many fractions you include in your series.
quickly becomes a very small amount and after only a few fractions of the series are calculated, the rest become so very tiny that you don’t need to bother with them. Once a mathematician gets the idea of using a formula that gives rapidly-shrinking series of fractions, there are several different ways to do it. My own favourite uses the diagram shown on Figure 5. Please look at it…
Again the circle is the graph of , or but now we consider only the coloured area, not the whole quarter-circle.
In calculus the formula for this area is an integral expressed as
On the diagram it divides into two natural parts. The red part is a segment of a circle, like a wedge of pie. You can see that twelve such segments would make up the whole circle and you know that the area of the circle is π, so the area of the red part has to be .
The green part is half an equilateral triangle (which is one with all of its sides and angles equal). From Pythagoras’ Theorem, we know that its height is half the square-root of three , so the green area
SO… from the geometrical diagram of Figure 5 we arrive at the algebraic equation:
= +
(the whole coloured area) (the red area) (the green area)
By rearranging, we get:
so π = 12 x the difference between those two numbers
SO… to find π we need to calculate two irrational numbers:
## Calculation of the numbers
The first one, , is another series of fractions, related to the one above but modified to shrink away quickly, namely:
which simplifies to:
As before, we convert the common fractions to decimals and add them up, noting how quickly they shrink to almost zero.
.0208333… .0007812… .0000697… .0000084… .0000012… + .0000002… .0216940… and now subtract the total from the “1/2” at the beginning: 0.5000000 - 0.0216940 = 0.4783060
For the second one, we have to calculate the square root of three to as many places of decimals as we like, let’s say seven. When you know how, this is quite straightforward, and the answer is:
So to get the number we want,
Now we’re ready. If we subtract the second decimal from the first one, the answer should be one-twelfth of π. Let’s see!
0.4783060…
- 0.2165063…
0.2617997…
E U R E K A!!!!! That agrees with the accepted value of π to the fifth place of decimals, which is more accurate than is ever normally necessary.
So here we are! We set out looking for an accurate value for π and we found it. In ancient Egypt, when the measurement of land (= geo + metry) was a serious matter for Euclid and his colleagues, they used the approximation π = 3. We can see now that their usage was reasonably close but not precise.
* The line of dots means that the decimal goes on and on, but we choose to stop here.
The numbers used for finding each point are called its coordinates. In geography the location of anywhere in the world can be identified by its coordinates of longitude and latitude. For instance, Ottawa is roughly 45° North, 76° West. You may find similar use of the word in a more local sense, on maps and atlases.
* Here is a new symbol. That little 2 indicates that you multiply the number by itself . If you multiplied three, or four, or five, or as many as you like of a number together, you would write it as x3, x4, x5, and so on. (You will see some examples before we get to π.)
When you talk about a number multiplied by itself several times, all except for twice and for three times, you use the phrase “x to the 4th power,” “x to the 5th power,” and so on. Very often the word “power” is omitted (understood) and “x to the fourth” is enough.
In the two exceptions, x2 ( = x multiplied by itself) gives the square of x, as we have already discussed. In speaking, x2 is called “x squared.” Then if you do it once again, , we call the result the cube of the number. = 4-cubed = the cube of 4. (The reason for using a geometric word ties in with the use of “square” for second-power numbers.)
Also the ratio of the circumference to the diameter = π
* The reason is that in the formulas which you will shortly see, to multiply or divide by one makes no differences.
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HW 2 solns
# HW 2 solns - Networks Spring 2008 David Easley and Jon...
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Networks: Spring 2008 Homework 2 Solutions David Easley and Jon Kleinberg February 13, 2008 Grading: Part A is out of 10, Part B is out of 14. 1.a [2 points] D is a dominant strategy for player A and R is a dominant strategy for player B . Thus the only Nash equilibrium is ( D,R ). 1.b [2 points] U is a dominant strategy for player A and L is player B ’s unique best response to U . Thus the only Nash equilibrium is ( U,L ). 1.c [2 points] This game has two pure strategy Nash equilibria: ( D,L ) and ( U,R ). It also has a Nash equilibrium in which both players use mixed strategies. Let Pr A ( U ) = p and Pr B ( L ) = q . To have a Nash equilibrium p,q , with both p and q greater than 0 and less than 1, A ’s expected payoffs from U and D must be equal; and B ’s expected payoffs from L and R must be equal. So we have q + 4(1 - q ) = 3 q + 2(1 - q ) and p + 3(1 - p ) = 2 p + 2(1 - p ) . Solving we see that p = q = 1 / 2. 2.a [ 2 points] This game has two pure strategy Nash equilibria ( U,L ) and ( D,R ). Strategy L is a weakly dominated strategy for player B . So the equilibrium ( U,L ) uses a weakly dominated strategy. 2.b [2 points] Here we are just looking for a discussion of why a player might or might not use a weakly dominated strategy. There are at least two equally good answers and an argument along either line received full credit for this part: One way to reason about the game is that B knows that if he has an opportunity to make a move that matters, then A must have chosen D . As R is the unique best response to D , player B should chose R . A second possibility is to note that, no matter what A does, B has nothing to lose from choosing R rather than L . So he should chose R .
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# Find the first term and the common difference of the arithmetic sequence if a2-a6+a4=-7 and a8-a7=2a4
The nth term of an arithmetic sequence can be written as a + (n - 1)*d, where a is the first term and d is the common difference.
We have a2-a6+a4=-7
=> a + d - a - 5d + a + 3d = -7
=> a - d = -7
=> d = a + 7
a8 - a7 = 2*a4
=> a + 7d - a - 6d = 2*(a + 3d)
=> d = 2a + 6d
=> 2a + 5d = 0
Now substitute d = a + 7 inĀ 2a + 5d = 0
=> 2a + 5(a + 7) = 0
=> 2a + 5a + 35 = 0
=> 7a + 35 = 0
=> a = -35/7
=> a = -5
d = a + 7 = -5 + 7 = 2
The first term is -5 and the common difference is 2
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Sample bartleby Q&A Solution
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Question
Draw shear force and bending moment diagrams for the given simply supported beam.
Draw a free body diagram converting the UDL into a point load.
As UDL is givens to be 5kN and it is acting on a length of 2m .
Point load will act on the center of the UDL at 1 m towards the left from point "B" and shall be calulated as : 5 x 2 = 10 kN.
Now, Lets caluclate the resulting reactions on this beam that is RA and RB.
Condition for equilibrium is Anticlockwise moments = Clockwide moments
Taking moments about "A"(moment = Force x Perpendicular Distance
RB x 5 = ( 10 x 1) + ( 10 x 4)
RB x 5 = 50
RB = 50/5 = 10 kN.
Summation of upward force = Summation of downward force
RA+ RB = 10 + 10
RA+10 = 20
RA = 20 -10
RA = 10 kN
Shear force calculations:
Shear force is the algebraic sum of unbalanced vertical forces to left or right of the section.
So, if we start calculating from left side then upward force will be positive and downward force will be negative and if we start calculating from the right side of the section downward force will be positive and downwards force will be negative.
In this case we will start calculating from left side.
Shear force @ A = + 10
Shear force @ C = +10-10 = 0
Shear force @ D = 0
Shear force @ B = 0 -10 = -10
Note: Shear force on UDL shall be represented as a inclined line.
Bending moment calculations:
Bending moment @ A = Ma=0
Bending moment @ B = Mb= 0
Bending moment @ C = Mc= ( 10 x 1)= 10 kN.m
Bending moment @ D = Md = (10 x 3) – (10 x 2) = 30-20 = 10 kN.m
Note: Bending moment on UDL shall be represented as a parabolic curve.
Hence, "Shear force" and "Bending moment" diagram's are drawn for the given simply supported beam.
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# Does this weird sequence have a limit?
Yesterday I was trying to come up with an example of the weirdest sequence I could think of, and I came up with this. I'm not even sure if this could be called a sequence, but here it goes:
We'll define the sequence $a_n$ in the following way. We'll keep a list of previously calculated terms. Whenever we want to know $a_k$ for some $k$, we'll check if $a_k$ is in the list. If it is, we're done: that's its value. If it isn't, we roll a die, assign the number we get to $a_k$, and write that down in the list. This way, we can find out $a_k$ for any $k$ we want. For example, I just calculated some terms of it: $a_1 =1, a_2=4, a_{27}=1, a_{googol} = 5$.
There is an obvious problem with this definition, namely, that the sequence will be different each time someone different calculates it. I think a way to get past this would be to say that there will be a file on the Internet or something that keeps a list of all previously calculated terms of $a_n$. Another way would be, maybe, to talk not about the sequence itself (because it isn't unique, in a way) but of the probability of it having certain properties each time someone different calculates it.
My main question is: does this sequence have a limit? Or, to put it in the second way I just mentioned: what is the probability it will have a limit? What would happen if, instead of rolling a die, we pick a random real number for each term?
I'm sorry if this question doesn't make sense; I know practically nothing about the kind of math that would be needed to answer it. I don't even know how one would approach a problem like this, and I'm not sure this even counts as a sequence. This is something I randomly thought of, and I'm wondering if there is any previous work on problems like this.
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"Random" sequences are interesting and useful. It is not clear how $a_k$ is defined if "$a_k$ is in the list." – André Nicolas Jun 30 '12 at 14:38
The way you've defined the sequence is needlessly complicated. Instead of forming it in chronological order (i.e. by increasing index), you seem to suggest a scenario where we iteratively pick $k$ randomly out of the naturals, and if we already know $a_k$ for the given $k$ we move on to the next choice of $k$, otherwise we roll a die and record the outcome as $a_k$. In addition to being superfluous there are two complications: what does it mean to randomly pick a natural number, and what of the logical possibility that there are terms in the sequence that never get defined! – anon Jun 30 '12 at 14:43
(There's no need for the sequence to be "constructed" outside of chronological order, because the die rolls are independent anyway, and there's no point in allowing ourselves to pick the same index more than once if we're only going to do something with it the first time.) – anon Jun 30 '12 at 14:45
@AndréNicolas: What I mean if that $a_k$ has already been calculated, there is no need to roll the die again. We just look at the list and check what was the value of $a_k$. – Javier Jun 30 '12 at 15:01
@anon: Making a needlessly complicated definition was sort of the point. Also, you don't necessarily have to choose $k$ randomly. You can start from $1$ and work your way up if you want; the point is not so much in what order the terms are calculated, but that you can calculate $a_k$ for any $k$ you want. – Javier Jun 30 '12 at 15:04
In particular, since your sequence $(a_k)$ only takes on values from a discrete set, in order to converge it would have to be constant from some point $k_0 \in \mathbb N$ onwards. But that can only happen if $a_k = a_{k_0}$ for all $k > k_0$, which is the intersection of infinitely many independent events, each occurring with probability less than $1-\epsilon$ for some fixed $\epsilon > 0$, and thus their intersection only occurs with probability 0.
That's interesting. Does this change if instead of picking a number from ${1,2,3,4,5,6}$ we choose a random real number, or maybe one from the interval $[0,1]$? – Javier Jun 30 '12 at 15:05
@Javier: Just to answer an old question, no it doesn't. If the real-valued sequence converged to some $x \in \mathbb R$, then for any $\epsilon>0$, there would have to be some $k_0$ such that, for all $k>k_0$, $a_k\in[x-\epsilon,x+\epsilon]$. As long as we can choose an $\epsilon>0$ such that each $a_k$ has a non-zero probability of not belonging to $[x-\epsilon,x+\epsilon]$ (i.e. as long as the probability distribution is not concentrated at $x$), the conclusion still follows. – Ilmari Karonen Aug 29 '14 at 1:22
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# Centrifugal force/ pseudo force in circular motion [closed]
As an example, I have considered earth's circular motion around the sun and an observer revolving around the sun too, both at different distances $$r_2$$ and $$r_1$$ respectively. The angular velocity for both is the same, say $$\omega$$.
So with respect to the observer, there are two forces on the earth, first the centripetal force $$Mr_2\omega^2$$ and the other is the pseudo force due to the acceleration of observer which is $$Mr_1\omega^2$$. ($$M$$ is mass of earth)
Now, the observer would see the earth at rest since their angular velocities are same. This implies that magnitude of centripetal force is equal to the magnitude of pseudo force or $$Mr_1\omega^2=Mr_2\omega^2$$. But this can't be true as then $$r_1$$ will be equal to $$r_2$$ which is absurd.
the other is the pseudo force due to the acceleration of observer which is M(r1)w². (M is mass of earth)
Here is the mistake. The pseudo force is $$M r_2 \omega^2$$. The magnitude of the pseudo force on an object depends on that object’s position in the non-inertial frame. Not on the position of the observer.
... the observer would see the earth at rest since their angular velocities are same ...
Their angular velocities about the Sun are the same. However, if the observer is revolving around the Sun but not rotating, then the observer sees both the Earth and the Sun revolve around them with angular velocity $$\omega$$. Since the earth is a distance $$r_2-r_1$$ from them they infer that there must be a net centripetal force $$M(r_2-r_1)\omega^2$$ acting on the Earth. They know that the gravitational force acting on the Earth is $$Mr_2\omega^2$$ towards the Sun. So they infer the existence of a pseudo force $$Mr_1\omega^2$$ acting away from themselves.
If, on the other hand, the observer is revolving around the Sun and rotating so that the Earth and the Sun appear to stand still, then the pseudo force depends on distance from the observer (as another answer points out). If an object of mass $$M$$ is at a distance $$r$$ from the observer then the pseudo force is $$Mr_1\omega^2$$ away from the Sun due to revolving around the Sun plus $$Mr\omega^2$$ away from the observer due to the observer's rotation. For the Earth at distance of $$r=r_2-r_1$$ from the observer we have
$$Mr_1\omega^2 + M(r_2-r_1)\omega^2 = Mr_2\omega^2$$
as expected.
So with respect to the observer, there are two forces on the earth, first the centripetal force $$Mr_2\omega^2$$ and the other is the pseudo force due to the acceleration of observer which is $$Mr_1\omega^2$$
In cases like this, where and object appears to be stationary in an accelerating reference frame, the pseudo centrifugal force is equal in magnitude and opposite in direction to the centripetal force and we use the radial distance of the object ($$r_2$$) from the centre of rotation to determine the centrifugal force and not the radial location of the observer.
As, Biophysicist points out, the centrifugal force observed in a rotating reference frame is not always a reaction to an opposite centripetal force. For example, if an observer is standing on the perimeter of a rotating turntable, and releases a ball, the ball will appear to accelerate outward in a curved path and the observer will attribute this to the pseudo centrifugal force and in this case the centrifugal force appears to be present even without the corresponding centripetal force. In this latter case the a n accelerometer mounted on the ball would measure no proper acceleration and the ball is actually moving inertially.
This pseudo force is usually inferred and is not directly measurable. When cornering hard in a car we can feel radial forces acting on us. The force we feel is the inward acting centripetal force and an accelerometer measures only this inward force. For a long time it was assumed that centrifugal forces provided the force required to counter the centripetal force of gravity acting on an orbiting body. When Einstein formulated General Relativity, gravity was no longer considered a force and orbiting bodies just follow geodesics. Since there is no centripetal force provided by gravity, there is no centrifugal force required to balance it in the case of orbiting bodies. An accelerometer attached to a small satellite would not measure any proper radial forces. (This is why astronauts appear to be weightless in International Space Station.) So for your case of the Earth orbiting the Sun, there is actually no centrifugal force acting on the Earth as a result of it orbiting the Sun. (but there is centrifugal force experienced by a non inertial observer standing on the the surface of the Earth as a result of Earth's own spin and this slightly reduces the apparent force of gravity and gives the Earth its oblate shape).
• "The pseudo centrifugal force is defined to be be equal in magnitude and opposite in direction to the centripetal force" The centrifugal force can be present in a rotating reference frame even if there are no centripetal forces present, and even if there are centripetal forces, they do not need to cancel the centrifugal force. Commented Aug 4 at 11:59
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# What is 77 over 140 as a percentage?
Updated: 8/20/2019
Wiki User
11y ago
% rate:
= 77/140 * 100%
= 0.55 * 100%
= 55%
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11y ago
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Q: What is 77 over 140 as a percentage?
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### What is 77 is 140 percent?
7 is what percent of 140= 77 / 140= 0.55Converting decimal to a percentage: 0.55 * 100 = 55%
### What is the simplified fraction for 77 over 140?
It is: 77/140 = 11/20
### 77 over 140 in its simplest form?
77/140 = 11/20 simplified
77%
It is: 77%
100*77/140 = 55%
### What is 126 over 140 as a percentage?
Expressed as a percentage, 126/140 x 100 = 90 percent.
### What is 140 over 2000 as a percentage?
To convert a fraction (or any number) to a percentage multiply it by 100% → 140/2000 = 140/2000 × 100% = 140/20% = 7%
### What is 47 over 77 as a percentage?
Expressed as a percentage, 47/77 x 100 = 61.038961 recurring (that is, 61.038961038961...) percent.
### What is 68 over 77 as a percentage?
Expressed as a percentage, 68/77 x 100 = 88.311688 recurring (that is, 88.311688311688...) percent.
### How do you change 7 over 5 into a percentage?
7/5 into percentage = 140%
### What is the LCM of 140 and 77?
LCM x GCD = 140 x 77 140 = 2 x 2 x 5 x 7 77 = 7 x 11 GCD = 7 LCM = 140 x 77 / 7 = 140 x 11 = 1540
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6.5.1 Proving Trigonometric Identities Using Addition Formula and Double Angle Formulae (Examples)
Example 2:
Prove each of the following trigonometric identities.
$\begin{array}{l}\text{(a)}\frac{1+\mathrm{cos}2x}{\mathrm{sin}2x}=\mathrm{cot}x\\ \text{(b)}\mathrm{cot}A\mathrm{sec}2A=\mathrm{cot}A+\mathrm{tan}2A\\ \text{(c)}\frac{sinx}{1-cosx}=\mathrm{cot}\frac{x}{2}\end{array}$
Solution:
(a)
$\begin{array}{l}LHS\\ =\frac{1+\mathrm{cos}2x}{\mathrm{sin}2x}\\ =\frac{1+\left(2{\mathrm{cos}}^{2}x-1\right)}{2\mathrm{sin}x\mathrm{cos}x}\\ =\frac{\overline{)2}{\mathrm{cos}}^{\overline{)2}}x}{\overline{)2}\mathrm{sin}x\overline{)\mathrm{cos}x}}\\ =\frac{\mathrm{cos}x}{\mathrm{sin}x}\\ =\mathrm{cot}x=RHS\text{(proven)}\end{array}$
(b)
$\begin{array}{l}RHS\\ =\mathrm{cot}A+\mathrm{tan}2A\\ =\frac{\mathrm{cos}A}{\mathrm{sin}A}+\frac{\mathrm{sin}2A}{\mathrm{cos}2A}\\ =\frac{\mathrm{cos}A\mathrm{cos}2A+\mathrm{sin}A\mathrm{sin}2A}{\mathrm{sin}A\mathrm{cos}2A}\\ =\frac{\mathrm{cos}A\left({\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A\right)+\mathrm{sin}A\left(2\mathrm{sin}A\mathrm{cos}A\right)}{\mathrm{sin}A\mathrm{cos}2A}\\ =\frac{{\mathrm{cos}}^{3}A-\mathrm{cos}A{\mathrm{sin}}^{2}A+2{\mathrm{sin}}^{2}A\mathrm{cos}A}{\mathrm{sin}A\mathrm{cos}2A}\\ =\frac{{\mathrm{cos}}^{3}A+\mathrm{cos}A{\mathrm{sin}}^{2}A}{\mathrm{sin}A\mathrm{cos}2A}\\ =\frac{\mathrm{cos}A\left({\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A\right)}{\mathrm{sin}A\mathrm{cos}2A}\\ =\frac{\mathrm{cos}A}{\mathrm{sin}A\mathrm{cos}2A}←\overline{){\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A=1}\\ =\left(\frac{\mathrm{cos}A}{\mathrm{sin}A}\right)\left(\frac{1}{\mathrm{cos}2A}\right)\\ =\mathrm{cot}A\mathrm{sec}2A\end{array}$
(c)
$\begin{array}{l}LHS\\ =\frac{sinx}{1-cosx}\\ =\frac{2sin\frac{x}{2}\mathrm{cos}\frac{x}{2}}{1-\left(1-2si{n}^{2}\frac{x}{2}\right)}←\overline{)\begin{array}{l}\mathrm{sin}x=2sin\frac{x}{2}\mathrm{cos}\frac{x}{2},\\ \mathrm{cos}x=1-2{\mathrm{sin}}^{2}\frac{x}{2}\end{array}}\\ =\frac{\overline{)2}\overline{)sin\frac{x}{2}}\mathrm{cos}\frac{x}{2}}{\overline{)2}si{n}^{\overline{)2}}\frac{x}{2}}=\frac{\mathrm{cos}\frac{x}{2}}{sin\frac{x}{2}}\\ =\mathrm{cot}\frac{x}{2}=RHS\text{(proven)}\end{array}$
Example 3:
(a) Given that such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).
(b) Given that such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A
B).
Solution:
(a)
$\begin{array}{l}\mathrm{sin}P=\frac{3}{5},\text{}\mathrm{cos}P=\frac{4}{5}\\ \mathrm{sin}Q=\frac{5}{13},\text{}\mathrm{cos}Q=-\frac{12}{13}\\ \\ \mathrm{cos}\left(P+Q\right)\\ =\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B\\ =\left(\frac{4}{5}\right)\left(-\frac{12}{13}\right)-\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)\\ =-\frac{48}{65}-\frac{15}{65}\\ =-\frac{63}{65}\end{array}$
(b)
$\begin{array}{l}\mathrm{sin}A=-\frac{3}{5},\text{}\mathrm{cos}A=-\frac{4}{5}\\ \mathrm{sin}B=-\frac{5}{13},\text{}\mathrm{cos}B=\frac{12}{13}\\ \\ sin\left(A-B\right)\\ =sinA\mathrm{cos}B-cosA\mathrm{sin}B\\ =\left(-\frac{3}{5}\right)\left(\frac{12}{13}\right)-\left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right)\\ =-\frac{36}{65}-\frac{20}{65}\\ =-\frac{56}{65}\end{array}$
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Class 6 Maths Understanding Elementary Shapes Three Dimensional Shapes
Three Dimensional Shapes
The shapes which can be measured in 3 directions are called three-dimensional shapes. These shapes are also called solid shapes. Length, width, and height (or depth or thickness) are the three measurements of the three-dimensional shapes. Sphere, cone, cylinder, cuboid, cube are examples of three-dimensional shape.
Faces, edges and vertices:
Let us consider cube.
Each side of the cube is a flat surface called a flat face or simply a face.
When two faces meet at a line segment then it is called an edge.
Three edges meet at a point called a vertex.
Problem: What shape is:
(a) Your instrument box? (b) A brick? (c) A match box?
Solution:
(a) An instrument box has shape of cuboid.
(b) A brick has shape of cuboid.
(c) A match box has shape of cuboid.
(d) A road-roller has shape of cylinder.
(e) A sweet laddu has shape of Sphere.
.
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# Slope, Mon., Jan. 9th, 7th Grade
## Presentation on theme: "Slope, Mon., Jan. 9th, 7th Grade"— Presentation transcript:
Slope, Mon., Jan. 9th, 7th Grade
GLE Conceptualize the meanings of slope using various interpretations, representations, and contexts. Assessments: Understand that the graph of a linear function f is the set of points on a line representing the ordered pairs (x, f(x)). SPI Interpret the slope of a line as a unit rate given the graph of a proportional relationship. Essential Question: How do we graph data, find slope and explain what slope means.
Graph the data; Find the slope; Explain what the slope represents. 1
Graph the data; Find the slope; Explain what the slope represents. 1. Envelopes The table shows the number of envelopes stuffed for various times. To graph the data, use the graph next to problem 1; go to the 5 line at the bottom, from there go up to 30 (the second line); place a dot where the 5 and 30 intersect. Slope formula: slope= y2-y1/x2-x1 : Pick any 2 ordered pairs for ex. (5, 30) and (10, 60); Plug in those numbers into the slope formula 60-30/10-5= 30/5=6 The slope shows how steep the line is. It is the ratio of the vertical change to the horizontal change in the straight line. Time (min) X 5 10 15 20 Envelopes Stuffed Y 30 60 90 120
Yards (X) 1 (X1) 2 (X2) Feet (Y) 3 (Y1) 6 (Y2)
2. Measurement There are 3 feet for every yard Make a chart similar to the one in #1. It should have at least two ordered pairs or sets of numbers for X and Y. For example, the first ordered pair would be (1, 3) where 1 is the yard and 3 is the feet. The second ordered pair could be (2, 6) where 2 is the yards and 6 represents the number of feet. Graph as you did in #1 Find the slope using the slope formula as you did in #1. Slope = Y2-Y1/X2-X1; or Explain what slope represents as I did in #1. Yards (X) 1 (X1) 2 (X2) Feet (Y) 3 (Y1) 6 (Y2)
3. Use the graph that shows the number of laps completed over time
3. Use the graph that shows the number of laps completed over time. Find the slope of the line. a) Time in minutes is X with 10 as X1 and 20 or 30 as X2; b) Number of laps is Y with 5 as Y1 and 10 or 15 as Y2. c) Decide which ordered pairs you want to use The easiest to use is (10,5) and (20, 10). d) Plug the numbers into the formula for Slope which is Slope = Y2-Y1/X2-X1. Which line is the steepest? Explain using the slopes of lines l, m, and n. Slope = Y2-Y1/ X2-X1 a. Slope for l= 2- 1/ 2-0= b. Slope for m= 2- -4/2-0= c. Slope for n= 3- -3/ 0- 3=
Page 108, 1-4 Go-Karts The graph shows the average speed of two go-karts in a race. What does the point (2,20) represent on the graph? If the point was (1,10), you could say that 1 is for 1 hour; 10 is for 10 miles, so the go-kart Y was traveling 10 miles per hour at the beginning of the race. Be sure to include which go-kart the point is referring to. What does the point (1,12) represent on the graph? Follow the clues above. What does the slope of each line represent? Look in your notes to find what slope represents. Which car is traveling faster? Which car goes more distance in the same amount of time as the other car?
2. Skateboarding The line represents the length and height of a skateboard ramp. Find the slope of the ramp. Put two points anyplace on the line. Write down the ordered pairs of the points you selected. Decide what numbers are Y2, Y1, X2 and X1. Then plug them into the slope formula, Y2-Y1/ X2-X1
3. Raisins The table shows the number of packages of raisins per box.
Graph the data. Hint: Boxes are X; Packages are Y Find the slope of the line. Write the formula first, then plug in the numbers for any two ordered pairs. Explain what the slope represents. Look in your notes. 4. Resort The Snells can spend 4 nights at a resort for \$500 or 6 nights at the same resort for \$750. Graph the data. Hint: number of nights is the input or X; cost is the output or Y Then find the slope. Write the formula first, then plug in the numbers for these two ordered pairs.
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Systems of Equations
# Systems of Equations - Name WordProblems Date...
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CHAPTER 7 TEST REVIEW ANSWER KEY 1 PART I: Answer all questions in this part. Circle the number representing your response. 2.) How many solutions does the system of equations below have? x 2y = 4 4x 8y = 12 a.) One Solution b.) Two Solutions c.) No Solution d.) Infinitely Many Solutions 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 y x 1.) When drawn on the same set of axes, the graph of the equations y = 3x 4 and y = x 7.5 intersect at the point whose coordinates are a.) (1, 7) c.) (7, 1) b.) (7, 1) d.) (1, 7) y=3x4 y=x7.5 1 2 1 2 -4(1x - 2y = -4) 4x - 8y = -12 -4x + 8y = 16 4x - 8y = -12 0 = 4 No! 3(2x + 1y = 8) 1x - 3y = -3 6x + 3y = 24 1x - 3y = -3 7x = 21 7 7 x = 3 2x + y = 8 2(3) + y = 8 6 + y = 8 - 6 - 6 y = 2 3.) What is the value of y in the following system of equations? 2x + y = 8 x 3y = 3 (-1, -7) Name ____________________________ Date _____________ Test Review Solving Systems Of Equations, Int. Algebra Class ______ Word Problems
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https://www.codeproject.com/Articles/4959/Encryption-RSA-implemented-through-Java-Script-to/?fid=18183&df=90&mpp=10&sort=Position&select=2466021&tid=2931687
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Posted 9 Sep 2003
# Encryption - RSA implemented through Java Script to Encryt/Decrypt data
, 9 Sep 2003
Rate this:
Demonstrate the use of RSA algorithm in HTML pages
## Introduction
RSA is a public key cryptosystem for both encryption and authentication; it was given by three scientists viz. Ron Rivest, Adi Shamir and Leonard Adleman. This algorithm is much secure than any other algorithm. The latest key size used for this encryption technique is 512 bits to 2048 bits.
With the advent of computers it has become possible to perform computations at teraflop speeds so such algorithms could easily be cracked. But RSA encryption uses the concept of two large prime numbers, such that, their product could not be easily factorized. Let us see how this algorithm works and understand its implementation using Java Script.
## Mathematical Background
### Modular Arithmetic
RSA uses modular arithmetic. This is similar to conventional arithmetic, but only uses positive integers that are less than a chosen value, called the modulus. Addition, subtraction and multiplication work like regular maths, but there is no division. You can use any value for the modulus; the diagram uses 13, so counting goes 0, 1, 2, ..., 11, 12, 0, 1, 2 ... The notation used for expressions involving modular arithmetic is:
x = y (mod m)
Which reads as "x is equivalent to y, modulo m". What this means is that x and y leave the same remainder when divided by m. For example, 7 = 23 (mod 8) and 22 = 13 (mod 9). The following statement is a basic principle of modular arithmetic:
a + kp = a (mod p)
You can visualize this on the diagram - each time you add p you go round the circle, back to where you started. It doesn't matter where you start, how big the circle is, or how many times you do it, it's always true.
### Primality and Coprimality
• A number is prime if the only numbers that exactly divide it are 1 and itself. e.g. 17 is prime, but 15 isn't, because it's divisible by 3 and 5.
• A pair of numbers are coprime if the largest number that exactly divides both of them is 1. The numbers themselves don't have to be prime. e.g. 8 and 9 are coprime, but 8 and 10 are not, because they're both divisible by 2.
• If you have a pair of distinct prime numbers, they will always be coprime to each other.
### Chinese Remainder Theorem
This theorem provides a way to combine two modular equations that use different moduli.
Theorem x = y (mod pq) Proof x = y + kp x - y = kp p divides (x - y) by a similar route, q divides (x - y) as p and q are coprime, pq divides (x - y) x - y = l(pq) x = y (mod pq)
### Fermat/Euler Theorem
This theorem is a surprising identity that relates the exponent to the modulus.
Theorem p-1 = 1 (mod p) if p is prime and x 0 (mod p) Proof as p is prime, these numbers are coprime to p 0 is not coprime to p Q includes all the numbers in (mod p) coprime to p now consider the set U, obtained by multiplying each element of Q by x (mod p) each element of U is coprime to p i = xQj (mod p) with i j Qi = Qj (mod p) as x 0 elements of U are distinct U is a permutation of Q U1.U2 ... Up-1 = Q1.Q2 ... Qp-1 (mod p) xQ1.xQ2 ... xQp-1 = Q1.Q2 ... Qp-1 (mod p) 1.Q2 ... Qp-1 xp-1 = 1 (mod p)
## Using the code
This implementation of RSA uses 32 bit key. There are two files: input.htm and output.htm. The code for the input.htm file is as follows:
<html>
<title>Input</title>
<script language="JavaScript">
<!-- hide from old browsers
function gcd (a, b)
{
var r;
while (b>0)
{
r=a%b;
a=b;
b=r;
}
return a;
}
function rel_prime(phi)
{
var rel=5;
while (gcd(phi,rel)!=1)
rel++;
return rel;
}
function power(a, b)
{
var temp=1, i;
for(i=1;i<=b;i++)
temp*=a;
return temp;
}
function encrypt(N, e, M)
{
var r,i=0,prod=1,rem_mod=0;
while (e>0)
{
r=e % 2;
if (i++==0)
rem_mod=M % N;
else
rem_mod=power(rem_mod,2) % N;
if (r==1)
{
prod*=rem_mod;
prod=prod % N;
}
e=parseInt(e/2);
}
return prod;
}
function calculate_d(phi,e)
{
var x,y,x1,x2,y1,y2,temp,r,orig_phi;
orig_phi=phi;
x2=1;x1=0;y2=0;y1=1;
while (e>0)
{
temp=parseInt(phi/e);
r=phi-temp*e;
x=x2-temp*x1;
y=y2-temp*y1;
phi=e;e=r;
x2=x1;x1=x;
y2=y1;y1=y;
if (phi==1)
{
y2+=orig_phi;
break;
}
}
return y2;
}
function decrypt(c, d, N)
{
var r,i=0,prod=1,rem_mod=0;
while (d>0)
{
r=d % 2;
if (i++==0)
rem_mod=c % N;
else
rem_mod=power(rem_mod,2) % N;
if (r==1)
{
prod*=rem_mod;
prod=prod % N;
}
d=parseInt(d/2);
}
return prod;
}
function openNew()
{
var subWindow=window.open(
"Output.htm", "Obj","HEIGHT=400,WIDTH=600,SCROLLBARS=YES");
var p=parseInt(document.Input.p.value);
var q=parseInt(document.Input.q.value);
var M=parseInt(document.Input.M.value);
var N=p * q;
var phi=(p-1)*(q-1);
var e=rel_prime(phi);
var c=encrypt(N,e,M);
var d=calculate_d(phi,e);
subWindow.document.Output.N.value=N;
subWindow.document.Output.phi.value=phi;
subWindow.document.Output.e.value=e;
subWindow.document.Output.c.value=c;
subWindow.document.Output.d.value=d;
subWindow.document.Output.M.value=decrypt(c,d,N);
}
// end scripting here -->
</script>
<body>
<p><font size="6">Input Form</font></p>
<hr>
<form name="Input">
<table border="0" width="100%" height="109">
<tr>
<td width="24%" height="23">
<font color="#0000FF">Enter P</font></td>
<td width="76%" height="23">
<input type="text" name="p" size="20"></td>
</tr>
<tr>
<td width="24%" height="23"><font color="#0000FF">
Enter Q</font></td>
<td width="76%" height="23">
<input type="text" name="q" size="20"></td>
</tr>
<tr>
<td width="24%" height="20">
<font color="#0000FF">Enter any Number ( M )</font></td>
<td width="76%" height="20"><input type="text" name="M" size="20">
<font size="1" color="#FF0000">(1-1000)</font></td>
</tr>
<tr>
<td width="24%" height="19"><input type="button"
value="Submit" name="Submit" onClick="openNew()"></td>
<td width="76%" height="19"><input type="reset"
value="Reset" name="Reset"></td>
</tr>
</table>
</form>
<p> </p>
</body>
</html>
The code for the output.htm file is as follows:
<html>
<title>Output</title>
<body>
<p><font size="6">Output Form</font></p>
<hr>
<p><font color="#FF0000">1. N = p * q
</font></p>
<p><font color="#FF0000">2.
phi = ( p - 1 ) * ( q - 1
) </font></p>
<p><font color="#FF0000">3. GCD
( phi , e ) = 1</font></p>
<p><font color="#FF0000">4.
Encrypted Text ( c ) = M<sup>e</sup>
* ( mod N )</font></p>
<p><font color="#FF0000">5.
e * d = 1 * ( mod phi )</font></p>
<p><font color="#FF0000">6.
Decrypted Text = c<sup>d</sup> * (
mod N )</font></p>
<form name="Output">
<table border="0" width="100%">
<tr>
<td width="22%"><font color="#0000FF">N
</font></td>
<td width="78%"><input type="text" name="N"
size="20"></td>
</tr>
<tr>
<td width="22%"><font color="#0000FF">Phi</font></td>
<td width="78%"><input type="text"
name="phi" size="20"></td>
</tr>
<tr>
<td width="22%"><font color="#0000FF">e
</font></td>
<td width="78%">
<input type="text" name="e" size="20"></td>
</tr>
<tr>
<td width="22%"><font color="#0000FF">Encrypted Text
</font></td>
<td width="78%">
<input type="text" name="c" size="20"></td>
</tr>
<tr>
<td width="22%"><font color="#0000FF">d
</font></td>
<td width="78%"><input type="text" name="d" size="20">
</td>
</tr>
<tr>
<td width="22%"><font color="#0000FF">
Decrypted Text</font></td>
<td width="78%"><input type="text" name="M" size="20"></td>
</tr>
<tr>
<td width="22%">
<input type="button" value="Close" name="Close"
onClick="window.close()"></td>
<td width="78%"> </td>
</tr>
</table>
</form>
<p> </p>
</body>
</html>
Now, after creating the files you can run the input.htm file from your browser and provide the necessary values. Clicking the Submit button will open the output.htm file with the necessary outputs.
## Compatibility of Code
The code has been tested on IE 4.0 and above as well as on Netscape Navigator.
## Working of RSA Algorithm
Suppose that B wants to send a message to A. A and B have exchanged their public keys. Let us try to understand how this works:
1. Person A selects two prime numbers. Say p = 53 and q = 61.
2. Person A calculates p * q = 3233. This is the public key which he sends to B.
3. Person A calculates the value of e such that GCD (( p – 1 ) * ( q – 1 ), e) = 1. This is also send to B.
4. Suppose person B wants to send message M = 999 to A.
5. Person B encrypts the message, c = Me (mod N) = 9997 (mod 3233) = 3026.
6. Person B sends c to person A.
7. Person A decodes c = 3026. Firstly, he finds d such that e * d = 1 (mod ( ( p – 1 ) * ( q - 1) ).
8. This equation is solved using Extended Euclidean Algorithm. Hence d = 1783.
9. Secondly, person A decodes the encrypted message c using: cd (mod N) = 30261783 (mod 3233) = 999.
## Points of Interest
1. The factors of the public key N, that is, p and q should be large enough so that its not easy to factorize N.
2. In general, the order of the primes should be 160 (512 bits) digits to 640 (2048 bits) digits.
3. No algorithm is available that could factorize a number of the mentioned order in reasonable amount of time.
4. So the RSA algorithm is defended by the non-availability of such algorithms.
## Conclusion
1. One has to use brute-force to factorize N.
2. The algorithms to factorize N have a running time exponential with respect to the length of N.
3. Still the existence of a faster algorithm, to factorize N, is very remote.
## Further Suggestions
1. Possible attacks on RSA
2. Algorithm to find whether a number is prime or not (less time complexity algorithm).
3. Largest prime numbers in use.
I would suggest that the readers should try to work on these topics so as to learn more about the RSA encryption scheme. I am having the necessary content, but I don't want to complicate the things write now.
## Implementations of RSA
• S.S.L. (Secure Sockets Layer)
• Firewalls
• ATM machines
• Digital Signatures
• Certificates
## Contact Info
Send your recommendations, suggestions and problems at [email protected]
## Share
Over 10 years of experience in designing & development of Java/J2EE standalone and web based systems for financial & commercial institutions. Over 4 years of experience in enhancing and maintenance of multi-threaded systems. Solid background in Object-Oriented analysis and design, web site development and maintenance. Proficient in developing thorough design documentation of system interfaces, data model and application components. Profound knowledge of design patterns (GOF & J2EE patterns), coding standards and best practices. Deep knowledge in creation of standalone batch based data extraction & loader systems.
Had been to onsite (NY, USA) for 2 years (2009-2011) and worked with technical, implementation team, customer support team and business team for creation, design, implementation, deployment and post deployment support of new web based and batch based systems.
View All Threads First Prev Next
rsa encryption algo Rahul Shahane19-Feb-09 6:11 Rahul Shahane 19-Feb-09 6:11
Re: rsa encryption algo Gaurav Saini4-Apr-09 22:44 Gaurav Saini 4-Apr-09 22:44
Last Visit: 31-Dec-99 19:00 Last Update: 17-Jan-18 14:56 Refresh 1
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# Stopping Distance (Friction and Kinematics)
1. Oct 16, 2009
### Chandasouk
1. The problem statement, all variables and given/known data
If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 27.9 m/s?
Take the free fall acceleration to be g = 9.80 m/s^2.
On wet pavement the coefficient of kinetic friction may be only 0.250. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)
For the first part, I did
Weight = mg -> m(-9.80) = w
And Fn = 9.80m
I found the frictional force = $$\mu$$k*Fn
Frictional Force = (.800)*9.80m = 7.84m
From here what should I do? Fnet = ma so therefore
ma=7.84m
a = 7.84m/s^2
Then using Vfinal^2=Vinitial^2+2a$$\Delta$$x
$$\Delta$$x= 49.6 m ?
2. Oct 16, 2009
### Dick
Sure. That's fine. If you are writing it up to turn in I'd suggest putting units on numbers that have units. Like '9.8' should be '9.8*m/s^2'. To avoid confusion with m=mass and m=meters you might want to use a different symbol for mass.
3. Oct 16, 2009
### Chandasouk
For part B, is the answer 15.6m/s?
Frictional Force = (.250)(9.80m) = 2.45m
ma=2.45m
a=2.45m/s^2
Vf = 0
and plugging in the values
Vfinal^2=Vinitial^2+2a/\ x
Vintial = 15.6m/s ?
The thing is, if I solved it using that equation, I ended up having to square root a negative number which is not possible to do. I had to square root -243.04 but instead just squared 243.04 to get my answer. Is it even correct? If so, where did I go wrong with ending up with a negative sign?
4. Oct 16, 2009
### Dick
Your answer is right again. Vfinal should be 0. And in both cases 'a' should correctly speaking, be negative. You are decelerating. You should have run into the same problem in the first case, if you were really paying close attention to signs. I wasn't.
Last edited: Oct 16, 2009
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THREE WAYS TO SHARE
Beginner's Guide: Make G-Code and CNC Coordinates Easy
CNCCookbook's G-Code Tutorial
Quickly learn how CNC coordinate systems work for CNC mills and lathes from 2 to 5-axis. It's so much more fun to know what's going on, and mastering coordinates is one of the first steps. Plus, it's just not that hard--we'll show you how right now.
Does CNC use a Right or Left Handed Coordinate System?
Hold up your right hand with the index finger extended and the thumb up, like you're simulating a gun. Now extend your second finger at right angles to the index finger. Those three fingers are now pointing in the directions of positive X (second finger), Y (index finger) and Z (thumb). In other words, the Z coordinate will get smaller as your spindle moves down towards the table of a mill. Now here is the tricky part:
Even if the table moves instead of the spindle, the handedness is based on the assumption it is the spindle moving!
What that means is that for motions of the table, we reverse directions. Hence even though the diagram shows what looks like a left-handed system, if we consider how to get the same motion if the spindle moved instead of the table, we'd see it is really right handed.That's why we say CNC uses a Right Handed Coordinate System, and the whole hand thing is just an easy way of remembering. Even though most machines are right handed, it is not a requirement, and you need to check out how your machine is actually set up.
This whole handedness thing is just a way to remind yourself which way the coordinates go. It's not a big deal otherwise and you'll get used to what your machine uses pretty fast.
Each machine will have its own specific axis orientation which you should become familiar with. Here are some common types:
Mill Axes for a Typical Vertical Machining Center. Note: arrows show table motion in positive g-code direction. Handedness is spindle motion and reversed!
Lathe Axes for a Typical 2-Axis Lathe...
The cylinders in each drawing represent the spindle of the machine. Be sure to have a look at exactly how the axes are laid out on your machine. For example, horizontal mills are turned around considerably from the drawing I've shown. Lathes can get a lot more complicated than the simple 2-axis version I've shown.
4-Axis, 5-Axis, and More
Much more complex configurations are possible when you have more axes. For example, here is a 5-axis setup:
5-Axis Mill With Trunion Table...
Note that we have added two rotational axes to the basic mill diagram to provide an A-Axis and a B-Axis. In general, A, B, and C are rotational axes that rotate around axes formed by the X, Y, and Z respectively.
Expressing Coordinates in G-Code
Now that we know what the coordinate systems are, how do we express coordinates in G-Code?
It's pretty simple: just take the axis letter and add the value. Spaces between the letter and its value are optional.
For example, a position that is 1 inch from 0 along X, 2 inches along Y, and 3 inches along Z is written as:
X1Y2Z3
You get used to reading them all run together like that quickly, but you can format them with spaces to make them more readable:
X1 Y2 Z3
or
X 1 Y 2 Z 3
Again, you get used to keeping the letters with the numbers, so I wouldn't add more spaces than just between the axes:
X1 Y2 Z3
That's actually the easiest to read once you get used to it.
The example I just gave used inches, but in actuality the controller can be set to use either metric or Imperial. It's up to you to know which default the system comes up in and to change the units as needed. Try not to change units in the middle of a program, do so at the very beginning and then stay with the same units. It's too confusing otherwise. The G-Codes to change units only affect how the machine interprets the numbers. They don't change your program. We'll talk more about changing units in a future article, but for now, just be aware.
For rotational axes (which you'll only be using on a 4 or 5 axis machine), we don't use dimensions for the units, we use angles, typically in degrees. Rotating the 4th axis to the 90 degree position might be done as A90, for example.
Relative Versus Absolute Coordinates
Sometimes, it is very convenient to refer to Relative instead of Absolute coordinates. Let's assume the tool tip on my mill is at X0 Y0 Z0 and I want to move it to X1 Y2 Z3 (I dropped the commas, which are not used in G-Code, because I'm just trying to get you used to the switch from how you learned coordinates in school, e.g. (0, 0, 0), to how it's done in G-Code X0 Y0 Z0). I can make the move absolute or relative and it doesn't matter. "X1 Y2 Z3" does the trick since in either case we started from X0 Y0 Z0.
But, suppose your cutter is positioned at some point and you need to cut a 1" square with the corner aligned to that point. Perhaps you've used your edgefinder to locate the cutter precisely on some feature of the part. This is easily done with relative moves:
X1
Y1
X-1
Y-1
In essence, move 1" right, 1" up in Y, 1" left, and then 1" down in Y. Now we have a 1" square whose lower left corner is the initial point.
There are lots of cases where relative moves are handy so the ability to switch back and forth comes up a lot. We'll show you how to make that switch when we talk about how to move with G-Code, but for now, just be aware that there are both Relative and Absolute Coordinates.
Sometimes, we refer to relative coordinates with special axis letters. For example, IJK may be relative XYZ when defining arc centers. On some controllers, UVW may be used alongside XYZ to refer to relative coordinates without needing to change back and forth between relative and absolute modes. In other words, XYZ is used always as absolute and UVW is always relative.
For now, it is enough to be aware that relative coordinates exist. A little later, we have an entire chapter just on the subject of relative versus absolute coordinates.
Offsets
The last Coordinate System concept I want to cover is that of Offsets. Offsets are another fancy way to think about relative motions. Let's suppose you want to machine 2 identical parts. Each is held in a vise on your table at the same time. How do you make one program that can do both parts without having to change the program for the position of each part?
The answer is that we use a Work Offset. More detail on those later, but for now, imagine that Work Offsets let us position the X0 Y0 Z0 origin in more than one place. We can put one on the first vise and another on the second vise. Now just by changing the work offset the same program can work to make the part on either vise.
There are lots of different kinds of offsets in CNC, and the skilled CNC operator/machinist finds that offsets are an extremely handy way to nudge the behavior of a G-Code Program without having to change that program. Most CNC controllers have an offsets screen where you do that. I mention this because any time you get a chance to learn about offsets, take the time to do so. They're digital power tools for the CNC machinist and are very handy. We'll cover them in greater detail later.
Planes
It's convenient to refer to planes for various purposes. A plane is a flat 2 dimensional space defined by two axes. For example, the default plane on most mills is XY. If you draw an arc without specifying a change in the plane, it will be drawn on the XY plane. There is a plane for each combination of the linear axes XYZ:
• XY
• YZ
• XZ
The G17, G18, and G19 G-Codes select which plane is active. More on G17-G19 when we talk more about arcs.
Conclusion
You've now got the basics:
- You know how to visualize the coordinate systems relative to your machine using the left handed rule.
- You know how to express coordinates in G-Code.
- You know what units are used to measure the coordiantes.
- You know there is the possibility of both relative and absolute coordinates.
- You know that offsets let you shift the coordinate system around for various handy purposes.
We'll shortly introduce the notion of MDI, which is a simple way to use G-Code as though you are still a manual machinist. It's a good introduction to the basics of moving your CNC's axes. But first, we need to get you set up on G-Wizard Editor so you'll have a CNC Simulator to use for practice during these tutorial lessons.
Exercises
1. Form the left-handed and right-handed coordinate systems with your own hands and visualize which way the axes run on your machine. Which direction is the positive (increasing positive coordinates) for each axis? Which direction do increasingly positive coordinates in the g-code move the table and spindle?
2. Get out the manual for your machine and find the diagram that shows how its coordinate system works. Make sure to leave the manual handy, whether it is paper or online. We'll refer back to it a number of times as we go through the various exercises.
3. Bring up G-Wizard G-Code Editor. By default, you're in Mill Mode. There are views for Perspective, Top, Front, and Right. Download the sample engraving file from our download page. You want the file called HomeSwitchRearPanelEngrave. Start up GWE and do a File Open to load the downloaded file. Have a look at it in each view.
- Top is a view from the XY plane
- Front is a view from the XZ plane
- Right is a view from the YZ plane
Try the Free Trial Version of G-Wizard G-Code Editor...
No credit card required--just your name and email.
Next Article:G-Code Dialects, Post Processors, and Setting Up GWE
All material © 2010-2016, CNCCookbook, Inc.
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# Meaning of Equality
Often we teach the meaning of equality as the equal sign indicating “put the answer.” To prepare students for future mathematics, we need to rethink the way we communicate this representation to students. The equal sign is too important to attach such a limited meaning, especially when students are moving to abstraction. When moving to abstraction, it is important to use this symbol as a “reader” versus a “speller.” What do I mean by that? Let me explain. First, a speller sounds foreign because the student is trying to make sense of abstract symbols that they don’t fully understand. A reader connects meaning that they can comprehend while connecting the story to the symbol.
For example, when teachers record the following equation,
A speller would say, “Four plus five equals nine”.
A reader would say, “Four apples and five apples is the same as saying ‘I have nine apples’” .
In the journal prompt below, we want students to see the statement as 5 and 4 is the same as 2 less than 11. If we focus too much as a speller, students see symbols as performing actions rather than relating ideas, which leads to misconceptions of the equal sign.
Less is more. Hesitate in thinking math becomes harder by creating larger numbers, rather provide depth early on using children’s intuitive understanding of basic number operations. A simple journal entry can help assess what students understand about operations and the meaning of equality. Once you find that a student understands the meaning of addition and subtraction, it is time to create tasks that dive deeper in understanding equality. These task can be as simple as asking questions such as, 4 + 5 = ____ + 6. Additional equality tasks can be found in the Algebraic Thinking and Reasoning with Numbers books (Groundwork Series) by Greenes, Carole, et al., (https://www.mheducation.com/prek-12/program/MKTSP-O8302M06.html).
In the book, “Thinking Mathematically,” Carpenter, Thomas P., et al. suggest some benchmarks to keep in mind while moving towards the conceptual understanding of equality.
1. Establish a starting point. What are students’ initial ideas regarding the equal sign?
2. Mix it up. Don’t always write equations in the form a + b = c, rather c = b + a.
3. See the equal sign as a relational symbol. Emphasize students proving that each side is the same as the other.
4. Compare sides without calculations. Encourage students to look for relationships without performing calculations.
Listen to students and be aware of where they are in the process. Encourage them to question if it is true. If so, why is it true? By listening and looking we may be pleasantly surprised at what we find out.
Below is student work demonstrating various levels of understanding within the concept of equality.
Spiky says that 5 + 4 = 11 – 2 is false.
Curly says that it is true.
Who do you agree with and why?
This student understands how to simplify expressions but lacks applying the meaning of equality.
This student is confused in understanding what it means to have expressions on each side of the equal sign.
This student understands the meaning of equality. What can we ask next?
This student demonstrates a shallow understanding of equality. What can we do next?
This student can solve for an answer but needs practicing in mixing up the form of an equation.
# How can we differentiate instruction without putting more work on our plate?
Teachers do not have time to prepare multiple tasks and lessons to meet the needs of each student in the classroom. The key is giving an anchor task that all students can enter and knowing how to adjust the task based on our observations.
In this class, students were guided to use “Katell’s” method (aka-left to right strategy) to solve the problem 65 – 12. Notice how the student recorded their thinking process.
How can we differentiate with one task?
During the guided structure component of the lesson, teachers are observing and questioning students. This student demonstrates and can explain the left to right strategy. What comes next for this student? During our planning phase, we need to ask ourselves, “What can I do for the student that already knows the answer”? Looking at this student’s work, how can we deepen the mathematical understanding?
After listening to the student explain his thinking, the teacher commented, I can understand your verbal explanation, but your written work is confusing to me. I wonder why I am confused? The teacher then walked away, and the student grappled with the idea of equality. Differentiation is meeting the student where they are at and extending the learning.
Do not try to create more problems for students to practice. Plan your task thinking of what mistakes students typically make and how you can help extend the thinking. Use your observations to go deeper and differentiate!
What are other ways you could take the task 65 – 12 and extending the task? Share your thought below.
# How can we help student refocus on understanding versus just an answer?
Too often when we pose a problem and students shout out an answer. We need to ask ourselves do students understand the concept or are they obtaining the correct answer by fitting the symbols and numbers into a structure they know?
In this example, the anchor task was to subtract 34 from 87. The teacher wanted to screen the children first wondering, Do students know what it means to subtract one number from another? To find out more the teacher posted the following problem, removing the numbers.
Do students have the conceptual understanding or just fit the numbers into a given structure?
Students were asked to set up the expression that could represent the situation. The student A on the far right was the only student in the class that seemed to understand. When asked to explain his thinking, many observing teachers felt he understood the concept and that he gave us a platform to generate a discussion.
Do students understand the part-whole relationship here?
Following this analysis, the numbers were inserted into the problem. Subtract 34 from 87. It was interesting to see student A’s work. Much to our surprise, Student A who seemed to understand the concept had a hard time transferring that knowledge to another situation. Notice his equation is not correct but he gets the final answer.
Can student A transfer knowledge from one setting to the next? Look and listen to help guide students to conceptual understanding. Looking at answers does not tell us the whole story.
Less is more. Spend more time on conceptual understanding and listening and watching students versus solving more problems. The answer does not help assess student reasoning or how we can extend or guide the learning process.
At 2017 Mathodology Fall Institute participants worked on a task from think!Mathematics which required them to use the given shapes to make a composite figure. Figure C posed a problem for many. Is it possible? Participants worked and worked and did not find a way to solve the problem with the given shapes. The teacher allowed the task to move on without closure for a purpose. When students work to become confident problem solvers we want them stating, “Figure C is not possible with the given shapes. I would need an additional… to complete this shape.” We have built perseverance and another way to assess them. Do they know what to ask? “I need two more pieces of one shape.”
When we do not provide immediate closure, we allow students to continue to explore. Many participants emailed after the institute because they would not quit until they figured it out. Below are a few of the pictures they sent.
Try giving a task where there is not enough information and see how your students handle it. Will they be confident to state this is not possible, and know what to ask?
Great perseverance by participants who refused to give up?
## Grade 2 begining Journal Entry
Journal Entry: Given a number write four things you know about that number.
How can we use this to help assess where a student is before we even begin a formal lesson?
Student 1:
Student 2:
Student 3:
## What strategy is Victoria using?
Can you follow Victoria’s thinking? Help me understand.
## Is this correct?
Take a look at this problem. What was she thinking? Do you agree or disagree? Explain.
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# Mode in Python
In statistics, the mode is the value that appears most frequently in a dataset. In Python, you can calculate the mode of a list or an array using the statistics module or the numpy module.
Here’s an example using the statistics module:
import statistics
data = [1, 2, 2, 3, 3, 3, 4, 4, 5]
mode = statistics.mode(data)
print(mode)
Output:
3
And here’s an example using the numpy module:
import numpy as np
data = [1, 2, 2, 3, 3, 3, 4, 4, 5]
mode = np.mode(data)
print(mode)
Output:
ModeResult(mode=array([3]), count=array([3]))
Note that np.mode returns an object that contains both the mode value and its frequency count.
### An Introduction to Statistics Mode:
In statistics, the mode is a measure of central tendency, which represents the most frequently occurring value in a dataset. It is one of the three main measures of central tendency, along with the mean and median. The mode is a useful metric for describing the distribution of data in a dataset, especially when the data is categorical or discrete.
To find the mode, you need to identify the value that occurs most frequently in a dataset. This can be done by sorting the data and counting the number of times each value appears. The value with the highest count is the mode.
It is important to note that a dataset may have multiple modes. If two or more values occur with equal frequency, then they are both considered modes. This is called a bimodal distribution. For example, if a dataset contains the values {1, 2, 2, 3, 4, 4}, then both 2 and 4 are modes, since they both occur twice.
The mode is often used in descriptive statistics to summarize the most common value or category in a dataset. It can also be used to describe the shape of a distribution. For example, a unimodal distribution has one mode, a bimodal distribution has two modes, and a multimodal distribution has three or more modes.
It is important to note that the mode is not always a good measure of central tendency. In some cases, it may not be a representative value for the dataset. For example, in a skewed distribution, the mode may not be a good measure of the central tendency since it is influenced by the extreme values. In such cases, the median or the mean may be a better measure of central tendency.
In summary, the mode is a measure of central tendency that represents the most frequently occurring value in a dataset. It is a useful metric for describing the distribution of data in a dataset, especially when the data is categorical or discrete. However, it is not always a representative measure of central tendency and should be used in conjunction with other measures, such as the mean and median, to fully describe the dataset.
### The mode() function in Python:
In Python, the mode() function is used to find the mode of a list or an array. The function is available in the statistics module, which provides a collection of functions for calculating basic statistical properties of data.
Here’s the syntax for using the mode() function:
import statistics
data = [1, 2, 2, 3, 3, 3, 4, 4, 5]
mode_value = statistics.mode(data)
In this example, the mode() function is called with a list data as its argument. The function returns the mode of the list, which is assigned to the variable mode_value.
If the list has multiple modes, the mode() function will return the first one it encounters.
It’s important to note that the mode() function raises a StatisticsError if the list is empty or if there is no mode, which occurs when all the values in the list occur with equal frequency. To avoid this error, you can check if the list is empty or use a try-except block to handle the exception:
import statistics
data = []
try:
mode_value = statistics.mode(data)
except statistics.StatisticsError:
mode_value = None
print(mode_value)
In this example, the try block attempts to calculate the mode of an empty list. Since an empty list has no mode, the mode() function raises a StatisticsError, which is caught by the except block. In this case, mode_value is set to None to indicate that there is no mode.
### Some Applications of mode() function:
The mode() function in Python is useful for a variety of applications, especially when working with categorical or discrete data. Here are a few examples of how the mode() function can be used:
1. Analyzing survey results: The mode() function can be used to find the most common response to a survey question. For example, if a survey asks respondents to select their favorite color from a list of options, the mode of the responses can be used to determine the most popular color.
2. Quality control: The mode() function can be used to identify the most common defect in a manufacturing process. By analyzing the frequency of defects, the mode can be used to identify the most common problem and prioritize corrective actions.
3. Marketing research: The mode() function can be used to analyze customer preferences and behavior. For example, the mode of customer purchase history can be used to identify the most popular products and optimize marketing strategies.
4. Educational assessment: The mode() function can be used to analyze the results of multiple-choice tests. The mode of the responses can be used to identify the most common answer and evaluate the effectiveness of the test questions.
5. Analyzing financial data: The mode() function can be used to analyze the frequency of stock prices or exchange rates in financial data. The mode can be used to identify the most common price or rate and inform investment decisions.
These are just a few examples of the many applications of the mode() function in Python. By identifying the most common value in a dataset, the mode can provide valuable insights and inform decision-making in a variety of fields.
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# Real Analysis HW 2 Solutions
Problem 18: Let E have finite outer measure. Show that there is a Gδ set G ⊇ E with
m(G) = m∗ (E). Show that E is measurable if and only if there is an Fσ set F ⊂ E with
m(F ) = m∗ (E).
Solution: Let n be a positive integer. By the definition of outer measure we may find a
collection of open intervals {Ik,n }∞
k=1 such that
X
l(Ik,n ) < m∗ (E) + 1/n.
k=1
Define the Gδ set G =
T∞ S∞
k=1 Ik,n
n=1
and note that
m (G) ≤ m (
[
Ik,n ) < m∗ (E) + 1/n
k=1
for every integer n ≥ 1. Therefore m∗ (G) ≤ m∗ (E). However, since E ⊆ G, by monotonicity
we have the reverse inequality m∗ (E) ≤ m∗ (G). Therefore m∗ (E) = m∗ (G).
Note that by Theorem 11 (iv) E is measurable if and only if there is an Fσ set F ⊆ E
such that m∗ (E ∼ F ) = 0. However, since F is measurable, by the excision property this is
equivalent to m∗ (E) = m∗ (F ).
Problem 19: Let E have finite outer measure. Show that if E is not measurable, then
there is an open set O containing E that has finite outer measure and for which
m∗ (O ∼ E) > m∗ (O) − m∗ (E).
Solution: Since E is not measurable, we know by Theorem 11 that there exists an 0 > 0
such that for any open set O containing E we have m∗ (O ∼ E) ≥ 0 . By the definition of
outer measure, we know that there exists
S∞ a countable collection of bounded open intervals
{Ik }k=1 , whose union we denote O ≡ k=1 Ik , such that
m∗ (O) − m∗ (E) < 0 ≤ m∗ (O ∼ E).
1
Note that we may write E1 ∪ E2 as a 2 . k=1 Therefore m∗ (O ∼ E) < . Problem 24: Show that is E1 and E2 are measurable. b). Solution: Note that if either E1 or E2 have infinite measure then the inequality trivially holds. k=1 Since Ik is open and bounded. then m(E1 ∪ E2 ) + m(E1 ∩ E2 ) = m(E1 ) + m(E2 ). Show that E is measurable if and only if for each open. b) ∼ E). b). we have m∗ (Ik ) = m∗ (Ik ∩ E) + m∗ (Ik ∼ E). since E has finite outer measure we know that there is a countable collection of bounded open intervals Ik such that ∞ X m∗ (Ik ) < m∗ (E) + . m∗ (Ik ) ≥ m∗ (E) + m∗ (O ∼ E).b) ∼ E) holds for every bounded interval (a.Problem 20: (Lebesgue) Let E have finite outer measure. then E is measurable. Let > 0. b)) = m∗ ((a. b − a = m∗ ((a. k=1 and using the fact that E = ∗ ∞ [ S∞ k=1 Ik ∩E and S∞ k=1 (Ik ∼ E) = O ∼ E. bounded interval (a. Suppose that m∗ ((a.b) ∩ E) + m∗ ((a. Assume that E1 and E2 have finite measure. b) ∩ E) + m∗ ((a. which by Theorem 11 is equivalent to the measurability of E. we simply need to show that if the measurability condition is satisfied on all open bounded intervals. Solution: Since the (Carath´eodory) definition of measurability is immediately satisfied for open bounded intervals. Upon summing both sides and using countable sub-additivity ∞ X ∗ m (Ik ) = ∞ X ∗ m (Ik ∩ E) + ∗ ≥m ∞ [ m∗ (Ik ∼ E) !k=1 k=1 k=1 ∞ X (Ik ∩ E) ∗ +m k=1 Denoting O ≡ we find S∞ k=1 Ik m (E) + > ∞ X ! (Ik ∼ E) .
Therefore by finite additivity m0 (A) = m0 (A ∼ B) + m0 (B).disjoint union E1 ∪ E2 = (E1 ∩ E2 ) ∪ (E1 ∼ (E1 ∩ E2 )) ∪ (E2 ∼ (E1 ∩ E2 )). Using countable additivity and monotonicity ! ! ∞ ∞ ∞ ∞ [ [ X X 0 0 0 0 m (E) ≤ m Ek = m Ak = m (Ak ) ≤ m0 (Ek ). m(E1 ∪ E2 ) = m(E1 ∩ E2 ) + m(E1 ∼ (E1 ∩ E2 )) + m(E2 ∼ (E1 ∩ E2 )) = m(E1 ∩ E2 ) + m(E1 ) − m(E1 ∩ E2 ) + m(E2 ) − m(E1 ∩ E2 ) = m(E1 ) + m(E2 ) − m(E1 ∩ E2 ). which relies on the properties of monotonicity. To prove excision and monotonicity. and the excision property. (i) Show that m0 is finitely additive. 0 To prove countable monotonicity we let {Ek }∞ k=0 ⊆ M be a collection of sets that covers some set E ∈ M0 . countably monotone. ∞]. monotone. suppose A. This proves both monotonicity and excision since m0 (A ∼ B) ≥ 0. Solution: (i) Finite additivity follows trivially from countable additivity. By finite additivity and excision. An = En ∼ n−1 [ Ek . As usual we define A0 = E0 and for n ≥ 1. since we may consider collections of sets for which only finitely many are non-empty. 3 . k=1 S S∞ so that {An } are disjoint. Since we can write A as a disjoint union A = (A ∼ B) ∪ B. is countably additive. k=0 k=0 k=0 k=0 (ii) The proof is identical to the proof of in the case of Lebesgue. Problem 27: Let M0 be any σ-algebra of subsets of R and m0 a set function on M0 which takes values in [0. B ∈ M0 with B ⊆ A. and such that m0 (∅) = 0.An ⊆ En and ∞ k=0 Ak = k=0 Ek . and possesses the excision property (ii) Show that m0 possesses the same continuity properties as Lebesgue measure. countable additivity.
S∞ S∞ ∞ and note that {An }k=1 is ascending and that k=1 An = k=1 En . Problem 28: Show that continuity of measure together with finite additivity of measure implies countable additivity of measure. Sn of measurable sets. Define A = Solution: Let {Ek }∞ n k=1 be a disjoint collection k=1 Ek . k=1 ∞ X m(Ek ). Continuity of measure then implies ! ! ∞ ∞ [ [ m Ek = m Ak = lim m(An ). k=1 n→∞ k=1 However by finite additivity lim m(An ) = lim n→∞ n→∞ Therefore m ∞ [ n X m(Ek ) = k=1 ! Ek = k=1 ∞ X m(Ek ). k=1 4 .
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# In an air motor cylinder.. 450 kJ/kg. solved problem
#### In an air motor cylinder the compressed air has an internal energy of 450 kJ/kg at the beginning of the expansion and an internal energy of 220 kJ/kg after expansion. If the work done by the air during the expansion is 120 kJ/kg, calculate the heat flaw to and from the cylinder.
In an air motor cylinder the compressed air has an internal energy of 450 kJ/kg at the beginning of the expansion and an internal energy of 220 kJ/kg after expansion. If the work done by the air during the expansion is 120 kJ/kg, calculate the heat flaw to and from the cylinder.
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# Digital Electronics | BCD Adder
BCD stand for binary coded decimal. Suppose, we have two 4-bit numbers A and B. The value of A and B can varies from 0(0000 in binary) to 9(1001 in binary) because we are considering decimal numbers.
The output will varies from 0 to 18, if we are not considering the carry from the previous sum. But if we are considering the carry, then the maximum value of output will be 19 (i.e. 9+9+1 = 19).
When we are simply adding A and B, then we get the binary sum. Here, to get the output in BCD form, we will use BCD Adder.
Example 1:
Input :
A = 0111 B = 1000
Output :
Y = 1 0101
Explanation: We are adding A(=7) and B(=8).
The value of binary sum will be 1111(=15).
But, the BCD sum will be 1 0101,
where 1 is 0001 in binary and 5 is 0101 in binary.
Example 2:
Input :
A = 0101 B = 1001
Output :
Y = 1 0100
Explanation: We are adding A(=5) and B(=9).
The value of binary sum will be 1110(=14).
But, the BCD sum will be 1 0100,
where 1 is 0001 in binary and 4 is 0100 in binary.
Note – If the sum of two number is less than or equal to 9, then the value of BCD sum and binary sum will be same otherwise they will differ by 6(0110 in binary).
Now, lets move to the table and find out the logic when we are going to add “0110”.
We are adding “0110” (=6) only to the second half of the table.
The conditions are:
1. If C’ = 1 (Satisfies 16-19)
2. If S3′.S2′ = 1 (Satisfies 12-15)
3. If S3′.S1′ = 1 (Satisfies 10 and 11)
So, our logic is
C' + S3'.S2' + S3'.S1' = 1
Implementation :
Read related articles: BCD to 7 Segment Decoder, BCD(8421) to/from Excess-3
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Chapter 1 - Matter: Its Properties and Measurement - Exercises - Units of Measurement - Page 28: 33
h= 1.5 m
Work Step by Step
h = 15 hands 1 hand = 4 inches, so 15 hands = 15 * 4 inches = 60 inches. So h = 60 inches. 1 inch = 2.54 cm, so 60 inches = 60 * 2.54 cm = 152.4 cm. So h = 152.4 cm = 1.524 m Rounding to 2 significant figures gives 1.5 m
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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# Substrate, Flourite in a cylindrical aquarium
## cylindrical aquarium sand, gravel and substrate calculator (10)
### The weight of Substrate, Flourite, 2 inches high, in a cylindrical aquarium with the diameter of a base of 20 inches
gram 10 347.78 ounce 365.01 kilogram 10.35 pound 22.81
### The volume of Substrate, Flourite, 2 inches high, in a cylindrical aquarium with the diameter of a base of 20 inches
centimeter³ 10 296.3 inch³ 628.32 foot³ 0.36 meter³ 0.01
• Substrate, Flourite weighs 1.01 gram per cubic centimeter or 1 005 kilogram per cubic meter, i.e. density of substrate, Flourite is equal to 1 005 kg/m³. In Imperial or US customary measurement system, the density is equal to 62.74 pound per cubic foot [lb/ft³], or 0.58 ounce per cubic inch [oz/inch³] .
• Substrate, Flourite weighs 1 005 kg/m³ (62.7401 lb/ft³) with specific gravity of 1.005 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
• A cylindrical tank, aquarium or a pond has a shape of a cylinder, with two circles of the same diameter (D) at the base and at the top of the tank. The cylindrical aquarium gravel calculator allows you to specify the diameter (D) and desired height of a selected gravel or substrate. The number in square brackets, to the right of the gravel name, is the gravel specific gravity or relative density, as compared to pure water.
The calculator computes the weight (1) and volume (2) of gravel using the following formulas:
(1) m = ρ × V and (2) V = π×r²×h = π×D²×h ⁄ 4, where ρ is the density, V — volume, m — weight (or mass), and h is the height of gravel; π is approximately equal to 3.14159265359, r and D are the radius and diameter of the base of an aquarium respectively.
See also our gravel calculators for quarter cylinder and rectangular shaped aquariums and ponds.
#### Foods, Nutrients and Calories
CAJUN CHEDDAR CRACKERS, UPC: 763027508156 contain(s) 393 calories per 100 grams (≈3.53 ounces) [ price ]
209328 foods that contain Total lipid (fat). List of these foods starting with the highest contents of Total lipid (fat) and the lowest contents of Total lipid (fat)
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Indo-Pacific Black weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Freon 13B1 [CBrF3] weighs 1 580 kg/m³ (98.63618 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 30 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
The millisievert beta-radiation [mSv β] is a derived unit of ionizing radiation dose in the International System of Units (SI) and is a measure of the effective biological damage of low levels of ionizing radiation on the human body caused by exposure to beta (β) radiation in thousandths of a sievert.
The electric resistance of a conductor determines the amount of current, that flows through the conductor for a given voltage at the ends of the conductor.
t/µ² to oz t/ha conversion table, t/µ² to oz t/ha unit converter or convert between all units of surface density measurement.
#### Calculators
Calculate gravel and sand coverage in a cylindrical aquarium
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February 28, 2024
# How to convert mg/l to meq/l:
In hydrological research, most of the time we need to convert the mg/l to meq/l. This milligram per liter (mg/L) is also a unit. The milliequivalent per liter (meq/L) is also a unit to measure the Cation or Anion concentration.
To know the milliequivalent, we need to know, what is milliequivalent. This is the atomic mass divided by the valency.
Suppose, Ca++ has a valency of 2+
and Ca++ atomic mass is 40.08 gm/mole
So, milliequivalent is the [40.08 (gm/mole)]/2 [(valency/mole)]= 20.04 gm/equivalent
Now, 33 mg/L of Ca++ need to convert to the milliequivalent/L
is simply divide by 20.04
Hence, (33/ 20.04) meq/L= 1.65 meq/l
Still Difficult to understand?
## How to convert mg/l to meq/l? Let’s explain more;
To convert milligrams per liter (mg/L) to milliequivalents per liter (meq/L), you need to know the molar mass of the substance in question. The molar mass is the mass of one mole of the substance, expressed in grams per mole (g/mol).
The formula to convert mg/L to meq/L is as follows:
meq/L = (mg/L) / (molar mass in mg/mol) * (valence factor)
The valence factor represents the number of positive or negative charges contributed by each molecule of the substance in the solution. It depends on the ionic nature of the substance.
Here’s an example to illustrate the conversion:
Let’s say you have a substance with a molar mass of 50 mg/mol and a valence factor of 2. The concentration of the substance is 100 mg/L. To convert this to meq/L:
meq/L = (100 mg/L) / (50 mg/mol) * (2)
meq/L = 2 meq/L
Therefore, 100 mg/L is equal to 2 meq/L for this particular substance.
Remember to adjust the molar mass and valence factor based on the specific substance you are working with.
## How to convert meq/l to mg/l:
To convert milliequivalents per liter (meq/L) to milligrams per liter (mg/L), you need to know the equivalent weight of the substance in question. The equivalent weight represents the amount of substance that corresponds to one mole of electrons transferred or one mole of protons donated or accepted.
Once you have the equivalent weight, you can use the following conversion:
1 meq/L = Equivalent weight in milligrams per liter (mg/L)
For example, let’s say you want to convert 10 meq/L of sodium ions (Na+) to mg/L. The equivalent weight of sodium is 23 g/mol, which means the equivalent weight in milligrams would be 23 mg/mol. Therefore:
10 meq/L = 10 × 23 mg/L = 230 mg/L
Keep in mind that the specific equivalent weight varies depending on the substance you are converting. So you’ll need to determine the appropriate equivalent weight for the specific ion or molecule you are working with.
## Some times “me/l to mg/l” is misspelled as meq/l to mg/l:
“me/L” is commonly used as an abbreviation for milliequivalents per liter, and “mg/L” stands for milligrams per liter. To convert milliequivalents per liter (meq/L) to milligrams per liter (mg/L), you need to know the molar mass or equivalent weight of the substance in question.
So the conversion factor is the same as before as I have already explained.
1 meq/L = Equivalent weight in milligrams per liter (mg/L)
Let’s give another example to convert 10 meq/L of calcium ions (Ca2+) to mg/L. The equivalent weight of calcium is 20.04 g/mol, which means the equivalent weight in milligrams would be 20.04 mg/mol. Therefore:
10 meq/L = 10 × 20.04 mg/L = 200.4 mg/L
Now Let’s discuss another bigger term:
## How to convert eq/l to mg/l?
To convert equivalents per liter (eq/L) to milligrams per liter (mg/L), you need to know the molar mass of the substance in question. The molar mass represents the mass of one mole of the substance in grams.
Once you have the molar mass, you can use the following conversion:
1 eq/L = Molar mass in milligrams per liter (mg/L)
For example, let’s say you want to convert 5 eq/L of sulfuric acid (H2SO4) to mg/L. The molar mass of sulfuric acid is 98.09 g/mol, which means the molar mass in milligrams would be 98.09 mg/mol. Therefore:
5 eq/L = 5 × 98.09 mg/L = 490.45 mg/L
Now I Hope you understand How to convert mg/l to meq/l or vice versa.
How to convert mg/l to meq/l: Learn more from the AL-Mustaqbal University College Pharmacy department and Get the PDF
How to convert mg/l to meq/l: Learn more From Research Gate a Published article
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Calculate ytm with coupon rate
It's easy to calculate the coupon rate on a plain-vanilla bond – one that pays a fixed coupon at equal intervals. For example, you might buy directly from the U.S. Treasury a 30-year bond with a face value of \$1,000 and a semiannual coupon of \$20. You'll collect \$20 of interest twice a year, or \$40 annually.
Dividing the \$40 annual interest by the \$1,000 face value gives a coupon rate of 4 percent. Some bond types, called floaters, have variable coupon payments that adjust to current prevailing interest rates and therefore do not have a defined coupon rate. Figuring the Current Yield Do not confuse the coupon rate with the current yield. P = I PV a (YTM/2, 29) + M PV s (YTM/2, 29) where I is the periodic bond income (= M * semi-annual coupon rate), M is the maturity value (I would assume \$1000), PV a is the Present Value for an annuity. Bond Yield to Maturity Calculator for Comparing Bonds with Different Prices and Coupon Rates This free online Bond Yield to Maturity Calculator will calculate a bond's total annualized rate of return if held until its maturity date, given the current price, the par value, and the coupon rate. The price of the bond is \$1,101.79 and the face value of the bond is \$1,000. The coupon rate is 7.5% on the bond. Based on this information, you are required to calculate approximate yield to maturity on the bond. Solution: Use the below-given data for calculation of yield to maturity.
how to calculate the yield to maturity of the bond if coupon is paid semi-annually problem gold inc Calculate the bond's YTM (expected rate of return) if coupon.
Coupon Rate: Annual payout as a percentage of the bond's par value To calculate it, we need to satisfy the same condition as with all composite payouts:. Jun 8, 2015 The formula for calculating YTM is as follows. Let's work it out with an example: Par value (face value) = Rs 1,000 / Current market price = Rs 920 / A tutorial for calculating and comparing bond yields: nominal and current yield, A simplification of the YTM formula can be made if the bond has no coupon Yield to maturity (YTM) calculator is an online tool for investment calculation, to invest in, Bond face value, Bond price, Coupon rate and years to maturity.
This calculator is designed to calculate the duration of a bond based on the YTM, coupon rate and remaining term of the bond. It also calculates modified
YTM = yield to maturity, as a decimal (multiply it by 100 to convert it to percent); M = maturity value; P = price; n = years until maturity. Let's say a zero coupon bond Calculating Reinvested Interest at YTM Rate. The interest-on-interest formula for reinvested coupon payments assumes the reinvested payments grow at an Given the YTM and a bond's cash flows, we can calculate the bond's price. Say a 10-year bond pays an annual \$50 coupon and has a 3% YTM. Then the bond's
The formula for calculating a bond's price uses the basic present value (PV) formula If the YTM is less than the bond's coupon rate, then the market value of the
Yield to Maturity Calculator - The rate of return anticipated on a bond if it is held Articles of Interest How To Properly Research For The Best Mortgage Rate. The speculative rate of return or interest rate of a fixed-rate security The YTM calculation is structured to show – based on compounding – the effective yield a The formula for calculating a bond's price uses the basic present value (PV) formula If the YTM is less than the bond's coupon rate, then the market value of the This calculator automatically assumes an investor holds to maturity, reinvests coupons, and all payments To calculate a bond's yield to maturity, enter the face value (also known as "par value"), the coupon rate, the number of years to maturity, the frequency of
On this bond, yearly coupons are \$150. The coupon rate for the bond is 15%, and the bond will reach maturity in 7 years. The formula for determining approximate YTM would look like below: The approximated YTM on the bond is 18.53%.
Yield to Maturity Calculator (YTM Calculator) - calculate the annual return rate for a bond when it is held until maturity. Bond yield Bond Coupon Rate: %. Bond Coupon Rate (% p.a.). %. Years to Maturity. Payment. Annually Semi- This calculator is designed to calculate the duration of a bond based on the YTM, coupon rate and remaining term of the bond. It also calculates modified When interest rates rise the value of an existing bond falls. the assumption that coupon payments will be reinvested at the YTM rate, for the life the bond. Yield = bndyield( Price , CouponRate , Settle , Maturity ) given NUMBONDS CouponRate — Annual percentage rate used to determine coupons payable on a The calculation of YTM takes into account the current market price, par value, coupon interest rate and time to maturity. It is also assumed that all coupons are When you buy a bond at par, yield is equal to the interest rate. YTM is a more advanced yield calculation that shows the total return you will receive if you hold
Without calculations: When the YTM increases, the price of the bond decreases. Without calculations: a longer time to maturity and a lower coupon rate make Learn how formulas are used to calculate rates of return - including interest rates, As noted earlier, the YTM, or simply the yield, is the rate that equates the how to calculate the yield to maturity of the bond if coupon is paid semi-annually problem gold inc Calculate the bond's YTM (expected rate of return) if coupon. This calculator calculates implied yield of a Zero Coupon Bond; It calculates Excel's XIRR equivalent yield to maturity of a discounted bond. Suppose we know the current price of a bond, its coupon rate, and its time to maturity. How do we calculate the YTM? • We can use the straight bond formula,
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# If 2^x = 1/2^y then x+y=??
+1 vote
100 views
If 2^x = 1/2^y then x+y=??
posted Aug 17, 2015
zero .........
we have
2^x = 1/2^y
2^x * 2^y =1
2^(x+y) =2^0
on comparing both sides,
we get,
x+y=0
Similar Puzzles
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
If 2^x = 3^y = 6 then which of the following statements is true?
1) x - y = xy
2) y - x = xy
3) x + y = xy
4) xy = x/y
+1 vote
If
x^4 + 5y = 6
x^2 y^2 + 5x = 6
Then
x,y=??
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Difference between revisions of "1962 AHSME Problems/Problem 37"
Problem
$ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE = AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(D)}\ \frac{5}{8}\qquad\textbf{(E)}\ \frac{2}3$
Solution
Let $AE=AF=x$ $[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}$ Or $[CDFE]=\frac{\frac{5}{4}-(x-\frac{1}{2})^2}{2}\le \frac{5}{8}$ As $(x-\frac{1}{2})^2\ge 0$ So $[CDFE]\le \frac{5}{8}$ Equality occurs when $AE=AF=x=\frac{1}{2}$ So the maximum value is $\frac{5}{8}$
Solution 2
Let us first draw a unit square. $[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("1",(0,0)--(0,1),W); [/asy]$ We will now pick arbitrary points $E$ and $F$ on $\overline{AB}$ and $\overline{AD}$ respectively. We shall say that $AE = AF = x$ $[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("1",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label("E",(0.5,1),N); label("F",(0,0.5),SE); [/asy]$ Thus, our problem has been simplified to maximizing the area of the blue quadrilateral. If we drop an altitude from $E$ to $\overline{DC}$, and call the foot of the altitude $G$, we can find the area of $EFDC$ by noting that $[EFDC] = [EGDF] + [EGC]$. $[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label("A",(0,1),NW); label("B",(1,1),NE); label("C",(1,0),SE); label("D",(0,0),SW); label("1",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label("E",(0.5,1),N); label("F",(0,0.5),SE); draw((0.5,1)--(0.5,0),red); draw((0.5,0.1)--(0.4,0.1),red); draw((0.4,0.1)--(0.4,0),red); label("G",(0.5,0),S); [/asy]$ We can now finish the problem.
Since $EG = 1$ and $FD = 1-x$, we have: $$[EFDC] = [EGDF] + [EGC] = DG\cdot\frac{EG + FD}{2} + \frac{(CG)(EG)}{2} = \frac{x(2-x)}{2} + \frac{1-x}{2} = \frac{1+x-x^2}{2}$$
To maximize this, we compete the square in the numerator to have: $$[EFDC] = \frac{-(x-\frac{1}{2})^2 + \frac{5}{4}}{2} = \frac{5}{8} - \frac{(x-\frac{1}{2})^2}{2}$$
Finally, we see that $\frac{(x-\frac{1}{2})^2}{2}\geq0$, as: $$\left(x-\frac{1}{2}\right)^2\geq0$$ So, $$\frac{(x-\frac{1}{2})^2}{2}\geq0,$$
where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by $2$, which does not change the inequality sign. Thus, the maximum area is $\boxed{\frac{5}{8}}$ or $\boxed{\textbf{(D)}},$ when $\frac{(x-\frac{1}{2})^2}{2} = 0.$
Solution 3(Calculus)
Let $AE = AF = x$.
The area of the quadrilateral $CDFE$ can be expressed as $A = 1 - \frac{x^2}{2} - \frac{(1-x)}{2}$.
By taking the derivative of $A$, we get $A' = -x + \frac{1}{2}$.
We make $A' = 0$ and get the critical point $x = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ , the maximum area is $\boxed{\frac{5}{8}}$ or $\boxed{\textbf{(D)}}$.
~belanotbella
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# The Simplest Transformations of Arithmetic Root (Radical)
When we transform the arithmetic root we should use their properties.
Example 1. Simplify the following: sqrt(45a^5).
Using property 1, we will obtain sqrt(45a^5)=sqrt(9a^4*5a)=sqrt(9)*sqrt(a^4)*sqrt(5a)=3a^2sqrt(5a).
Such transformation is called factoring out of root.
Example 2. Simplify the following: (root(3)(a^2))^5.
Using property 3, we will obtain (root(3)(a^2))^5=root(3)((a^2)^5)=root(3)(a^10). Let′s simplify radical expression, for this we will factor out of root.
Then root(3)(a^10)=root(3)(a^9*a)=root(3)(a^9)*root(3)(a)=a^3root(3)(a).
Example 3. Simplify the following: root(4)(x^2root(3)(x)).
Let′s transform expression x^2root(3)(x), for this we will factor in of root:
x^2root(3)(x)=root(3)((x^2)^3)*root(3)(x)=root(3)(x^6)*root(3)(x)=root(3)(x^6*x)=root(3)(x^7).
According to property 4, we have root(4)root(3)(x^7)=root(12)(x^7).
Example 4. Simplify the following: root(30)(2^9) .
According to property 5, we can divide the index of radical and exponent of radicand by the same natural number. Dividing these indicators by 3, we will obtain root(30)(2^9)=root(10)(2^3)=root(10)(8).
Example 5. Simplify the following: root(5)(a)*root(5)(a^2).
According to property 1, if we need to multiply the same degree we should multiply the radicands and extract the root of the same degree from obtained result. So, root(5)(a)*root(5)(a^2)=root(5)(a*a^2)=root(5)(a^3).
Example 6. Simplify the following: root(3)(a)*root(6)(a).
At first we should reduce the radicals to the same index. According to property 5, we can divide the index of radical and exponent of radicand by the same natural number. That is why root(3)(a)=root(6)(a^2). Then we have root(6)(a^2)xxroot(6)(a)=root(6)(a^3). Dividing the index of radical and exponent of radicand by 3, we will obtain root(6)(a^3)=sqrt(a).
When we perform the operations with radicals, we often transform into fractional exponents. For example,
root(8)(x^3)*root(12)(x^7)=x^(3/8)*x^(7/12)=x^(3/8+7/12)=x^(23/24)=root(24)(x^23).
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# Problem: An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?
🤓 Based on our data, we think this question is relevant for Professor Caceres' class at TEXAS.
###### FREE Expert Solution
Lens equation:
$\overline{)\frac{\mathbf{1}}{{\mathbit{s}}_{\mathbit{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbit{s}}_{\mathbit{i}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbit{f}}}$
Magnification:
$\overline{){\mathbf{m}}{\mathbf{=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{o}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{s}}_{\mathbf{i}}}{{\mathbf{s}}_{\mathbf{o}}}}$
The separation between the object and the image is 2.4m
Therefore, so + si = 2.4
si = 2.4 - so
Substituting to the lens equation:
1/so + 1/(2.4 - so) = 1/(55 × 10-2)
###### Problem Details
An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?
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# How do you graph the parabola x^2 + 10x +8y +17 = 0 using vertex, intercepts and additional points?
Jul 21, 2018
Vertex: $\left(- 5 , 1\right)$, symmetry: $x = - 5$, y intercept: $\left(0 , - 2.125\right)$, x intercepts: $\left(- 7.83 , 0\right) \mathmr{and} \left(- 2.17 , 0\right)$, additional points : $\left(3 , - 7\right) \mathmr{and} - 13 , - 7$
#### Explanation:
${x}^{2} + 10 x + 8 y + 17 = 0 \mathmr{and} 8 y = - \left({x}^{2} + 10 x\right) - 17$ or
$y = - \frac{1}{8} \left({x}^{2} + 10 x\right) - \frac{17}{8}$ or
$y = - \frac{1}{8} \left({x}^{2} + 10 x + 25\right) + \frac{25}{8} - \frac{17}{8}$ or
$y = - \frac{1}{8} {\left(x + 5\right)}^{2} + 1$ This is vertex form of equation ,
y=a(x-h)^2+k ;(h,k) is vertex , $h = - 5 , k = 1 , a = - \frac{1}{8}$
Since $a$ is negative, parabola opens downward.
Therefore vertex is at $\left(- 5 , 1\right)$ Axis of symmetry is
x= h or x = -5 ; y-intercept is found by putting $x = 0$
in the equation ${0}^{2} + 10 \cdot 0 + 8 y + 17 = 0 \mathmr{and} 8 y = - 17$ or
$y = - \frac{17}{8} = - 2.125$ or at $\left(0 , - 2.125\right)$
x-intercepts are found by putting $y = 0$
in the equation ${x}^{2} + 10 x + 17 = - 8 y \mathmr{and} {x}^{2} + 10 x = - 17$ or
${x}^{2} + 10 x + 25 = 25 - 17 \mathmr{and} {\left(x + 5\right)}^{2} = 8 \mathmr{and} x + 5 = \pm \sqrt{8}$
$\therefore x = - 5 \pm \sqrt{8} \mathmr{and} x \approx - 7.83 , x \approx - 2.17$ , therefore ,
x-intercepts are at $\left(- 7.83 , 0\right) \mathmr{and} \left(- 2.17 , 0\right)$
Additional points: $x = 3 , y = - \frac{1}{8} {\left(3 + 5\right)}^{2} + 1$ or
$y = - \frac{1}{8} \cdot {8}^{2} + 1 \mathmr{and} y = - 7 \therefore \left(3 , - 7\right)$
$x = - 13 , y = - \frac{1}{8} {\left(- 13 + 5\right)}^{2} + 1$ or
$y = - \frac{1}{8} \cdot {\left(- 8\right)}^{2} + 1 \mathmr{and} y = - 7 \therefore \left(- 13 , - 7\right)$
graph{x^2+10x+8y+17=0 [-20, 20, -10, 10]}[Ans]
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# Parametrization of a curve
• DougUTPhy
In summary, the problem involves finding a parametrization of a curve that follows a corkscrew pattern, starting at (\sqrt{2\pi},0,0) and ending at (0,0,2\pi), along the surface defined by z= 2\pi- x^2- y^2. The attempt at a solution involved using x=r \cos t and y=r \sin t, but the problem lies in finding the correct parametrization for the curve.
## Homework Statement
I'm doing a line integral and can't seem to figure out the parametrization of this curve:
$x^2+y^2+z=2\pi$
## Homework Equations
Looking to get it to the form:
$\textbf{c}(r,t)=(x(r,t),y(r,t),z(r,t))$ (I don't even know if this is right though).
## The Attempt at a Solution
Trying to use $x=r \cos t$ and $y=r \sin t$ but I still can't get anywhere.
I have a feeling I'm totally in the wrong direction.
The $2\pi$ is killing me too!
Last edited:
Your basic problem is NOT the "$2\pi$". It is that $x^2+ y^2+ z= 2\pi$ does NOT define a line (or curve or path) in three dimensions. It can be written as $z= 2\pi- x^2- y^2$ which is a surface (specifically, a paraboloid). Essentially, given any x and y you can solve for z so this is a two dimensional figure, not one dimension.
Please tell us what the entire problem really is.
I realized this after thinking about for a while, the real parametrizaion I can't figure out is a curve that is a corkscrew getting narrower as it goes up around the parabolioid, starting at $(\sqrt{2\pi},0,0)$ and ending at the top of the paraboloid, $(0,0,2\pi)$
## 1. What is the purpose of parametrizing a curve?
The purpose of parametrizing a curve is to represent the curve as a set of equations in terms of a parameter. This allows for easier calculation and analysis of the curve's properties, such as its length, slope, and curvature.
## 2. How is a curve parametrized?
A curve can be parametrized by expressing its coordinates in terms of a parameter, typically denoted as t. This parameter can represent time, distance, or any other variable that helps define the curve.
## 3. What are the advantages of using a parametrized curve?
Using a parametrized curve allows for more flexibility in representing and analyzing the curve. It also simplifies the process of finding derivatives and integrals of the curve, as well as calculating arc length and surface area.
## 4. Can any curve be parametrized?
Yes, any curve can be parametrized as long as it is continuous and has a one-to-one correspondence between the parameter and the points on the curve. However, some curves may have more complex parametrizations than others.
## 5. How does parametrization affect the orientation of a curve?
The orientation of a curve can be affected by the direction of the parameter. If the parameter increases in the same direction as the curve, the orientation remains the same. However, if the parameter increases in the opposite direction, the orientation of the curve will be reversed.
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# Symbolic Logic - Disjunctions
### statement true account doe
Another word used in both ordinary English and in logic is "or." Someone who says, "Either he did not hear me, or he is being rude," is saying that at least one of those two possibilities is true. By connecting the two possibilities about which he or she is unsure, the speaker can make a statement of which he or she is sure.
In logic, "or" means "and/or." If p and q are statements, p V q is the statement, called a "disjunction," formed by connecting p and q with "or," symbolized by "V."
For example if p is the statement, "Mary Doe may draw money from this account," and q is the statement, "John Doe may draw money from this account," then p V q is the statement, "Mary Doe may draw money from this account, or John Doe may draw money from this account."
The disjunction p V q is true when p, q, or both are true. In the example above, for instance, an account set up in the name of Mary or John Doe may be drawn on by both while they are alive and by the survivor if one of them should die. Had their account been set up in the name Mary and John Doe, both of them would have to sign the withdrawal slip, and the death of either one would freeze the account. Bankers, who tend to be careful about money, use "and" and "or" as one does in logic.
One use of truth tables is to test the equivalence of two symbolic expressions. Two expressions such as p and q are equivalent if whenever one is true the other is true, and whenevet one is false the other is false. One can test the equivalence of ~(p V q) and ~pΛ ~q (as with the minus sign in algebra, "~" applies only to the statement which immediately follows it. If it is to apply to more than a single statement, parentheses must be used to indicate it):
p q ~p ~q pVq ~(pVq) ~pΛ~q T T F F T F F T F F T T F F F T T F T F F F F T T F T T
The expressions have the same truth values for all the possible values of p and q, and are therefore equivalent.
For instance, if p is the statement "x > 2" and q the statement "x < 2," p V q is true when x is any number except 2. Then (p V q) is true only when x = 2. The negations p and q are "x 2" and "x 2" respectively. The only number for which ~ p Λ ~ q is true is also 2.
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# How do you find the domain and range of g(x) = -11/(4 + x)?
Jul 17, 2018
The domain is $x \in \left(- \infty , 4\right) \cup \left(4 , + \infty\right)$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$
#### Explanation:
The function is
$f \left(x\right) = - \frac{11}{4 + x}$
The denominator must be $\ne 0$
Therefore,
$4 + x \ne 0$
$x \ne - 4$
The domain is $x \in \left(- \infty , 4\right) \cup \left(4 , + \infty\right)$
To find the domain, Let
$y = - \frac{11}{4 + x}$
$y \left(4 + x\right) = - 11$
$y x + 4 y = - 11$
$y x = - 11 - 4 y$
$x = \frac{- 11 - 4 y}{y}$
The denominator must be $\ne 0$
$y \ne 0$
The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$
graph{-11/(4+x) [-33.13, 18.18, -13.24, 12.43]}
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1. The price of a computer system at an electronics store is \$1099. Determine the percent of change in price if the computer system sold for \$1349 a month earlier. A. 22.7% increase B. 18.5% increase C. 22.7% decrease D. 18.5% decrease Hint 2. During the first half of the year there were 54 students in the school's bicycling club. In the second half of the year the club had 68 members. Find the percent of increase in club membership to the nearest whole percent. A. 26% B. 14% C. 21% D. 34% Hint 3. After changing its format, a radio station lost 34% of its listeners. If there were an estimated 83,000 listeners originally, how many listeners does the radio station have now? A. 125,758 B. 49,000 C. 54,780 D. 28,220 Hint 4. Al's Quick Mart sells milk for \$1.99 per gallon, but they used to sell it for \$2.79. What is the percent decrease in the price? A. 28.70% B. 40.20% C. 71.30% D. 39.70% Hint 5. The table shows how many people attended each football home game at a certain university. Which statement is supported by the information in the table? A. The percent of increase from the first to the last game is about 25%. B. Fewer people came to each game. C. The percent of decrease from the first to the last game is about 25%. D. The greatest percent of increase was from game 2 to game 3. Hint
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https://www.edplace.com/worksheet_info/maths/keystage1/year1/topic/89/4941/count-and-write:-0-10-1
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Count and Write: 0-10 (1)
In this worksheet, students count in the number range 0-10 and write the answer.
Key stage: KS 1
Curriculum topic: Number: Number and Place Value
Curriculum subtopic: Count to 100
Difficulty level:
QUESTION 1 of 10
Counting is something that we do everyday of our lives. For example, we have to count when we go shopping, when we cook and when we play hide and seek!
In this worksheet, you will have to count the number of objects that you see. You can use the number line to help you.
How many peppers are there?
How many butterflies are there?
How many pineapples are there?
How many peaches are there?
How many sheep are there?
How many melons are there?
How many apples are there?
How many butterflies are there?
How many cars are there?
Count the cars given below and write the number.
• Question 1
How many peppers are there?
7
EDDIE SAYS
There are 7 peppers.
Try using the number line to count how many peppers there are. Starting at 1, count each number for each car. You should reach seven peppers.
• Question 2
How many butterflies are there?
2
EDDIE SAYS
There are 2 butterflies.
• Question 3
How many pineapples are there?
5
EDDIE SAYS
There are five pineapples.
• Question 4
How many peaches are there?
3
EDDIE SAYS
There are 3 peaches.
• Question 5
How many sheep are there?
3
EDDIE SAYS
There are 3 sheep.
• Question 6
How many melons are there?
9
EDDIE SAYS
There are 9 watermelons.
• Question 7
How many apples are there?
8
EDDIE SAYS
There are 8 apples.
• Question 8
How many butterflies are there?
3
EDDIE SAYS
There are 3 butterflies.
• Question 9
How many cars are there?
6
EDDIE SAYS
There are 6 cars.
• Question 10
Count the cars given below and write the number.
16
EDDIE SAYS
There are 16 cars.
---- OR ----
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https://www.8822.in/01406504965.html
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Information about the 01406504965 number. | 8822.in
01406504965
My Own Way
Numbers Look-up:
Curiosities and FactsSite Maps
• 01406504965 seconds correspond to 44 years, 218 days, 23 hours, 49 minutes and 25seconds.
• Let's do a mathematical trick with the number 0650496:
Take the first three digits: 065.
Multiply it by 80 (065 * 80 = 5200)
Now add 1 (5200 + 1 = 5201)
Multiply by 250 (5201 x 250 = 1300250)
Add in the last four digits twice (1300250 + (0496 * 2) = 1301242)
Substract 250 (1301242 - 250 = 1300992)
Divide by 2 and... you've got the original number!: 650496
• Calculating the determinant associated to the number 01406504965:
| 1 4 0 | The determinant of the matrix: | 6 5 0 | = -114 | 4 9 6 |
• Someone born on 0/4/96 would be probably 23 years old.
People's age matters! People act and are targeted differently depending on their age. For instance: any insurance company in UK could cover you with this age. Low insurance quotes. At the same time, this kind of people is the most active in technology terms, and they usually have a cellphone. Mobile users under 30 use also mobile services like ringtone and game downloads.
• You could write that number using Roman numerals this way: XIV(014) LXV(065) CDXCVI(0496)
... or maybe this other way: XIV (014) LXV (065) IV (04) XCVI (96)
The Roman Empire did not have insurance companies, so romans didn't have life insurances or health insurances either. No insurance quotes to be paid or whatever. Undoubtely, those were better times... despite the existence of roman lawyers.
They didn't use cell phones either. Much quieter life. Communications was done though alive messengers, such as slaves and carrier pidgeons, writing down in parchment paper and conveniently sealed.
• The number converted to binary is: 1010011110101011001000000000101
• 01 + 40 + 65 + 04 + 96 = 206
2 + 0 + 6 = 8
8 + = 8
The Number Seven is a powerful source of magic and has been related with religions since the start of time. It talks about planning, research, and power.
• 01406504965 GB pounds is equal to 1637312429.7565eur with current conversion rate of 1.1641(GBP to eur)
• Read with me! one milliard and four hundred six million and five hundred four thousand nine hundred SIXTY-five DOLLARS
Numbers Cloud Who Called Me?
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| 3.671875 | 4 |
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https://sciforums.com/threads/dividing-a-number-by-zero.133926/page-4
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Dividing a number by zero
I would say that it is mathmatical proof that there is something wrong with claiming a=b. I never seen any algebra books that claimed that $$a^{2} = ab$$ from then on it is just nonsense.
When you substitute equations into other equations, then it makes the variable have to agree with all of those equations. So then by saying that a=b, and then factoring out a zero from both sides, the solution to "a" and "b" would then have to be zero. It is the only number that you could factor out a zero and not change its value. Everything multiplied by zero is zero, so if it had a zero multiplied in it and you factored that out, then that number would have to be zero.
I think this is why they say "a" does not equal "b", and they cannot be equal to zero. The algebra just wouldn't work out anymore, as shown. I think this is a good example of that. It is never said that "a" can equal "b" in any fundemental math principals.
It's simple.
A = B (FINE)
Then multiply both sides by A, so we get A^2 = AB (FINE)
Subtract B^2 from both sides, so we A^2 - B^2 = AB - B^2 (FINE)
Divide both sides by (A-B), so we get A + B = B (NOT FINE!!)
If a = b, then a[sup]2[/sup] - b[sup]2[/sup] = 0. The rest of emil's equations skip over this; the equation a + b = b is true if a and b are both 0.
Right, zero is the only number that can satisfy all of the equations. So then just by saying that a=b, and then putting them into a quadradic formula, makes the value of "a" and "b" become zero. So then in quadradic formula, "a" cannot equal "b" and they cannot be zero, zero has the ability to make equations like a+b=b become true when any other number wouldn't be able to satisfy the equation.
It's simple.
A = B (FINE)
Then multiply both sides by A, so we get A^2 = AB (FINE)
Subtract B^2 from both sides, so we A^2 - B^2 = AB - B^2 (FINE)
Divide both sides by (A-B), so we get A + B = B (NOT FINE!!)
That is not fine. Since, a=b then you cannot put it into a quadradic formula. If you did then $$a^{2} - b^{2}$$ would equal zero, and then $$ab - b^{2}$$ would equal zero. So then at that time you just have 0 = 0. You then factor a zero from both sides and divide by zero. If you wanted to find the solution to "a" and "b", you could have stopped right when you found $$a^{2} - b^{2} = ab - b^{2}$$, the solution is 0 = 0.
That is not fine. Since, a=b then you cannot put it into a quadradic formula. If you did then $$a^{2} - b^{2}$$ would equal zero, and then $$ab - b^{2}$$ would equal zero. So then at that time you just have 0 = 0. You then factor a zero from both sides and divide by zero. If you wanted to find the solution to "a" and "b", you could have stopped right when you found $$a^{2} - b^{2} = ab - b^{2}$$, the solution is 0 = 0.
...or 1=1, or 2=2, or 3=3, or -100=-100. It's almost like there exists a solution for every time the first number equals the second number!
The problem resides in the step that I pointed out. I don't know what you're talking about regarding the quadratic formula, etc.
...or 1=1, or 2=2, or 3=3, or -100=-100. It's almost like there exists a solution for every time the first number equals the second number!
The problem resides in the step that I pointed out. I don't know what you're talking about regarding the quadratic formula, etc.
I may have said quadratic formula by mistake, the point is that in the equations $$a^{2} - b^{2}$$ and $$ab - b^{2}$$, if "a" and "b" are equal to each other then they could be no other value other than zero. That is why "a" and "b" cannot be equal to each other in this kind of situation.
Think about it, one squared minus one squared would be zero. One times one minus one squared would be zero. Zero doesn't play by the same rules as other numbers, so then it could only lead to wrong answers. A variable that is a zero would not act the same as any other variable. Both sides of the equation would already be nothing, so anything more than that would just be working on nothing even more still. It would just be a lot to do about nothing. I think that is why they put those mathmatical rules into place, if a = b, then $$a^{2} - b^{2} = 0$$ and $$ab - b^{2} = 0$$. Then neither of them could really say anything meaningful about something else.
Those are not "equations" those are "expressions."
$$a^2 - b^2 \; = \; ab-b^2$$ is an equation which is true if a = b OR if a = 0. Either condition is sufficient.
Those are not "equations" those are "expressions."
$$a^2 - b^2 \; = \; ab-b^2$$ is an equation which is true if a = b OR if a = 0. Either condition is sufficient.
Ah yes, that was the word I was looking for, it has really been a long time since I heard that one. I don't know what you mean by, "either condition is sufficient". I would think that it is the only values that could satisfy those conditions, but zero and a=b is said to not be valid in either of these expressions.
Prof., you are wrong. There is no need for "a" to equal zero. "a" can equal any number. It is only in the (a-b)/(a-b) step that things become undefined.
Prof., you are wrong. There is no need for "a" to equal zero. "a" can equal any number. It is only in the (a-b)/(a-b) step that things become undefined.
"a" can equal any number, but like I said, zero and a=b are the only conditions where the equations would be valid, and those expressions are no longer valid when a=0 and b=0 and a=b.
They say any variables can be any number, but they can only be certain numbers in certain equations in order for them to be true. So all I did really is think what numbers the equations are true for, so then you can know what the value of "a" and "b" actually are even without constants in this case. The equations simply do not allow "a" and "b" to be any number when they are put together in this fashion. They could have been any number when it was said $$a^{2} = ab$$ and a = b, but then when it was put into the next expressions, the values of "a" and "b" could only be zero. The expression would limit the values of "a" and "b" since $$a^{2} - b^{2} = 0$$ and $$ab - b^{2} = 0$$, so then from then on "a" and "b" would have to be zero. They could no longer be any number once these expressions where set equal to each other.
Remember the difference between an expression and an equation.
This is an expression:
$$a^2 + b^2$$
This expression can be evaluated for all values of a and b.
This is an equation. It equates two expressions:
$$a^2 - b^2 \; = \; 0$$
It is true for some values of a and b. In this case, for any value of a, there are two values of b that make the equation true.
They could have been any number when it was said $$a^{2} = ab$$ and a = b, but then when it was put into the next expressions, the values of "a" and "b" could only be zero.
No. The next equation is:
$$a^2 - b^2 \; = \; ab-b^2$$
In this equation, a and b can be anything, as long as a=b.
(considered in isolation, the equation is also true for all values of b when a=0, but it's already given that a=b.
Layman said:
The expression would limit the values of "a" and "b" since $$a^2 - b^2 \; = \; 0$$ and $$ab-b^2 \; = \; 0$$, so then from then on "a" and "b" would have to be zero.
The two expressions both evaluate to zero, but that does not imply that a=0 or b=0.
The two expressions both evaluate to zero, but that does not imply that a=0 or b=0.
It does if you factor it, you would be saying that zero times whatever is left is equal to zero. You could only take a zero out of zero, so then it couldn't be any number at that point. You wouldn't factor out a zero from one and say that it is still one. If you did anything with it from that point "a" and "b" would have to equal zero. The entire expression is equal to zero, so then any other operation would force it to become zero from that condition, since the expression itself is equal to zero.
Otherwise it would be like saying that you could factor zero out of a whole number, that is something that you cannot do.
Layman,
One of the factors of the expressions on both sides of the equation is (a-b), which we know is zero.
Right?
And I'm sure you'll agree that if a=10 and b=10 (for example), then (a-b) is still zero.
Right?
And this will make both sides of the equation equal to zero.
Right?
I'm not sure exactly where your confusion is, but perhaps you'll see if you plug some numbers in.
Plug in a=10 and b=10 into $$a^2-b^2 = ab-b^2$$.
You'll end up with 0=0, which I'm sure you'll agree is a valid equation.
I missed this from yesterday.
$$\lim_{a \to 0} \frac{0}{a} = 0$$
$$\lim_{a \to 0} \frac{a}{0} = undefined$$
$$\lim_{a \to 0} \frac{2a}{a} = 2$$
In the first example, I think 0 divided by any number is 0, so then the answer to that one would be zero. This seems to show that the zero comes from a different variable than "a", so then it would seem that zero's coming from different sources cannot be canceled.
That's correct. This is why it's important to be clear what the limit expression is.
You can't just say "The limit of 0/0" because it doesn't tell you where those zeros came from.
In the secound example, any number divided by zero is undefined, so then I don't think it would have an answer since the limit would never say "a" is actually 0. That does seem odd that you would get a different answer in this limit even though it seems to really be saying that it is searching for the limit of the same division problem $$\frac {0}{0}$$. In this case it is undefined and in the other it is 1. But since this example is undefined, I think the real solution for $$\frac {0}{0}$$ would be 1. The problem has only been rearranged so that it is undefined. In the example $$\lim_{a \to 0} \frac {a}{a}$$ you never actually have to divide by zero.
The point is that different limit expressions that approach 0/0 have different answers. There is no single answer to "The limit of 0/0".
In the third, any number divided by itself would be one, so every answer would then be 2. It is as simple as canceling out the "a" in the equation even if "a" approuches zero. You shouldn't allow yourself to think that just because "a" approaches zero, that it would give you a different answer rather than just canceling them out, you can cancel $$\frac {0}{0}$$ in limits and that should always be 1
No, this example show that approaching a limit of 0/0 does not always cancel to 1. You need to know how the numerator and denominator approach zero. "The limit of 0/0" is just not clearly defined enough to give it a definite value. You need to spell out the limit expression in full. No shortcuts.
What happens if you divide infinity by zero?
What happens if you divide infinity by zero?
:spank: Division by zero is forbidden!
it appears if you try to divide infinity by 0 it will still be infinity since there's nothing ( 0 ) to divide by
What happens if you divide infinity by zero?
$$\frac{0}{8}=0 \frac{8}{0}=8 \frac{0}{0}=R \frac{8}{8}=R (0)(8)=R$$
where R is some real number. I may be wrong though.
I used 8 in place of ∞.
Ha, pretty funny.
I'm not sure exactly where your confusion is, but perhaps you'll see if you plug some numbers in.
Plug in a=10 and b=10 into $$a^2-b^2 = ab-b^2$$.
You'll end up with 0=0, which I'm sure you'll agree is a valid equation.
I see where I got you confused now. So then it would follow that, $$(10 + 10 ) (10 - 10 ) = 10 ( 10 - 10 )$$ , by doing this would have factored out a zero out of each side of the equation. It would be the same as saying $$(10 + 10) 0 = 10 ( 0 )$$, but you cannot factor out a zero from a 10! So then "a" and "b" can no longer be 10, because the only time this would be true is when "a" and "b" is equal to zero. So then you would have $$(0 + 0 ) (0 - 0 ) = 0 ( 0 - 0 )$$, the value of "a" and "b" being zero is the only number that would be true by factoring out a zero.
The point here I am trying to make as that the mathmatical rules that say "a" is not equal to "b", and "a" and "b" are not equal to zero are in the right place where you would end up having these expressions. Not sooner, or later, when you have those expression they are no longer valid. Any type of multiplication or division or factoring would force them to become zero, and they wouldn't work for any other number other than zero.
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| 4.375 | 4 |
CC-MAIN-2024-26
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http://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-r-section-r-7-rational-expressions-r-7-assess-your-understanding-page-70/13
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## College Algebra (10th Edition)
$\dfrac{y+5}{2(y+1)}; y\ne-1, 5$
Factor the numerator and the denominator completely to obtain: $=\dfrac{(y-5)(y+5)}{2(y^2-4y-5)} \\=\dfrac{(y-5)(y+5)}{2(y-5)(y+1)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{\cancel{(y-5)}(y+5)}{2\cancel{(y-5)}(y+1)} \\=\dfrac{y+5}{2(y+1)}; y\ne-1, 5$ ($y$ cannot be $-1$ and $5$ because they will make the original expression undefined.)
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| 4.0625 | 4 |
CC-MAIN-2018-17
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latest
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| 0.702295 |
https://setscholars.net/year-six-math-worksheet-for-kids-understanding-and-solving-one-variable-inequalities/
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text/html
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crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00118.warc.gz
| 488,490,301 | 22,989 |
# Year Six Mathematics Worksheets
Mathematics is a crucial subject that provides children with the foundation for critical thinking and problem-solving skills. Understanding the basics of mathematics from an early age is important for success in the subject and in life. One of the important concepts in mathematics is the understanding and solving of one-variable inequalities.
An inequality is a mathematical expression that shows the relationship between two values, indicating that one value is either greater than, less than, or equal to the other value. For example, the inequality “x < 5” means that the value of x is less than 5. In one-variable inequalities, there is only one variable in the expression.
Solving one-variable inequalities involves finding the values of the variable that make the inequality true. For example, if we have the inequality “x < 5,” we want to find the values of x that are less than 5. The solution to the inequality is called the solution set and is represented by a range of values that satisfy the inequality.
There are several ways to solve one-variable inequalities. One common method is to isolate the variable on one side of the inequality by using the properties of inequalities, such as adding or subtracting the same value from both sides, multiplying or dividing both sides by the same value, or multiplying or dividing both sides by a negative number. Once the variable is isolated, the solution set can be found by testing values of the variable and checking if they make the inequality true.
It is important to remember that when solving one-variable inequalities, the direction of the inequality symbol may change when multiplying or dividing both sides by a negative value. For example, if we have the inequality “x < 5,” and we multiply both sides by -1, the inequality becomes “x > -5.”
Math worksheets for kids can be a great tool for practicing the understanding and solving of one-variable inequalities. These worksheets provide kids with a variety of problems to solve and allow them to check their answers and get feedback on their work. This can help them understand the concept better and improve their skills.
In conclusion, understanding and solving one-variable inequalities is an important concept in mathematics that helps us describe relationships between values. By solving one-variable inequalities, we can find the solution set of values that satisfy the inequality. Math worksheets for kids are a great tool for practicing this concept and improving skills.
# Year Six Math Worksheet for Kids – Understanding and Solving One-Variable Inequalities
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MAT 116 UOP Course Tutorial/Shoptutorial
Posted by admin on December 4, 2017 in Articles
MAT 116 Entire CourseFor more course tutorials visit
www.shoptutorial.comMAT 116 Week 1 Quiz (New)
MAT 116 Week 2 Quiz (New)
MAT 116 Week 3 Quiz (New)
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MAT 116 Week 5 Quiz (New)
MAT 116 Week 6 Quiz (New)
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MAT 116 Final Exam **New** (3 Sets)————————————————————————
MAT 116 Final Exam (New) (3 Sets)For more course tutorials visit
www.shoptutorial.comThis tutorial contains 3 sets of finals with different values
1. Calculate the sum of 1/7 and 1/3.
1. Calculate the product of 1/3 and 1/4.
1. Select the expression that represents the statement “22 more than a number”.
1. Determine the value of 5(x+7)-3 when x=9.
1. Determine the value of x(8-x)+9 when x= -4.
1. Select the expression that represents the statement “3 times than a number”.
1. There are 16 ounces in a pound. How many ounces are in 7 pounds
1. In a given training week, a swimmer completes 1/4 mile on day one, 2/5 mile on day two, and 3/7 mile on day three. How many total miles does the athlete swim in the given week
1. Select the simplified expression for 8(x-6)-6(x-1) +9x.
1. The temperature on one day in January for Phoenix, AZ, is 52 degrees Fahrenheit. On the same day in Barrow, AK, the temperature is -17 degrees Fahrenheit. What is the difference between these two temperatures
1. Solve 8x+6=1.
1. Select the expression that is equivalent to -2(x-4).
1. A clothing store offers a 60% discount on all purchases. What is the discount in dollars (before tax)for an item originally costing \$ 20.00 dollars
1. A taxi company charges passengers \$2.40 for the first mile and \$1.20 per mile for each additional mile .How much would the bill is for a 15 mile trip
———————————————————————————-MAT 116 Week 1 Quiz (New)For more course tutorials visit…
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This means that the binomial expansion will consist of terms related to odd numbers. This is called the general term, because by giving different values to r we can determine all terms of the expansion. Question: 3. b) Hence, deduce an expression in terms of a and b for a + b 4 + a - b 4 . Let us now look at the most frequently used terms with the binomial theorem. The binomial theorem states a formula for expressing the powers of sums. There are various important terms such as general term, middle, term, etc.
For Example (a+b)5 is a binomial. Let's say if you expand (x+y), therefore, the middle term results in the form the (2 / 2 + 1) which is equal to 2nd term. ( x + y) 0 = 1 ( x + y) 1 = x + y ( x + y) 2 = x 2 + 2 x y + y 2. and we can easily expand. It is usually represented as Tr+1. This calculators lets you calculate expansion (also: series) of a binomial.
By the binomial formula, when the number of terms is even, then coefficients of each two terms that are at the same distance from the middle of the terms are the same.
FREE Cuemath material for JEE,CBSE, ICSE for excellent results! T r + 1 = n C r x r. In the binomial expansion of ( 1 - x) n . T r+1 = general term = n C r a n-r b r . a. (Sec3 A Math) Binomial Theorem - The General Term ~ Solutions 6= @ 6 0 A()6(20+ @6 1 A)61(2)1+ @6 2 A()62(2)2) + @ 6 3 . The general term of a binomial expansion, also known as the (r+1)th term. (x +a)n = n k=0nCkxnkak ( x + a) n = k = 0 n n C k x n k a k. The First term would be = nC0xna0 n C 0 x n a 0.
The binomial theorem (or binomial expansion) is a result of expanding the powers of binomials or sums of two terms. General Term The general term in the binomial theorem can be referred to as a generic equation for any given term, which will correspond to that specific term if we insert the necessary values in that equation. What is General and middle term in a binomial expansion. Binomial Theorem For Rational Indices in Binomial Theorem with concepts, examples and solutions. The binomial theorem formula is (a+b) n = nr=0n C r a n-r b r, where n is a positive integer and a, b are real numbers, and 0 < r n. This formula helps to expand the binomial expressions such as (x + a) 10, (2x + 5) 3, (x - (1/x)) 4, and so on. Middle term of the expansion is , ( n 2 + 1) t h t e r m. When n is odd.
Properties of the Binomial Expansion (a + b)n There are This gives rise to several familiar Maclaurin series with numerous applications in calculus and other areas of mathematics. 3.
Taking the general term (4.5), we show that the left-hand side equals: [4.6] And the right-hand side is the Binomial Theorem! In this case ( n + 1 2) t h t e r m term and ( n + 3 2) t h t e r m are the middle terms.
The Third term would be = nC2xn2a2 n C 2 x n 2 a 2. (n2 + 1)th term is also represented . The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7:42.
When any term in any binomial expansion is to be found, the General Term must be used. . Any algebraic expression consisting of only two terms is known as a Binomial expression. There are (n + 1) terms in the expansion of , i.e., one more than the index; In the successive terms of the expansion the index of a goes on decreasing by unity. Here you will learn formula to find the general term in binomial expansion with examples.
. ( x + 3) 5. (ii) . To learn more about this interesting concept, watch our videos.
Lecture 3||Binomial Theorem 11||General Term and Middle Terms|| Exercise 8.2||Q1,Q2,Q4, Q6, Q7, Q9, Q10 (n2 + 1)th term is also represented . Each entry is the sum of the two above it. 7. a) Use the binomial theorem to expand a + b 4 . We do not need to fully expand a binomial to find a single specific term. Question: Binomial Theorem 1. a) Expand using Binomial theorem and write the general term of (2x-3) b) For the binomial ( x2 - 3)8 , Find i)The middle term ii)Term independent of x iii)Term containing x10 c) Use the binomial theorem to expand and simplify (3m2 )4 m d) Answer the following questions for the expansion of ( 2+kx). Binomial theorem for any index. (10pts - Binomial Theorem) Suppose that 90% of adults own a car. Fortunately, the Binomial Theorem gives us the expansion for any positive integer power . 3. So, counting from 0 to 6, the Binomial Theorem gives me these seven terms: (1) 3.
is called the binomial theorem. For the expansion $\left( 2x - {1 \over x} \right)^{10}$, find the coefficient of the term with ${1 \over x^$}\$ General Term of Binomial Expansion The General Term of Binomial Expansion of (x + y) n is as follows T r+1 is the General Term in the binomial expansion The General term expansion is used to find the terms mentioned in the above formula. Terms related to Binomial Theorem. The general term formula allows you to find a specific term inside a binomial expansion without the need to fully expand. Therefore, = = . In any term the sum of the indices (exponents) of ' a' and 'b' is equal to n (i.e., the power of the binomial). Let's say if you expand (x+y), therefore, the middle term results in the form the (2 / 2 + 1) which is equal to 2nd term. From the above pattern of the successive terms, we can say that the (r + 1) th term is also called the general term of the expansion (a + b) n and is denoted by T r+1.
General Term - Defination. ( 1.1)^(10000) is a larger .
Features of Binomial Theorem 1. Well, this is done using an interesting concept known as 'Binomial theorem'.
2. Factorial: This is discussed in finding factorial of a number in Java post. Binomial theorem The binomial coefficient appears as the k th entry in the n th row of Pascal's triangle (counting starts at 0 ). Each term in the sum will look like that -- the first term having k = 0; then k = 1, k = 2, and so on, up to k = n. Notice that the sum of the exponents (n k) + k, always equals n. (2) If n Middle Term in Binomial Expansion Read More
This is equal to choose multiplied by to the power of minus multiplied by to the power of . So, starting from left, the . Binomial Theorem Using the Binomial Theorem to Find a Single Term Expanding a binomial with a high exponent such as {\left (x+2y\right)}^ {16} (x+ 2y)16 can be a lengthy process. what holidays is belk closed; Powers of x and y in the general term: The index (power) of x in the general term is equal to the difference between the superscript n and the subscript r. The coefficients 1, 2, 1 that appear in this expansion are parallel to the 2nd row of Pascal's triangle. Let us now look at the most frequently used terms with the binomial theorem. It is n in the first term, n -1) in the second term, and so on ending with zero in the last term. Question 2. i) Find the general term in the expansion of (x + y) n.
Middle Terms in Binomial Expansion: When n is even. The Binomial Theorem is commonly stated in a way that works well for positive integer exponents. And what do we mean by this? The coefficients occuring in the binomial theorem are known as binomial coefficients. ( x + y) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3. Middle term in the expansion of (1 + x) 4 and (1 + x) 5. In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes-no question, and each with its own Boolean-valued outcome: success (with probability p) or failure (with probability q = 1 p).A single success/failure experiment is also .
In this condition, the middle term of binomial theorem formula will be equal to (n / 2 + 1)th term.
( x 2 + 2) k = m = 0 k 2 k m x 2 m ( k m) Hence you get a double sum in which the power of x is 2 m + k 7, setting this equal to 8 we get k = 15 2 m. This leaves this single sum over m. m = 0 7 2 15 3 m ( 7 15 2 m) ( 15 2 m m) Since, for n, m = 0, 1, 2,. the binomial coefficient ( n m) is zero . It shows how to calculate the coefficients in the expansion of ( a + b) n. The symbol for a binomial coefficient is . Using binomial theorem indicate which number is larger (1.1)^(10000) or 1000 Answer: :. The primary example of the binomial theorem is the formula for the square of x+y. Statistics and Probability questions and answers. In this condition, the middle term of binomial theorem formula will be equal to (n / 2 + 1)th term. a+b is a binomial (the two terms are a and b) Let us multiply a+b by itself using Polynomial . The first term in the binomial is "x 2", the second term in "3", and the power n for this expansion is 6. Solution. For example, when n = 5, each term in the expansion of ( a + b) 5 will look like . The following points analyze the significant terms related to the binomial theorem: The term which helps to represent or . Get Binomial Theorem Formulae Cheat Sheet & Tables. .
The Binomial Theorem - HMC Calculus Tutorial. (b) Given that the coefficient of 1 x is 70 000, find the value of d . Write down and simplify the general term in the binomial expansion of 2 x 2 - d x 3 7 , where d is a constant. xn-r . Find the coefficient of specific term. How to deal with negative and fractional exponents. The binomial theorem provides a simple method for determining the coefficients of each term in the series expansion of a binomial with the general form (A + B) n. A series expansion or Taylor series is a sum of terms, possibly an infinite number of terms, that equals a simpler function. The binomial theorem formula helps in the expansion of a binomial raised to a certain power. Binomial Theorem. (IMP-2013) ii) Find the middle term in the above expansion. The coefficients in the expansion follow a certain pattern known as pascal's triangle. We have therefore proved the Binomial Theorem for all real numbers, so we can legitimately use it with positive and negative and fractional r, and we are no longer limited to integers. Expand (4 + 2x) 6 in ascending powers of x up to the term in x 3.
The general term in the expansion of ( x + y ) n is n . Example 1. Second step.
So, counting from 0 to 6, the Binomial Theorem gives me these seven terms: In any term in the expansion, the sum of powers of \ (a\) and \ (b\) is equal to \ (n\). combinatorial proof of binomial theoremjameel disu biography. . Here, x = 1, y = a and n = 8. 7.0 k+. Find n.
Notice the following pattern: In general, the kth term of any binomial expansion can be expressed as follows: Example 2. Binomial Expression A binomial expression is an algebraic expression that contains two dissimilar terms such as a + b, a + b, etc. Use the binomial theorem to express ( x + y) 7 in expanded form. General Term in Binomial Theorem means any term that may be required to be found. \ (n\) is a positive integer and is always greater than \ (r\).
The result is in its most simplified form. . The binomial theorem provides us with a general formula for expanding binomials raised to arbitrarily large powers. The first term in the binomial is "x 2", the second term in "3", and the power n for this expansion is 6. Since n = 13 and k = 10, n. n n can be generalized to negative integer exponents. That is, there are an infinite number of terms in the expansion with the general term given by ${T_{r + 1}} = \frac{{n(n - 1)(n - 2)(n - r + 1)}}{{r! The coefficients of the terms in the expansion are the binomial coefficients (n k) \binom{n}{k} (k n ). Terms. The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus, and . In a sample of eight adults, what is the probability that exactly six adults will own a car? In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. This means that the binomial expansion will consist of terms related to odd numbers. (4x+y) (4x+y) out seven times. The terms in the above expansion become smaller and smaller. Terjemahan frasa DUA BINOMIAL dari bahasa indonesia ke bahasa inggris dan contoh penggunaan "DUA BINOMIAL" dalam kalimat dengan terjemahannya: Jumlah dari dua binomial adalah 5 x kuadrat minus. The upper index n is the exponent of the expansion; the lower index k indicates which term, starting with k = 0. Solution. Binomial theorem for any index. 1339560. Tr+1=Crn . As mentioned above, the binomial theorem is a type of theorem which helps to calculate or find the exponential value of an algebraic expression. The binomial theorem for positive integer exponents. 19:12. Chapter 8 Class 11 Binomial Theorem; Concept wise; General Term - Defination; Check sibling questions . The general term in the binomial expansion of plus to the th power is denoted by sub plus one. Binomial Theorem | General Term And Cofficient Of X^R. Therefore, substitute r = 4 in the binomial coefficient of the general term and evaluate. OnlineCalculator.Guru. Don't worry . Expansion of (1 + x) 4 has 5 terms, so third term is the . T r + 1 = ( 1) r n C r x n - r a r. In the binomial expansion of ( 1 + x) n, we have. It would take quite a long time to multiply the binomial. Evaluate (101)4 using the binomial theorem; Using the binomial theorem, show that 6n-5n always leaves remainder 1 when divided by 25. Answer: i) 11. 4. The general term in the expansion of (x + y) n is. A binomial is a polynomial that has two terms. Using binomial theorem ,Evaluate each of the following (99)^(5) Answer: 9509900499 View Text Solution 10. The most succinct version of this formula is shown immediately below. The term that has the fourth power of the variable a will be the fourth term in the expansion. Introducing your new favourite teacher - Teachoo Black, at only 83 per month. Every term in a binomial expansion is linked with a numeric value which is termed a coefficient. yr The Fourth term would be = nC3xn3a3 n C 3 x n 3 a 3. }}{x^r}$ We will now summarize the key points from this video. Let's begin - Middle Term in Binomial Expansion Since the binomial expansion of $$(x + a)^n$$ contains (n + 1) terms. Since r is not an integer, there is no independent term in this expansion, every term is dependent on x. A binomial expression that has been raised to a very large power can be easily calculated with the help of the Binomial Theorem. So the general term containing exponents of the form x^a will have the form COMB . We can test this by manually multiplying ( a + b ). To find the terms in the binomial expansion we need to expand the given expansion. A General Binomial Theorem. According to the theorem, it is possible to expand the power. Sometimes we are interested only in a certain term of a binomial expansion. Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem for (a+b)n : What follows the summation sign is the general term. For higher powers, the expansion gets very tedious by hand! Find the tenth term of the expansion ( x + y) 13. The expansion of (A + B) n given by the binomial theorem . In this example, a = 3x, b = - y, and n = 7. Example: * \$$(a+b)^n \$$ *
Binomial Theorem Formulas makes it easy for you to find the Expansion of Binomial Expression quickly. As we have seen, multiplication can be time-consuming or even not possible in some cases. This formula is known as the binomial theorem. General Term in Binomial Expansion. Binomial Theorem The theorem is called binomial because it is concerned with a sum of two numbers (bi means two) raised to a power. This formula is used to find the specific terms, such as the term independent of x or y in the binomial expansions of (x + y) n. This means use the Binomial theorem to expand the terms in the brackets, but only go as high as x 3. f ( x) = ( 1 + x) 3. f (x) = (1+x)^ {-3} f (x) = (1+x)3 is not a polynomial. It's expansion in power of x is known as the binomial expansion. 3. From the above formula, we have = To find the fourth term, , r = 3.
Use the binomial theorem to determine the general term of the expansion. The binomial theorem states the principle for extending the algebraic expression $$(x+y)^{n}$$ and expresses it as a summation of the terms including the individual exponents of variables x and y. Let's consider the properties of a binomial expansion first. If first term is not 1, then make first term unity in the following way, General term : Some important expansions. We use n =3 to best . (ii) In the successive terms of the expansion, the index of the first term is n and it goes on decreasing by unity. Equation 1: Statement of the Binomial Theorem. The Second term would be = nC1xn1a1 n C 1 x n 1 a 1. About . . e.g. This chapter covers topics like Binomial Theorem for Any Index, Binomial Theorem for Positive Integral Index, General Term, Middle Term and Greatest Term in Binomial Expansion, Multinomial Theorem, and Properties of Binomial Coefficients. Instead, I need to start my answer by plugging the binomial's two terms, along with the exterior power, into the Binomial Theorem.
The Binomial Theorem. The Binomial Theorem We use the binomial theorem to help us expand binomials to any given power without direct multiplication.
As a footnote it is worth mentioning that around 1665 Sir Isaac Newton came up with a "general" version of the formula that is not . The Binomial Theorem states that, where n is a positive integer: (a + b) n = a n + (n C 1)a n-1 b + (n C 2)a n-2 b 2 + + (n C n-1)ab n-1 + b n. Example. 2. Thus the general type of a binomial is a + b , x - 2 , 3x + 4 etc.
In an expansion of \ ( (a + b)^n\), there are \ ( (n + 1)\) terms. Here you will learn formula to find middle term in binomial expansion with examples.
Problems on approximation by the binomial theorem : We have, If x is small compared with 1, we find that the values of x 2, x 3, x 4, .. become smaller and smaller. Binomial Theorem Notes Class 11 Maths Chapter 8. Therefore, (1) If n is even, then $${n\over 2} + 1$$ th term is the middle term. 1339532. Lesson Explainer: General Term in the Binomial Theorem Mathematics In this explainer, we will learn how to find a specific term inside a binomial expansion and find the relation between two consecutive terms. Q8. Using Binomial theorem, expand (a + 1/b)11. The general term is also called as r th term. (10pts - Binomial Theorem) Suppose that 90% of adults own a car. Trolos doss 1o Eggo lo Sylioaxo Jog. We can expand the expression. 500+ 1.9 k+. e.g. it will all be explained! That pattern is summed up by the Binomial Theorem: The Binomial Theorem. Instead, I need to start my answer by plugging the binomial's two terms, along with the exterior power, into the Binomial Theorem. Write the general term in the expansion of (a2 - b )6. The Binomial Theorem is used in expanding an expression raised to any finite power. term, 1 in the second term and 2 in the third term and so on, ending with n in the last term. General Term; The general term in the binomial theorem can be referred to as a generic equation for any given term, which will correspond to that specific term if we insert the necessary values in that equation. The binomial theorem states that any non-negative power of binomial (x + y) n can be expanded into a summation of the form , where n is an integer and each n is a positive integer known as a binomial coefficient.Each term in a binomial expansion is assigned a numerical value known as a coefficient. Binomial Theorem Expansion According to the theorem, we can expand the power (x + y) n
Plus One Maths Binomial Theorem 3 Marks Important Questions. Binomial Theorem - As the power increases the expansion becomes lengthy and tedious to calculate. For example, when n =3: Equation 2: The Binomial Theorem as applied to n=3. \left (x+3\right)^5 (x+3)5 using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer. Where the sum involves more than two numbers, the theorem is called the Multi-nomial Theorem. Let's begin - General Term in Binomial Expansion We have, ( x + a) n = n C 0 x n a 0 + n C 1 x n - 1 a 1 + + n C r x n - r a r + + n C n x 0 a n We find that Read More Deductions of Binomial Theorem 8. North East Kingdom's Best Variety super motherload guide; middle school recess pros and cons; caribbean club grand cayman for sale; dr phil wilderness therapy; adewale ogunleye family. A polynomial consisting of two terms is termed as Binomial. General term (r + 1) th terms is called general term T r+1 = n C r x n-r a r. 3. A binomial theorem is a powerful tool of expansion, which is widely used in Algebra, probability, etc. Lecture 3||Binomial Theorem 11||General Term and Middle Terms|| Exercise 8.2||Q1,Q2,Q4, Q6, Q7, Q9, Q10 Ex 8.2, 3 - Chapter 8 Class 11 Binomial Theorem (Deleted) Last updated at Jan. 29, 2020 by Teachoo. Some observations : (i) Number of terms in binomial expansion = Index of the binomial + 1 = n + 1. The General Term: The general term formula is ( ( nC r)* (x^ ( n-r ))* (a^ r )). (i) a + x (ii) a 2 + 1/x 2 (iii) 4x 6y Binomial Theorem Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem. .
483627223. As per his theorem, the general term in the expansion of (x + y) n can be expressed in the form of px q y r, where q and r are the non-negative integers and also satisfies q + r = n. Here, ' p ' is called as the binomial coefficient. Exponents of (a+b) Now on to the binomial. 1. = - 2835 Hence, the fourth term in the expansion of = - 2835 3.3 k+. = . In this way we can calculate the general term in binomial theorem in Java. Binomial Theorem | General Term And Cofficient Of X^R. Chapter 8 BINOMIAL THEOREM Binomial Theorem 3.1 Introduction: An algebraic expression containing two terms is called a binomial expression, Bi means two and nom means term. The Binomial Theorem explains how to raise a binomial to certain non-negative power. 05:09. Question 1. i) The number of terms in the expansion of is _____.
n. n n. The formula is as follows: ( a b) n = k = 0 n ( n k) a n k b k = ( n 0) a n ( n 1) a n 1 b + ( n 2) a n 2 b .
The binomial theorem provides a short cut, or a formula that yields the expanded form of this expression. Consecutive terms in a binomial expansion are . Using binomial theorem ,Evaluate each of the following (101)^(4) Answer: 104060401 View Text Solution 9. 2. General and Middle Term Binomial Expansion for Positive Integral Index Example 1 Find the fourth term in the expansion of . (Because the top "1" of the triangle is row: 0) The coefficients of higher powers of x + y on the other hand correspond to the triangle's lower rows: In the binomial expansion of ( x - a) n, the general term is given by. 3. The Binomial Theorem was first discovered by Sir Isaac Newton. Solution. Use the binomial theorem to find the 18th term in the binomial expansion (2x - y square root 2)^2. We know that. (x+y)^n (x +y)n. into a sum involving terms of the form.
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Home >> AIMS >> Class11 >> Math >> Sets
# In a group of 760 persons 510 can speak Hindi, 360 can speak English then find i) how many can speak both Hindi and English ii) how many can speak Hindi only; iii) how many can speak English only.
(i) 110
(ii) 400
(iii) 250
Let OH be "only hindi" speaking and OE be "only english speaking".
Then 510 - OH is people who speak hindi + english.
Similarly 360 - OE speak hindi + english
Hence 510 - OH = 360 - OE
So OH - OE = 150
Also OH + OE + ( 510 - OH ) = 760 ( since 'only hindi' + 'only english' + 'hindi & english' is the total persons )
Hence OE = 250
Hence OH = 150 + 250 = 400
So persons who speak English & Hindi = 360 - OE = 360 - 250 = 110
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auxiliar10_MA1001_2012.pdf - Universidad de Chile Facultad De Ciencias F\u00b4\u0131sicas y Matem\u00b4aticas Prof Ra\u00b4 ul Uribe Prof Aux Braulio S\u00b4anchez Ib\u00b4an
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Universidad de Chile Facultad De Ciencias F´ ısicas y Matem´ aticas Prof.: Ra´ul Uribe Prof. Aux.: Braulio S´ anchez Ib´ a˜nez Introducci´on al C´ alculo Clase Auxiliar 10 - Sucesiones II 31 de mayo de 2012 Problema 1 [Sucesiones recursivas].- (a) Considere la sucesi´ on definida mediante la recurrencia a n +1 = a n 2 ( 1 + a 2 n - 1 ) , con a 0 , a 1 (0 , 1) (i) Demuestre que n N , a n (0 , 1) (ii) Muestre que ( a n ) es convergente. (iii) Calcule el l´ ımite de la sucesi´on. (b) Considere los reales 0 < a < b . Se definen las sucesiones ( x n ), ( y n ) mediante las siguientes recurrencias x 1 = a , x n +1 = x n y n y 1 = b , y n +1 = x n + y n 2 Demuestre que ambas sucesiones convergen, que poseen el mismo l´ ımite l y que ab < l < a + b 2 (c) Considere las sucesiones reales definidas por x n = n X k =0 1 k ! y n = x n + 1 n · n ! (i) Pruebe que ( x n ) e ( y n ) son sucesiones mon´ otonas. (ii) Demuestre que ambas sucesiones convergen, que lo hacen al mismo l´ ımite l , que n N * , x n < l < y n y que 2 , 5 < l < 2 , 75.
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# Conditional distribution of multivariate normal distribution
I'm doing some self studying this, I got stuck on this question:
Suppose $$(X,Y,Z)'$$ is normal with density $$\begin{equation*} \begin{split} C\cdot \text{exp}\{-\frac{1}{2}(4x^{2}+3y^{2}+5z^{2}+2xy+6xz+4zy)\}, \end{split} \end{equation*}$$ where $$C$$ is a normalizing constant. Determine the conditional distribution of $$X$$ given that $$X+Z=1$$ and $$Y+Z$$=0.
This is what I tried:
$$\begin{equation*} \begin{split} \mathbf{\mu}=\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right]\qquad \Lambda^{-1}=\left[\begin{array}{ccc} 4& 1& 3 \\ 1& 3& 2\\ 3& 2& 5 \end{array}\right],\qquad \\ \begin{array}{c} U=X \\ V=X+Z \\ W=Y+Z \end{array} \\ \mathbf{B}=\left[\begin{array}{c} U \\ V\\ W \end{array}\right]=\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \mathbf{B\mu}=\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right]\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right] =\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \Lambda=\frac{1}{|\mathrm{det} \Lambda^{-1}|} \left[\begin{array}{ccc} \mathrm{det} \left|\begin{array}{cc} 3 &2 \\ 2 &5 \end{array}\right|=11 &\mathrm{det}\left|\begin{array}{cc} 1 &2 \\ 3 &5 \end{array}\right|=(-1)& \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 3 &2 \end{array}\right|=(-7) \\ \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 2 &5 \end{array}\right|=(-1) & \mathrm{det}\left|\begin{array}{cc} 4 &3 \\ 3 &5 \end{array}\right|=11 &\mathrm{det}\left|\begin{array}{cc} 4 &1 \\ 3 &2 \end{array}\right|=5 \\ \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 3 &2 \end{array}\right|=(-7) & \mathrm{det}\left|\begin{array}{cc} 4&3 \\ 1 &2 \end{array}\right|=5 & \mathrm{det}\left|\begin{array}{cc} 4 &1 \\ 1 &3 \end{array}\right|=11 \end{array}\right]\\ \frac{1}{|\mathrm{det}\Lambda^{-1}|}=4\cdot 3\cdot 5-1\cdot 1\cdot 5-2\cdot 2\cdot 4-3\cdot 3\cdot 3+1\cdot 2\cdot 3+1\cdot 2\cdot 3\\ =60-5-16-27+6+6=24\\ \Lambda=\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ (-1) & 11 & 5\\ (-7) & 5 & 11 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \mathbf{B \Lambda}=\frac{1}{24}\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right]\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ (-1) & 11 & 5\\ (-7) & 5 & 11 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 1\cdot 11+0\cdot (-1)+0\cdot (-7) & 1\cdot (-1)+0\cdot 11+0\cdot 5& 1\cdot (-7)+0\cdot 5+0\cdot 11 \\ 1\cdot 11+0\cdot (-1)+1\cdot (-7) & 1\cdot (-1)+0\cdot 11+1\cdot 5& 1\cdot (-7)+0\cdot 5+1\cdot 11\\ 0\cdot 11+1\cdot (-1)+1\cdot (-7) & 0\cdot (-1)+1\cdot 11+1\cdot 5 &0\cdot (-7)+1\cdot 5+1\cdot 11 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ 4 & 4 & 4\\ (-8) & 16 & 16 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \mathbf{B\Lambda B^{T}}=\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ 4 & 4 & 4\\ (-8) & 16 & 16 \end{array}\right]\left[\begin{array}{ccc} 1 & 1&0 \\ 0 &0 &1\\ 0& 1&1 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11\cdot 1+(-1)\cdot 0+(-7)\cdot 0 & 11\cdot 1+(-1)\cdot 0+(-7)\cdot 1& 11\cdot 0+(-1)\cdot 1+(-7)\cdot 1 \\ 4\cdot 1+4\cdot 0+4\cdot 0 & 4\cdot 1+4\cdot 0+4\cdot 1& 4\cdot 0+4\cdot 1+4\cdot 1\\ (-8)\cdot 1+16\cdot 0+16\cdot 0& (-8)\cdot 1+16\cdot 0+16\cdot 1& (-8)\cdot 0+16\cdot 1+16\cdot 1 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11 & 4& (-8) \\ 4 & 8& 8\\ (-8)& 8& 32 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \mathrm{E}(U|V=0,W=1)=\mu_{x}+\Sigma_{12}\Sigma_{22}^{-1}\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ \Sigma_{12}=\left[\begin{array}{cc} 4&(-8) \end{array}\right]\\ \Sigma_{21}=\left[\begin{array}{c} 4 \\ (-8) \end{array}\right]\\ \Sigma_{22}=\left[\begin{array}{cc} 8 & 8 \\ 8 & 32 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \Sigma_{22}^{-1}=\frac{1}{8\cdot 32-8\cdot 8}\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right]=\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right] \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} \mathrm{E}(U|V=1,W=0)=0+\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 4&(-8) \end{array}\right]\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right]\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ =\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 4\cdot 32+(-8)\cdot (-8) & 4\cdot (-8)+(-8)\cdot 8 \end{array}\right]\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ =\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 192 &(-96) \end{array}\right]\left[\begin{array}{c} 1-0\\ 0-0 \end{array}\right]\\ =\frac{1}{192}\left[\begin{array}{cc} 192\cdot 1 + (-96)\cdot 0 \end{array}\right]=1 \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} 24\cdot \mathrm{Var}(U|V=1,W=0)=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}=11-\frac{1}{192}\left[\begin{array}{cc} 192 + (-96) \end{array}\right]\left[\begin{array}{c} 4 \\ (-8) \end{array}\right]\\ =11-\frac{1}{192}\left[\begin{array}{c} 192\cdot 4+(-96)\cdot (-8) \end{array}\right]\\ =11-\frac{1}{192}\left[\begin{array}{c} 1536 \end{array}\right]=11-8=3\\ \mathrm{Var}(U|V=1,W=0)=\frac{3}{24}=\frac{1}{8} \end{split} \end{equation*}$$ $$\begin{equation*} \begin{split} X|X+Y=1,Y+Z=0\in N(1,\frac{1}{8}) \end{split} \end{equation*}$$
Is this correct? My book gives me a different answer.
It looks like you're on the right track, but you're working way too hard on this one. Sooner or later algebraic and numerical mistakes will creep into even the best calculations. A good strategy is to minimize the amount calculation: the Principle of Mathematical Laziness. A key element of this principle is just-in-time computation: don't do any work until you have to. The following solution illustrates these ideas.
You have seen the virtue of changing variables. Keeping $$X$$ (whose conditional distribution we wish to compute), let the two new variables be
$$U = Y+Z,\ V = X+Z.$$
Consequently, looking ahead to the next step, note that the original variables can be expressed as
$$Y = U-V+X,\ Z=V-X.$$
You also recognized the need to compute the Jacobian of this transformation. Using the method I have described at https://stats.stackexchange.com/a/154298/919 this is almost trivial:
$$\left|\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\right| = \left|\mathrm{d}x\wedge \mathrm{d}(u-v+x)\wedge \mathrm{d}(v-x)\right| = \left|\mathrm{d}x\, \mathrm{d}u\, \mathrm{d}v\right|.$$
This leaves only the argument of the exponential, into which we need to substitute
$$y = u-v+x,\ z = v-x$$
and then set $$u=0$$ and $$v=1.$$ Focusing on the argument of the exponential (and ignoring the necessary division by $$2$$), this can be performed by visual inspection of the coefficients of $$x$$ and $$x^2$$ and then, as always with Normal distributions, completing the square:
\begin{aligned} 4x^{2}&+3y^{2}+5z^{2}+2xy+6xz+4zy\\ &= 4x^2 + 3(u-v+x)^2 + \cdots + 4(v-x)(u-v+x)\\ &= (4+3+5+2-6-4)x^2 \\&+ (0-6-10-2+6+8)x \\&+ \text{constants}\\ &= 4x^2 - 4x + \text{constants} \\ &= \frac{(x-1/2)^2}{(1/2)^2}+\text{some constant}. \end{aligned}
We know the conditional distribution will be Normal with some mean $$\mu$$ and some standard deviation $$\sigma$$, which means this quadratic part will take the form $$(x-\mu)^2/\sigma^2$$ plus some constant. Comparing with the foregoing, you can read off the values $$\mu=1/2$$ and $$\sigma=1/2.$$
You will, of course, wish to check this work: but I hope you find this to be much less effort than checking your original calculations.
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# NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise
### Access Exercises of Class 12 Maths Chapter 2 – Inverse Trigonometric Functions
Exercise 2.1 Solutions: 14 Questions (12 Short Answers, 2 MCQs)
Exercise 2.2 Solutions: 21 Questions (18 Short Answers, 3 MCQs)
Miscellaneous Exercise Solutions: 17 Questions (14 Short Answers, 3 MCQs)
Miscellaneous Exercise
Find the value of following functions :
$$\textbf{1.\space}\textbf{cos}^{\normalsize-1}\bigg(\textbf{cos}\frac{\textbf{13}\pi}{\textbf{6}}\bigg)\textbf{.}\\\textbf{Sol.}\space\text{cos}^{\normalsize-1}\bigg(\text{cos}\frac{13\pi}{6}\bigg)\\=\text{cos}^{\normalsize-1}\bigg[\text{cos}\bigg(2\pi + \frac{\pi}{6}\bigg)\bigg];\\\text{where}\frac{\pi}{6}\epsilon\lbrack0,\pi\rbrack$$
[Hence, the given angle does not lie in the principal interval i.e., [0, π], so we convert it such that it lies in [0, π]
$$= \text{cos}^{\normalsize-1}\bigg[\text{cos}\bigg(\frac{\pi}{6}\bigg)\bigg] = \frac{\pi}{6}\\\lbrack\because\space \text{cos}(2\pi +\theta) =\text{cos}\space\theta\rbrack$$
$$\textbf{2.\space tan}^{\normalsize-1}\bigg(\textbf{tan}\frac{\textbf{7}\pi}{\textbf{6}}\bigg)\\\textbf{Sol.\space}\text{tan}^{\normalsize-1}\bigg(\text{tan}\frac{7\pi}{6}\bigg)\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\bigg(\pi + \frac{\pi}{6}\bigg)\bigg];\\\text{where}\space\frac{\pi}{6}\epsilon\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\text{(Principal interval)}\\\therefore\space\text{tan}^{\normalsize-1}\bigg(\text{tan}\frac{7\pi}{6}\bigg)\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\frac{\pi}{6}\bigg]=\frac{\pi}{6}\\\lbrack\because\space \text{tan}(\pi +\theta) =\text{tan}\space\theta\rbrack$$
Prove the following functions :
$$\textbf{3. 2sin}^{\normalsize-1}\frac{\textbf{3}}{\textbf{5}} =\textbf{tan}^{\normalsize-1}\frac{\textbf{24}}{\textbf{7}}\textbf{.}\\\textbf{Sol.\space}\text{Given, 2 sin}^{\normalsize-1}\frac{3}{5} \\=\text{tan}^{\normalsize-1}\frac{24}{7}\\\text{LHS = 2 sin}^{\normalsize-1}\frac{3}{5}\\=\text{sin}^{\normalsize-1}\bigg[2×\frac{3}{5}\sqrt{1-\bigg(\frac{3}{5}\bigg)^{2}}\bigg]\\\lbrack\because\space \text{2 sin}^{\normalsize-1}y = \\\text{sin}^{\normalsize-1}(2y\sqrt{1-y^{2}})\rbrack\\=\text{sin}^{\normalsize-1}\bigg[2×\frac{3}{6}×\frac{4}{5}\bigg]$$
$$=\space\text{sin}^{\normalsize-1}\bigg(\frac{24}{25}\bigg)\\=\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{24}{25}}{\sqrt{1 - \bigg(\frac{24}{25}\bigg)^{2}}}\end{bmatrix}\\\bigg(\because\space\text{sin}^{\normalsize-1} y =\text{tan}^{\normalsize-1}\frac{y}{\sqrt{1-y^{2}}}\bigg)$$
$$=\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{24}{25}}{\sqrt{1 -\frac{576}{625}}}\end{bmatrix}\\=\text{tan}^{\normalsize-1}\bigg[\frac{24}{25}×\frac{25}{7}\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\frac{24}{7}\bigg]=\text{RHS.}\\\textbf{Hence proved.}$$
$$\textbf{4. sin}^{\normalsize-1}\frac{\textbf{8}}{\textbf{17}} \textbf{+}\textbf{sin}^{\normalsize-1}\frac{\textbf{3}}{\textbf{5}}\\\textbf{= tan}^{\normalsize-1}\frac{\textbf{77}}{\textbf{36}}\\\textbf{Sol.\space}\text{Given, sin}^{\normalsize-1}\frac{8}{17} + \text{sin}^{\normalsize-1}\frac{3}{5}\\=\text{tan}^{\normalsize-1}\frac{77}{36}\\\text{LHS} =\\ \text{sin}^{\normalsize-1}\bigg(\frac{8}{17}\bigg) + \text{sin}^{\normalsize-1}\bigg(\frac{3}{5}\bigg)\\=\text{sin}^{\normalsize-1}\begin{bmatrix}\frac{8}{17}\sqrt{1 -\bigg(\frac{3}{5}\bigg)^{2}} +\\ \frac{3}{5}\sqrt{1 -\bigg(\frac{8}{17}\bigg)^{2}}\end{bmatrix}$$
$$\lbrack\because\space\text{sin}^{\normalsize-1}x + \text{sin}^{\normalsize-1}y\\\text{sin}^{\normalsize-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})\rbrack\\=\text{sin}^{\normalsize-1}\bigg(\frac{8}{17}×\frac{4}{5} + \frac{3}{5}×\frac{15}{17}\bigg)\\=\text{sin}^{\normalsize-1}\bigg(\frac{77}{85}\bigg)\\=\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{77}{85}}{\sqrt{1-\bigg(\frac{77}{85}\bigg)^{2}}}\end{bmatrix}\\\bigg(\because\space\text{sin}^{-1}x = \text{tan}^{\normalsize-1}\frac{x}{\sqrt{1 - x^{2}}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg[\frac{77}{85}×\frac{85}{36}\bigg]$$
$$=\text{tan}^{\normalsize-1}\frac{77}{36}=\text{RHS.}\\\textbf{Hence proved.}$$
$$\textbf{5.\space cos}^{\normalsize-1}\frac{\textbf{4}}{\textbf{5}} \textbf{+}\textbf{cos}^{\normalsize-1}\frac{\textbf{12}}{\textbf{13}}\\=\textbf{cos}^{\normalsize-1}\frac{\textbf{33}}{\textbf{65}}\textbf{.}\\\textbf{Sol.\space}\text{Given,}\\\text{cos}^{\normalsize-1}\frac{4}{5}+\text{cos}^{\normalsize-1}\frac{12}{13}\\=\text{cos}^{\normalsize-1}\frac{33}{65}\\\text{LHS} = \\\text{cos}^{\normalsize-1}\bigg(\frac{4}{5}\bigg) + \text{cos}^{\normalsize-1}\bigg(\frac{12}{13}\bigg)\\=\text{cos}^{\normalsize-1}\\\bigg(\frac{4}{5}.\frac{12}{13}-\sqrt{1-\bigg(\frac{4}{5}\bigg)^{2}}\sqrt{1-\bigg(\frac{12}{13}\bigg)^{2}}\bigg)$$
$$\lbrack\because\space\text{cos}^{\normalsize-1}x + \text{cos}^{-1}y\\=\text{cos}^{-1}(xy -\sqrt{1-x^{2}}\sqrt{1 - y^{2}})\rbrack\\=\text{cos}^{\normalsize-1}\bigg(\frac{48}{65} -\sqrt{\bigg(\frac{3}{5}\bigg)^{2}}\sqrt{\bigg(\frac{5}{13}\bigg)^{2}}\bigg)\\=\text{cos}^{\normalsize-1}\bigg(\frac{48}{65} - \frac{3}{5}×\frac{5}{13}\bigg)\\=\text{cos}^{\normalsize-1}\bigg(\frac{48}{65} -\frac{3}{13}\bigg)\\=\text{cos}^{\normalsize-1}\bigg(\frac{33}{65}\bigg)\\=\text{RHS.}\space\textbf{Hence proved.}$$
$$\textbf{6. cos}^{\normalsize-1}\frac{\textbf{12}}{\textbf{13}} \textbf{+}\textbf{sin}^{\normalsize-1}\frac{\textbf{3}}{\textbf{5}}\\=\textbf{sin}^{\normalsize-1}\frac{\textbf{56}}{\textbf{65}}\\\textbf{Sol.\space}\text{Given,}\\\text{cos}^{\normalsize-1}\frac{12}{13} + \text{sin}^{\normalsize-1}\frac{3}{5}\\=\text{sin}^{\normalsize-1}\frac{56}{65}\\\text{Let\space}\text{cos}^{\normalsize-1}\frac{12}{13} = \theta\\\Rarr\space\text{cos}\space\theta =\frac{12}{13}\\\therefore\space\text{sin}\space\theta =\sqrt{1 - \bigg(\frac{12}{13}\bigg)^{2}}$$
$$\lbrack\because\space\text{sin}\space\theta =\sqrt{1 - cos^{2}}\theta\rbrack\\\Rarr\space\text{sin}\space\theta=\sqrt{\frac{25}{169}}=\frac{5}{13}\\\Rarr\space\theta = \text{sin}^{\normalsize-1}\frac{5}{13}\\\text{Now,\space LHS = cos}^{\normalsize-1}\frac{12}{13} +\text{sin}^{\normalsize-1}\frac{3}{5}\\=\text{sin}^{\normalsize-1}\frac{5}{13} + \text{sin}^{\normalsize-1}\frac{3}{5}\\=\text{sin}^{\normalsize-1}\bigg(\frac{5}{13}\sqrt{1 -\bigg(\frac{3}{5}\bigg)^{2}} + \\\frac{3}{5}\sqrt{1 -\bigg(\frac{5}{13}\bigg)^{2}}\bigg)$$
$$\lbrack\because\space\text{sin}^{\normalsize-1}x + \text{sin}^{\normalsize-1}y\\ = \text{sin}^{\normalsize-1}(x\sqrt{1 - y^{2}} + y\sqrt{1- x^{2}})\rbrack\\=\text{sin}^{\normalsize-1}\bigg(\frac{5}{13}\sqrt{\frac{16}{25}} + \frac{3}{5}\sqrt{\frac{144}{169}}\bigg)\\=\text{sin}^{\normalsize-1}\bigg(\frac{5}{13}×\frac{4}{5} + \frac{3}{5}×\frac{12}{13}\bigg)\\=\text{sin}^{\normalsize-1}\bigg(\frac{56}{65}\bigg)=\text{RHS}\\\textbf{Hence proved.}$$
$$\textbf{7.\space}\textbf{tan}^{\normalsize-1}\frac{\textbf{63}}{\textbf{16}} =\\\textbf{sin}^{\normalsize-1}\frac{\textbf{5}}{\textbf{13}} \textbf{+}\textbf{cos}^{\normalsize-1}\frac{\textbf{3}}{\textbf{5}}.\\\textbf{Sol.\space}\text{Given, tan}^{\normalsize-1}\frac{63}{16}\\=\text{sin}^{\normalsize-1}\frac{5}{13} + \text{cos}^{\normalsize-1}\frac{3}{5}\\\text{RHS = sin}^{\normalsize-1}\frac{5}{13} + \text{cos}^{\normalsize-1}\frac{3}{5}\\\text{Let}\space\text{sin}^{\normalsize-1}\bigg(\frac{5}{13}\bigg) = x\\\Rarr\space \text{sin x =}\frac{5}{13}$$
∵ 1 – sin2x = cos2 x
$$\Rarr\space1 -\bigg(\frac{5}{13}\bigg)^{2} =\text{cos}^{2}x\\\Rarr\space \frac{169-25}{169}=\text{cos}^{2}x\\\Rarr\space \frac{144}{169} = \text{cos}^{2}x\\\Rarr\space \frac{12}{13}=\text{cos x}\\\therefore\space\text{tan x} =\frac{\text{sin x}}{\text{cos x}}\\\Rarr\space \text{tan}\space x =\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}\\\Rarr\space \text{x = tan}^{\normalsize-1}\frac{5}{12}$$
$$\text{Let\space cos}^{\normalsize-1}\frac{3}{5} = y\\\Rarr\space\text{cos y} =\frac{3}{5}\\\because\space\text{1 - cos}^{2}y = \text{sin}^{2}y\\\Rarr\space 1-\bigg(\frac{3}{5}\bigg)^{2} - \text{sin}^{2}y\\\Rarr\space\frac{25-9}{25} = \text{sin}^{2}y\\\Rarr\space\frac{16}{25} =\text{sin}^{2}y\\\Rarr\space \frac{4}{5} = \text{sin y}\\\text{tan y} = \frac{\text{sin y}}{\text{cos y}}$$
$$\Rarr\space\text{tan y} =\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\\\Rarr y =\text{tan}^{\normalsize-1}\frac{4}{3}\\\therefore\space\text{Given equation becomes}\\\text{tan}^{\normalsize-1}\frac{63}{16} = x+y\\\Rarr\space \text{tan}^{\normalsize-1}\frac{63}{16} =\text{tan}^{\normalsize-1}\bigg(\frac{5}{12}\bigg) +\\ \text{tan}^{\normalsize-1}\bigg(\frac{4}{3}\bigg)\\\therefore\space\text{RHS = tan}^{\normalsize-1}\bigg(\frac{5}{12}\bigg) + \text{tan}^{\normalsize-1}\bigg(\frac{4}{3}\bigg)$$
$$= \text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12}×\frac{4}{3}}\end{bmatrix}\\\bigg[\because\space\text{tan}^{\normalsize-1}x + \text{tan}^{\normalsize-1}y\\=\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1-xy}\bigg)\bigg]$$
$$=\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{15 +48}{12×3}}{\frac{12×3-20}{12×3}}\end{bmatrix}\\=\text{tan}^{\normalsize-1}\bigg[\frac{63}{36}×\frac{36}{36-20}\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\frac{63}{16}\bigg]=\text{RHS}$$
∴ LHS = RHS Hence proved.
$$\textbf{8. tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{5}} \textbf{+}\textbf{tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{7}}\textbf{+}\\\textbf{tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{3}} \textbf{+}\textbf{tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{8}}\textbf{=}\frac{\pi}{\textbf{4}}\\\textbf{Sol.\space} \text{Given},\space\text{tan}^{-1}\frac{1}{5} + \text{tan}^{\normalsize-1}\frac{1}{7}+\\\text{tan}^{\normalsize-1}\frac{1}{3} + \text{tan}^{\normalsize-1}\frac{1}{8}=\frac{\pi}{4}\\\text{LHS}= \bigg(\text{tan}^{\normalsize-1}\frac{1}{5} + \text{tan}^{\normalsize-1}\frac{1}{7}\bigg) +\\\bigg(\text{tan}^{\normalsize-1}\frac{1}{3} + \text{tan}^{\normalsize-1}\frac{1}{8}\bigg)\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5}×\frac{1}{7}}\end{pmatrix} + \\\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{1}{3} + \frac{1}{2}}{1 -\frac{1}{3}×\frac{1}{8}}\end{pmatrix}$$
$$\bigg[\because\space\text{tan}^{\normalsize-1}x +\text{tan}^{\normalsize-1}y =\\\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1 - xy}\bigg)\bigg]$$
$$= \text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{7+5}{35}}{\frac{35-1}{35}}\end{pmatrix} + \text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{8+3}{24}}{\frac{24-1}{24}}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\bigg(\frac{6}{17}\bigg) + \text{tan}^{\normalsize-1}\bigg(\frac{11}{23}\bigg)\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{6}{17}+\frac{11}{23}}{1- \frac{6}{17}×\frac{11}{23}}\end{pmatrix}+\\\text{tan}^{\normalsize-1}\bigg(\frac{6×23 + 11×17}{17×23 - 6×11}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{325}{325}\bigg)\\=\text{tan}^{\normalsize-1}1 \frac{\pi}{4}=\text{RHS.}$$
Hence proved.
Prove that :
$$\textbf{9.\space tan}^{\normalsize-1}\sqrt{\textbf{x}}\space\textbf{=}\space\\\frac{\textbf{1}}{\textbf{2}}\textbf{cos}^{\normalsize-1}\bigg(\frac{\textbf{1-x}}{\textbf{1+x}}\bigg)\textbf{,}\space \textbf{x}\epsilon\lbrack\textbf{0,1}\rbrack\\\textbf{Sol.\space}\text{Given, tan}^{\normalsize-1}\sqrt{x} \\=\frac{1}{2}\text{cos}^{\normalsize-1}\bigg(\frac{\text{1-x}}{\text{1+x}}\bigg), x\epsilon\lbrack0,1\rbrack\\\text{LHS = tan}^{\normalsize-1}\sqrt{x}\\=\frac{1}{2}×(2×\text{tan}^{\normalsize-1}\sqrt{x})\\=\frac{1}{2}×\text{cos}^{\normalsize-1}\frac{1 - (\sqrt{x})^{2}}{1 + (\sqrt{x})^{2}}\\\begin{bmatrix}\because\space 2\text{tan}^{\normalsize-1}x = \text{cos}^{\normalsize-1}\bigg(\frac{\text{1 - x}^{2}}{\text{1 + x}^{2}}\bigg)\end{bmatrix}\\=\frac{1}{2}\text{cos}^{\normalsize-1}\bigg(\frac{1-x}{1 +x}\bigg)=\text{RHS.}$$
Hence proved.
$$\textbf{10. cot}^{\normalsize-1}\bigg(\frac{\sqrt{\textbf{1 + sin x}} + \sqrt{\textbf{1 - sin x}}}{\sqrt{\textbf{1 + sin x}} - \sqrt{\textbf{1 - sin x}}}\bigg)\\=\frac{\textbf{x}}{\textbf{2}}\textbf{,}\space \textbf{x}\epsilon\bigg(0,\frac{\pi}{\textbf{4}}\bigg)$$
Sol. Given,
$$\text{cot}^{\normalsize-1}\begin{pmatrix}\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} -\sqrt{1 - sin x}}\end{pmatrix}\\=\frac{x}{2}, x\epsilon\bigg(0,\frac{\pi}{4}\bigg)\\\text{LHS =}\\\text{ cot}^{\normalsize-1}\bigg(\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x}\space-\space\sqrt{1 - sinx}}\bigg)\\\text{...(i)}$$
Now, we can write
$$\sqrt{\text{1 + sin x}} =\\\sqrt{\text{sin}^{2}\frac{x}{2} + \text{cos}^{2}\frac{x}{2} + 2\text{sin}\frac{x}{2}\text{cos}\frac{x}{2}}\\=\sqrt{\bigg(\text{sin}\frac{x}{2} + \text{cos}\frac{x}{2}\bigg)^{2}} \\\begin{bmatrix}\because\space\text{sin}^{2}\frac{x}{2} + \text{cos}^{2}\frac{x}{2}=1\\\text{and \space sin x = 2sin}\frac{x}{2}\text{cos}\frac{x}{2}\end{bmatrix}\\\Rarr\space\sqrt{\text{1 + sinx}} \\=\text{sin}\frac{x}{2} + \text{cos}\frac{x}{2}$$
Similarly, we can get
$$\sqrt{\text{1 - sin x}} = \text{cos}\frac{x}{2} - \text{sin}\frac{x}{2}$$
On substituting above two values in Eq. (i), we get
$$\text{LHS }\\\text{= cot}^{\normalsize-1}\begin{pmatrix}\frac{\text{sin}\frac{x}{2} + \text{cos}\frac{x}{2} + \text{cos}\frac{x}{2} -\text{sin}\frac{x}{2}}{\text{sin}\frac{x}{2} + \text{cos}\frac{x}{2} + \text{sin}\frac{x}{2} -\text{cos}\frac{x}{2}}\end{pmatrix}\\=\text{cot}^{\normalsize-1}\begin{pmatrix}\frac{\text{2 cos}\frac{x}{2}}{\text{2 sin}\frac{x}{2}}\end{pmatrix}\\=\text{cot}^{\normalsize-1}\bigg(\text{cot}\frac{x}{2}\bigg)=\frac{x}{2}=\text{RHS}$$
Alternative method
$$\text{LHS} =\\ \text{cot}^{\normalsize-1}\frac{\sqrt{\text{1 + sinx}} + \sqrt{\text{1 - sin x}}}{\sqrt{\text{1 + sin x}}-\sqrt{\text{1 - sin x}}}\\=\text{cot}^{\normalsize-1}\\\begin{bmatrix}\frac{\sqrt{\text{1 + sin x}} + \sqrt{\text{1 - sin x}}}{\sqrt{\text{1 + sin x}} - \sqrt{\text{1 - sinx}}}×\\\frac{\sqrt{\text{1 + sin x}} + \sqrt{\text{1 - sin x}}}{\sqrt{\text{1 + sin x}} +\sqrt{\text{1 - sin x}}}\end{bmatrix}$$
(by rationalyzing denominator)
$$=\text{cot}^{\normalsize-1}\begin{bmatrix}\frac{(\sqrt{1 + sin \space x} + \sqrt{1 - sin x})^{2}}{(\sqrt{\text{1 + sinx}})^{2} - (\sqrt{\text{1 - sin x}})^{2}}\end{bmatrix}\\=\text{cot}^{\normalsize-1}\\\begin{bmatrix}\frac{\text{1 + sinn x +1 - sin x +2}\sqrt{\text{1 - sin}^{2}x}}{\text{1 + sin x - 1 + sin x}}\end{bmatrix}\\=\text{cot}^{\normalsize-1}\bigg[\frac{\text{2 +2 cos x}}{\text{2 sin x}}\bigg]\\\begin{bmatrix}\because\space \text{cos}^{2}x + \text{sin}^{2}x=1\\\Rarr\space\text{cos x} = \sqrt{\text{1 - sin}^{2}x}\end{bmatrix}\\=\text{cot}^{\normalsize-1}\bigg[\frac{\text{1 + cos x}}{\text{sin x}}\bigg]\\=\text{cot}^{\normalsize-1}\bigg[\frac{\text{2 cos}^{2}\space\text{x/2}}{\text{2 sin x/2 cos x/2}}\bigg]$$
$$\begin{pmatrix}\because\space \text{1 + cos x = 2 cos}^{2}\frac{x}{2}\\\text{ and sin x = 2 sin}\frac{x}{2}\text{cos}\frac{x}{2}\end{pmatrix}\\=\text{cot}^{\normalsize-1}\bigg[\text{cot}\frac{x}{2}\bigg]=\frac{x}{2} = \text{RHS}\\\textbf{Note :\space}\text{If x}\epsilon\bigg(0,\frac{\pi}{2}\bigg),\space\text{then}\\=\sqrt{\text{1 - sin x}} = \text{cos}\frac{x}{2} -\text{sin}\frac{x}{2}\\\text{and if}\space x\epsilon\bigg(\frac{\pi}{2},\pi\bigg),\text{then}\\\sqrt{\text{1 - sin x}} =\text{sin}\frac{x}{2}-\text{cos}\frac{x}{2}$$
$$\textbf{11.\space tan}^{\normalsize-1}\bigg(\frac{\sqrt{\textbf{1 + x}} \textbf{-}\sqrt{\textbf{1- x}}}{\sqrt{\textbf{1 + x}} \textbf{+} \sqrt{\textbf{1 - x}}}\bigg)\\=\frac{\pi}{\textbf{4}}-\frac{\textbf{1}}{\textbf{2}}\textbf{cos}^{\normalsize-1}\textbf{x.}$$
Sol. Let x cos θ so that cos–1 x = θ
$$\therefore\space \text{tan}^{\normalsize-1}\bigg(\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 +x} + \sqrt{1 - x}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{\sqrt{\text{1 + cos}\space\theta} - \sqrt{\text{1 - cos}\space \theta}}{\sqrt{\text{1 + cos}\space\theta} + \sqrt{\text{1 - cos}\theta}}\bigg)\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\sqrt{2}\text{cos}\frac{\theta}{2} -\sqrt{2}\text{sin}\frac{\theta}{2}}{\sqrt{2}\text{cos}\frac{\theta}{2} + \sqrt{2}\text{sin}\frac{\theta}{2}}\end{pmatrix}$$
$$\begin{pmatrix}\because\space \text{1 + cos}\space\theta \text{= 2 cos}^{2}\frac{\theta}{2}\\\text{and}\space\text{1 - cos}\space\theta \text{= 2 sin}^{2}\frac{\theta}{2}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\text{cos}\frac{\theta}{2} - \text{sin}\frac{\theta}{2}}{\text{cos}\frac{\theta}{2} + \text{sin}\frac{\theta}{2}}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\text{1 - tan}\frac{\theta}{2}}{\text{1 + tan}\frac{\theta}{2}}\end{pmatrix}$$
(inside the bracket divide numerator and denominator by cos θ/2)
$$= \text{tan}^{\normalsize-1}\begin{bmatrix}\text{tan}\bigg(\frac{\pi}{4} -\frac{\theta}{2}\bigg)\end{bmatrix}\\\begin{bmatrix}\because\space \text{tan}(\text{A-B}) =\\\frac{\text{tan A - tan B}}{\text{1 + tan Atan B}}\end{bmatrix}\\=\frac{\pi}{4} -\frac{\theta}{2}=\frac{\pi}{2}-\frac{1}{2}\text{cos}^{\normalsize-1}x\\\textbf{Hence proved.}$$
$$\textbf{12.\space}\frac{\textbf{ 9}\pi}{\textbf{8}}\textbf{-}\frac{\textbf{9}}{\textbf{4}}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{1}}{\textbf{3}}\bigg)\\\textbf{=}\space\frac{\textbf{9}}{\textbf{4}}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{2}\sqrt{\textbf{2}}}{\textbf{3}}\bigg)\\\textbf{Sol.\space}\text{Given,\space}\frac{9\pi}{8}-\frac{9}{4}\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}\bigg)\\=\frac{9}{4}\text{sin}^{\normalsize-1}\bigg(\frac{2\sqrt{2}}{3}\bigg)\\\Rarr\space \frac{9\pi}{8} =\frac{9}{4}\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}\bigg) +\\\frac{9}{4}\text{sin}^{\normalsize-1}\bigg(\frac{2\sqrt{2}}{3}\bigg)$$
$$\Rarr\space\frac{9\pi}{8} =\frac{9}{4}\begin{bmatrix}\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}\bigg) +\\\text{sin}^{\normalsize-1}\bigg(\frac{2\sqrt{2}}{3}\bigg)\end{bmatrix}\\\text{RHS} = \frac{9}{4}\begin{bmatrix}\text{sin}^{-1}\bigg(\frac{1}{3}\bigg) + \\\text{sin}^{\normalsize-1}\bigg(\frac{2\sqrt{2}}{3}\bigg)\end{bmatrix}\\=\frac{9}{4}\begin{bmatrix}\text{sin}^{\normalsize-1}\begin{Bmatrix}\frac{1}{3}\sqrt{1- \bigg(\frac{2\sqrt{2}}{3}\bigg)^{2}} +\\ \frac{2\sqrt{2}}{3}\sqrt{1 -\bigg(\frac{1}{3}\bigg)^{2}}\end{Bmatrix}\end{bmatrix}$$
$$\lbrack\because\space\text{sin}^{\normalsize-1} \text{x + sin}^{\normalsize-1}y\\=\text{sin}^{\normalsize-1}(x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}})\rbrack\\=\frac{9}{4}\begin{bmatrix}\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}×\frac{1}{3} + \frac{2\sqrt{2}}{3}×\frac{2\sqrt{2}}{3}\bigg)\end{bmatrix}\\=\frac{9}{4}\begin{bmatrix}\text{sin}^{\normalsize-1}\bigg(\frac{1}{9} +\frac{8}{9}\bigg)\end{bmatrix}\\=\frac{9}{4}\text{sin}^{\normalsize-1}(1)\\=\frac{9}{4}×\frac{\pi}{2} =\frac{9\pi}{8}=\text{LHS.}\\\textbf{Hence proved.}\\\textbf{Alternative method}\\\text{LHS} =\frac{9\pi}{8} -\frac{9}{4}\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}\bigg)$$
$$=\frac{9}{4}\bigg(\frac{\pi}{2} -\text{sin}^{\normalsize-1}\bigg(\frac{1}{3}\bigg)\bigg)\\=\frac{9}{4}\bigg(\text{cos}^{\normalsize-1}\bigg(\frac{1}{3}\bigg)\bigg)\\\bigg[\because\space\text{sin}^{\normalsize-1}x +\text{cos}^{\normalsize-1}x=\frac{\pi}{2}\bigg]\\=\frac{9}{4}\text{sin}^{\normalsize-1}\sqrt{1 - \bigg(\frac{1}{3}\bigg)^{2}}\\\lbrack\because\space\text{sin}^{\normalsize-1}x + \text{cos}^{\normalsize-1}x = \frac{\pi}{2}\rbrack\\=\frac{9}{4}\text{sin}^{\normalsize-1}\sqrt{1 - \bigg(\frac{1}{3}\bigg)^{2}}$$
$$\lbrack\because\space \text{cos}^{\normalsize-1}x =\text{sin}^{\normalsize-1}\sqrt{1 - x^{2}}\rbrack\\=\frac{9}{4}\text{sin}^{\normalsize-1}\sqrt{1 -\frac{1}{9}}\\=\frac{9}{4}\text{sin}^{\normalsize-1}\frac{2\sqrt{2}}{3}\space\text{ = RHS.}\\\textbf{Hence Proved.}$$
Solve the following equations :
13. 2 tan–1 (cos x) = tan–1 (2 cosec x).
Sol. Given, 2 tan–1 (cos x) = tan–1 (2 cosec x)
$$\Rarr\space\text{tan}^{\normalsize-1}\bigg(\frac{\text{2 cos x}}{\text{ 1 - cos}^{2}x}\bigg)\\=\text{tan}^{\normalsize-1}(\text{2 cosec x})\\\bigg[\because\space\text{2 tan}^{\normalsize-1}x = \text{tan}^{\normalsize-1}\bigg(\frac{2x}{1 - x^{2}}\bigg)\bigg]\\\Rarr\space \text{tan}^{\normalsize-1}\bigg(\frac{\text{2 cos x}}{\text{sin}^{2}x}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{2}{\text{sin x}}\bigg)\\\Rarr\space \frac{\text{2cos x}}{\text{sin}^{2}x} =\frac{2}{\text{sin x}}\\\Rarr\space\frac{\text{2 cos x}}{\text{sin x}}= 2\\\Rarr\space\text{cot x = 1}$$
$$\Rarr\space\text{cot x = cot}\frac{\pi}{4}\\\Rarr\space x =\frac{\pi}{4}$$
$$\textbf{14.\space}\textbf{tan}^{\normalsize-1}\frac{\textbf{1-x}}{\textbf{1+x}} \textbf{=}\\\frac{\textbf{1}}{\textbf{2}}\textbf{tan}^{\normalsize-1}\textbf{x,}\textbf{(x}\gt\textbf{0).}$$
Sol. Given,
$$\text{tan}^{\normalsize-1}\bigg(\frac{\text{1 - x}}{\text{1 + x}}\bigg)\\=\frac{1}{2}\text{tan}^{\normalsize-1}x\\\text{or \space 2 tan}^{\normalsize-1}\bigg(\frac{\text{1 - x}}{\text{1 + x}}\bigg)\\=\text{tan}^{\normalsize-1}x \\\Rarr\space\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{2\bigg(\frac{\text{1 - x}}{\text{1 + x}}\bigg)}{1 - \bigg(\frac{\text{1 - x}}{\text{1 + x}}\bigg)^{2}}\end{bmatrix}\\=\text{tan}^{\normalsize-1}x\\\bigg[\because\space\text{2 tan}^{\normalsize-1}y =\text{tan}^{\normalsize-1}\bigg(\frac{2y}{1 - y^{2}}\bigg)\bigg]$$
$$\Rarr\space\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\frac{2(1-x)}{(1 + x)}}{\frac{(1 + x^{2}) -(1 - x)^{2}}{(1 + x)^{2}}}\end{bmatrix}\\=\text{tan}^{\normalsize-1}x\\\Rarr\space\text{tan}^{\normalsize-1}\bigg[\frac{2(1-x)(1 + x)}{(1 + x)^{2} -(1 - x)^{2}}\bigg]\\=\text{tan}^{\normalsize-1}x\\\Rarr\space\text{tan}^{\normalsize-1}\bigg[\frac{2(1 - x)^{2}}{1^{2} + x^{2} + 2x - 1^{2} - x^{2} + 2x}\bigg]\\=\text{tan}^{\normalsize-1}x\\\Rarr\space\text{tan}^{\normalsize-1}\bigg[\frac{2(1 - x^{2})}{4x}\bigg]\\=\text{tan}^{\normalsize-1}x$$
$$\Rarr\space\text{tan}^{\normalsize-1}\bigg(\frac{(1 - x^{2})}{2x}\bigg)\\=\text{tan}^{\normalsize-1}x\\\Rarr\space\frac{1 - x^{2}}{2x} = x\\\Rarr\space 1-x^{2} = 2x^{2}\\\Rarr\space 1 = 3x^{2}\space\Rarr x^{2}=\frac{1}{3}\\\Rarr\space x=\pm\frac{1}{\sqrt{3}}\\\begin{bmatrix}\because\space x\gt 0\text{given, so we do not take}\\ x = -\frac{1}{\sqrt{3}}\end{bmatrix}\\\Rarr\space x=\frac{1}{\sqrt{3}}$$
15. sin (tan–1x), |x| < 1, is equal to :
$$\textbf{(a)}\frac{\textbf{x}}{\sqrt{\textbf{1 - x}^{2}}}\\\textbf{(b)\space}\frac{\textbf{1}}{\sqrt{\textbf{1 - x}^{\textbf{2}}}}\\\textbf{(c)\space}\frac{\textbf{1}}{\sqrt{\textbf{1 +x}^{\textbf{2}}}}\\\textbf{(d)\space}\frac{\textbf{x}}{\sqrt{\textbf{1 + x}^{2}}}\\\textbf{Sol.\space}\text{(d)}\space\frac{x}{\sqrt{1 + x^{2}}}\\\text{sin (tan}^{\normalsize–1}x) \\= \text{sin}\bigg[\text{sin}^{\normalsize-1}\frac{8}{\sqrt{1 + x^{2}}}\bigg]\\\bigg[\because\space \text{tan}^{\normalsize-1}x = \text{sin}^{\normalsize-1}\frac{x}{\sqrt{1 + x^{2}}}\bigg]$$
$$= \frac{x}{\sqrt{1 + x^{2}}}.$$
16. sin−1 (1 −x ) − 2sin−1
$$\textbf{x} = \frac{\pi}{\textbf{2}}\textbf{,}\\\textbf{then x is equal to}\\\textbf{(a) 0,}\frac{\textbf{1}}{\textbf{2}}\\\textbf{(b)\space 1,}\frac{\textbf{1}}{\textbf{2}}$$
(c) 0
$$\textbf{(d)}\space\frac{\textbf{1}}{\textbf{2}}$$
Sol. (c) 0
$$\text{Given, sin}^{\normalsize−1}(1− x) − 2sin^{\normalsize−1}x\\ = \frac{\pi}{2}\\\Rarr\space -\text{2 sin}^{-1}x = \frac{\pi}{2} -\text{sin}^{\normalsize-1}(1 - x)\\\Rarr\space\text{-2 sin}^{\normalsize-1}x = \text{cos}^{\normalsize-1}(1-x)\\\bigg[\because\space\text{sin}^{\normalsize-1}(1 -x) + \text{cos}^{\normalsize-1}(1 - x)=\frac{\pi}{2}\bigg]\\\Rarr\space\text{cos}\space(-2\text{sin}^{\normalsize-1}x) = 1-x$$
[Multiplying both sides by cos x]
$$\Rarr\space \text{cos}(2\space\text{sin}^{\normalsize-1}x)=1-x\\\lbrack\because\space\text{cos}(\normalsize-x) = +\space\text{cos x}\rbrack\\\Rarr\space[1 - 2\text{sin}^{2}(\text{sin}^{\normalsize-1}x)]=1-x\\\lbrack\because\space \text{cos 2x = 1-2\space\text{sin}}^{2}x\rbrack\\\Rarr\space 1 - 2x^{2}= 1 - x\\\Rarr\space 2x^{2}-x =0\\\Rarr\space x(2x-1) =0\\\Rarr\space x = 0$$
or 2x – 1 = 0
$$\Rarr\space x = 0 \space\text{or}\space\frac{1}{2}\\\text{But\space x = }\frac{1}{2}\space\text{does not satisfy}\\\text{the given equation, so x = 0.}\\ \textbf{Alternate method}\\\text{Given, sin}^{\normalsize-1}(1 - x) -\\ 2\text{sin}^{\normalsize-1}x=\frac{\pi}{2}$$
$$\text{Let x = sin}\space\theta\\\Rarr\space\theta = \text{sin}^{\normalsize-1}x,\space\text{then}\\\text{sin}^{\normalsize-1}(\text{1 - sin}\space\theta)- 2\theta =\frac{\pi}{2}\\\Rarr\space \text{sin}^{\normalsize-1}(\text{1 - sin}\space\theta) =\frac{\pi}{2} + 2\theta\\\Rarr\space \text{1 - sin}\space\theta =\text{sin}\bigg(\frac{\pi}{2} + 2\theta\bigg)\\\Rarr\space \text{1 - sin}\space\theta = \text{cos 2}\theta\\\bigg[\because\text{sin}\bigg(\frac{\pi}{2}+ \theta\bigg) =\text{cos}\theta\bigg]\\\Rarr\space\text{1 - sin}\space\theta = \text{1 - sin}^{2}\theta\\\lbrack\because\space \text{cos 2}\theta = 1- 2\text{sin}^{2}\theta\rbrack$$
$$\Rarr\space\text{2 sin}^{2}\theta -\text{sin}\space\theta =0\\\Rarr\space \text{sin}\space\theta(2 sin\theta-1)=0$$
Either sin θ = 0
or 2 sin θ – 1 = 0
$$\Rarr\space \text{x = 0}\space\text{or}\space\text{2x-1 =0}\\\lbrack\because\space\text{sin}\space\theta = x\rbrack\\\text{x =0 or x}=\frac{1}{2}\\\text{But\space x} =\frac{1}{2}\space\text{does not satisfy}$$
the given equation.
So, x = 0.
Put x = 0 in the given equation,
$$\text{sin}^{\normalsize-1}(1 -0)- 2\space\text{sin}^{\normalsize-1} 0 =\frac{\pi}{2}\\\Rarr\space\frac{\pi}{2}-2×0 = \frac{\pi}{2}\\\Rarr\space \frac{\pi}{2}=\frac{\pi}{2}\\\text{Put}\space x =\frac{1}{2}\space\text{in the given equation,}\\\therefore\space\text{sin}^{\normalsize-1}\bigg(1 -\frac{1}{2}\bigg)-\\ 2\text{sin}^{\normalsize-1}\frac{1}{2}=\frac{\pi}{2}\\\therefore\space\text{sin}{\normalsize-1}\frac{1}{2} - 2\text{sin}^{\normalsize-1}\frac{1}{2}\\=\frac{\pi}{2}\\\Rarr\space \frac{\pi}{6}-2×\frac{\pi}{6} =\frac{\pi}{2}$$
$$\Rarr\space \frac{\pi - 2\pi}{6}=\frac{\pi}{2}\\\Rarr\space \frac{-\pi}{6}\neq\frac{\pi}{2},\\\text{So \space x =}\frac{1}{2}\space\text{is not possible.}$$
Note : While solving the trigonometric equation, sometimes it may have some extra roots, so please take careful about it.
$$\textbf{17.\space tan}^{\normalsize-1}\bigg(\frac{\textbf{x}}{\textbf{y}}\bigg) \textbf{- tan}^{\normalsize-1}\bigg(\frac{\textbf{x-y}}{\textbf{x+y}}\bigg)\\\textbf{is equal to}$$
$$\textbf{(a)\space}\frac{\pi}{\textbf{2}}\\\textbf{(b)\space}\frac{\pi}{\textbf{3}}\\\textbf{(c)\space}\frac{\pi}{\textbf{4}}\\\textbf{(d)\space}\frac{\textbf{3}\pi}{\textbf{4}}\\\textbf{Sol.\space}(c)\frac{\pi}{4}\\\text{tan}^{\normalsize-1}\bigg(\frac{x}{y}\bigg) -\text{tan}^{\normalsize-1}\bigg(\frac{x-y}{x + y}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{x}{y}\bigg)-\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{x}{y}-1}{\frac{x}{y}+1}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\bigg(\frac{x}{y}\bigg) -\\\begin{Bmatrix}\text{tan}^{\normalsize-1}\frac{x}{y} - \text{tan}^{\normalsize-1}1\end{Bmatrix}$$
$$\bigg[\because\space\text{tan}^{\normalsize-1}x - \text{tan}^{\normalsize-1}y=\\\text{tan}^{\normalsize-1}\bigg(\frac{x-y}{\text{1 + xy}}\bigg)\bigg]\\=\text{tan}^{\normalsize-1}1 =\frac{\pi}{4}.$$
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# If $X$ and $Y$ are iid, is $E(X\mid X+Y)=E(Y\mid X+Y)$?
Just like question asked, my thought is they have the same mapping.
My classmate gives me a counterexample: if $$X$$ and $$Y$$ are equal, then $$X+Y=2X$$. However, $$\operatorname{Var}(2X) \neq \operatorname{Var}(X+Y)$$. So $$2X$$ and $$X+Y$$ don't have the same distribution? I am confused.
• Yes. This is essentially because the distribution of $(X, Y)$ is the same as $(Y, X)$. Apr 24, 2019 at 3:55
• Why would $\text{Var}(2X) \neq \text{Var}(X+Y)$ in your classmate's "counterexample"? It's not as though the variance of the sum is the sum of variances here. The answer to your question is yes, because you are conditioning two i.i.d. random variables on the same $\sigma$-algebra.
– snar
Apr 24, 2019 at 4:09
• @snar The conditional expectations of two iid variables on the same $\sigma$-algebra are not necessarily (or generally) equal. Apr 24, 2019 at 4:59
• @spaceisdarkgreen You are of course correct, as the example $X = E[X | X] \neq E[Y | X] = E[Y]$ shows.
– snar
Apr 24, 2019 at 15:11
If $$X$$ and $$Y$$ are exactly equal, the only way for them to be independent is if they're constant (i.e. have 0 variance). Hence we'd actually have $$\text{Var}(2X) = \text{Var}(X + Y) = 0$$ in this case.
Like Sangchul Lee said, the way to prove your claim is to note that $$(X ,Y)$$ and $$(Y, X)$$ have the same joint distribution.
• iid doesn't mean $X=Y$. Suppose you roll two identical dice. Apr 24, 2019 at 4:24
• @S.PhilKim That's my point, if $X$ and $Y$ are equal then they can't be i.i.d. unless they're constant. Hence his classmate's counterexample is not a counterexample. Apr 24, 2019 at 4:27
• Yes, right. I misread your answer. Apr 24, 2019 at 4:42
iid: Independent and Identical Distribution.
If you roll two identical dice, you don't get two identical number($$2X$$) every time.
$$2X$$ is not the same as $$X+Y$$(Sum of two dice). Not even in distribution.
$$2X$$ will be either 2,4,6,8,10,12, and the probabilities are all the same $$\frac{1}{6}$$.
However, $$X+Y$$ will be one of 2,3,4,5,6,7,8,9,10,11,12 and the probabilities are all different.
Pr($$X+Y=12$$) = $$\frac{1}{36}$$ but Pr($$X+Y=6$$) = $$\frac{5}{36}$$
Indeed this example is famous as an example of the central limit theorem.
Therefore, Var($$2X$$) $$\ne$$ Var($$X+Y$$) is what it is supposed to be.
In fact, Let Var($$X$$) = $$\sigma^2$$, then Var($$2X$$) = 4$$\sigma^2$$, and Var($$X+Y$$) = 2$$\sigma^2$$.
Since we don't distinguish these two dice (identical), $${E}(X|X+Y) ={E}(Y|X+Y)$$
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Description:
The apparatus and set up used demonstrates the magnitude and direction of centripetal forces.
Watch Video:
If you liked this video, check these ones out:
Possible Incorporated Topics:
• Centripetal Force
• Spring Force
Theory:
An object moving in a circle at a constant speed is said to be undergoing "uniform circular motion". Though the speed of the object is not changing, the direction of motion is which implies the object is experiencing acceleration. To cause this acceleration, there must be a force on the object. This force is directed toward the center of the circle, so it is called a "centripetal force".
The machine used in this demonstration (Fig.1) can be used to show some properties of centripetal force. It consists of a mass (M) hanging by a string from a bar on a rotating post (A). The vertical post (B) and the mass are joined by a spring (S), which is responsible for exerting the centripetal force that keeps the mass moving in a circle. Watching the video, one can see not only that the centripetal force is directed toward the center of the circle, but also that it is possible to calculate the magnitude of the centripetal force using this set-up. As the mass rotates faster and faster, it stretches the spring and moves in a circle with a larger and larger radius. Once the circle is large enough, the mass strikes a flexible post (P). Since the spring is exerting the centripetal force, the magnitude of the spring force is equal to the centripetal force. Calculating the spring force at this radius will therefore give the magnitude of the centripetal force.
Apparatus:Three metal rods; two that can interlock to form a "T" shape; one rod must hold a mass and the other must have a spring connected to it (A & B)Various hanging massesMass hangerSpring (S)Flexible metal rod (P)String which connects to mass M and the mass hangerPulley attached to a metal rod Figure 1: Labelled Apparatus
Procedure:
Attach a string to the side of the mass facing the pulley and place it over the pulley.
• Refer to Figure 1 for set-up. Make sure everything is securely attached since the apparatus may end up being spun fairly quickly! (In the filmed demonstration, the entire apparatus is clamped to the table to eliminate wobbling)
• Spin the rotating post faster and faster until the mass, M, just touches the flexible vertical post (P). (This demonstrates the direction of the centripetal motion)
To calculate the magnitude of the centripetal force at the radius of the post:
• Attach a mass hanger to the other end of the string.
• Place hanging masses onto the mass hanger until the spring is stretched enough that the mass, M, touches the vertical post.
• The spring force (and therefore the centripetal force) is then the total mass added to the mass hanger times the acceleration due to gravity, i.e.: Fcentripetal= mg
SAFETY WARNINGS:
• Ensure the mass attached to post A is screwed on tight; otherwise it will fly off when spinning.
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Content of present website is being moved to www.lukoe.com/finance . Registration of www.opentradingsystem.com will be discontinued on 2020-08-14.
I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
1 Black-Scholes formula.
2 Change of variables for Kolmogorov equation.
3 Mean reverting equation.
4 Affine SDE.
5 Heston equations.
A. Affine equation approach to integration of Heston equations.
B. PDE approach to integration of Heston equations.
6 Displaced Heston equations.
7 Stochastic volatility.
8 Markovian projection.
9 Hamilton-Jacobi Equations.
IV. Data Analysis.
V. Implementation tools.
VI. Basic Math II.
VII. Implementation tools II.
VIII. Bibliography
Notation. Index. Contents.
## Affine equation approach to integration of Heston equations.
he Heston equations are of the affine type. Hence, we proceed as discussed in the section ( section about Affine equation ).
We transform the equations to the -form ( Affine equation ). Let , hence In particular, According to the summary ( Affine characteristic function 1 )-( Affine boundary conditions ) the function is given by the expression where the functions and must satisfy the following system of ODEs: We transform the above relationships: The are to satisfy the final conditions: The next task is to solve the equation We introduce the convenience notation and write We perform the change of the unknown function as follows We perform the transformation as follows: In addition, Hence, and, consequently, We arrived to a linear equation. We look for solutions of the form It suffices to have We mean to integrate over because this is the argument of the characteristic function and we want to take the inverse Fourier or Laplace transform. Note that when the expression under the square root is positive. For large it is negative. Hence, we would rather change for real . Then the square root is never zero. Indeed, the real part would be , where the correlation is not greater then 1.
We perform the backward substitutions We introduce the quantity according to the relationship and transform the expression for to emphasize that the requirement uniquely identifies the .
We now consider the equation for : where the is a constant and the has just been calculated: We have hence It remains to evaluate the integral Since satisfies the final condition we have
Notation. Index. Contents.
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• Resource ID: M7M2L17
• Grade Range: 7
• Subject: Math
Using Theoretical and Experimental Probability to Make Predictions
Given an event to simulate, the student will use theoretical probabilities and experimental results to make predictions and decisions.
• Resource ID: K2KA103
• Grade Range: 8–10
• Subject: Math
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Kid2Kid videos on determining the meaning of slope and intercepts in English and Spanish
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Creating Nets for Three-Dimensional Figures
Given nets for three-dimensional figures, the student will apply the formulas for the total and lateral surface area of three-dimensional figures to solve problems using appropriate units of measure.
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Finding the Probabilities of Dependent and Independent Events
Given problem situations, the student will find the probability of the dependent and independent events.
• Resource ID: M7M2L20
• Grade Range: 7
• Subject: Math
Recognizing Misuses of Graphical or Numerical Information
Given a problem situation, the student will analyze data presented in graphical or tabular form by evaluating the predictions and conclusions based on the information given.
• Resource ID: M8M1L2*
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• Subject: Math
Approximating the Value of Irrational Numbers
Given problem situations that include pictorial representations of irrational numbers, the student will find the approximate value of the irrational numbers.
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Expressing Numbers in Scientific Notation
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Comparing and Ordering Rational Numbers
Given a problem situation, the student will compare and order integers, percents, positive and negative fractions and decimals with or without a calculator.
• Resource ID: M8M2L15*
• Grade Range: 8
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Determining if a Relationship is a Functional Relationship
The student is expected to gather and record data & use data sets to determine functional relationships between quantities.
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Graphing Dilations, Reflections, and Translations
Given a coordinate plane, the student will graph dilations, reflections, and translations, and use those graphs to solve problems.
• Resource ID: M8M2L3*
• Grade Range: 8
• Subject: Math
Graphing and Applying Coordinate Dilations
Given a coordinate plane or coordinate representations of a dilation, the student will graph dilations and use those graphs to solve problems.
• Resource ID: M8M2L4*
• Grade Range: 8
• Subject: Math
Developing the Concept of Slope
Given multiple representations of linear functions, the student will develop the concept of slope as a rate of change.
• Resource ID: M7M2L6
• Grade Range: 7
• Subject: Math
Using Multiplication by a Constant Factor
Given problems involving proportional relationships, the student will use multiplication by a constant factor to solve the problems.
• Resource ID: M7M2L2
• Grade Range: 7
• Subject: Math
Predicting, Finding, and Justifying Data from a Table
Given data in table form, the student will use the data table to interpret solutions to problems.
• Resource ID: M7M2L4b
• Grade Range: 7
• Subject: Math
Predicting, Finding, and Justifying Data from Verbal Descriptions
Given data in a verbal description, the student will use equations and tables to solve and interpret solutions to problems.
• Resource ID: M8M2L14*
• Grade Range: 8
• Subject: Math
Comparing and Contrasting Proportional and Non-Proportional Linear Relationships
Given problem solving situations, the student will solve the problems by comparing and contrasting proportional and non-proportional linear relationships.
• Resource ID: E8WrM1L1
• Grade Range: 8
• Subject: ELA & Reading
Writing Literary Text with an Engaging Story Line
You will learn how to write an imaginative story that sustains reader interest and includes well-paced action, an engaging story line, and a believable setting.
• Resource ID: E8WrM1L2
• Grade Range: 8
• Subject: ELA & Reading
Write Literary Text That Develops Interesting Characters
You will learn how to write an imaginative story that develops interesting characters and believable dialogue.
• Resource ID: E8WrM1L3
• Grade Range: 8
• Subject: ELA & Reading
Write Literary Text That Uses Literary Strategies/Devices to Enhance the Style and Tone
You will learn how to write an imaginative story that uses literary strategies/devices to enhance style and tone.
• Resource ID: E8WrM1L4
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• Subject: ELA & Reading
Write a Personal Narrative
You will learn how to write a personal narrative that has a defined focus and includes reflections about decisions, actions, and/or consequences.
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### Quaternions and Rotations
Find out how the quaternion function G(v) = qvq^-1 gives a simple algebraic method for working with rotations in 3-space.
### Quaternions and Reflections
See how 4 dimensional quaternions involve vectors in 3-space and how the quaternion function F(v) = nvn gives a simple algebraic method of working with reflections in planes in 3-space.
# Two and Four Dimensional Numbers
##### Stage: 5 Challenge Level:
This problem is about establishing the existence of complex numbers and of quaternions by providing two models.
Real numbers fill a line (1 dimension), complex numbers fill a plane (2 dimensions) and quaternions fill 4-dimensional space. If we assume that real numbers exist (that is, if we know about the arithmetic of real numbers) can we be sure that complex numbers and quaternions exist?
In part (1) we study a model where the arithmetic and algebra is isomorphic to the arithmetic and algebra of complex numbers.
In part (2) we extend the model to one where the arithmetic and algebra is isomorphic to that of quaternions.
To add and take scalar multiples of quaternions just treat them like 4 dimensional vectors, for example: $$(a_1 +b_1{\bf i} + c_1{\bf j} + d_1{\bf k}) + (a_2 +b_2{\bf i} + c_2{\bf j} + d_2{\bf k})= (a_1+a_2) + (b_1+b_2){\bf i} + (c_1+c_2){\bf j} + (d_1+d_2){\bf k}).$$ Multiplication is defined by the rules of ordinary algebra where $${\bf i}^2={\bf j}^2={\bf k}^2=-1,\quad {\bf i j} = {\bf k} = {\bf -j i}, \quad {\bf j k} = {\bf i} = {\bf -k j},\quad {\bf k i} = {\bf j} = {\bf -i k} .$$ We can look at quaternions in three different but equivalent ways.
(1) Quaternions are points $(a,b,c,d)$ in 4-dimensional space ${\bf R}^4$.
(2) Quaternions are ordered pairs of complex numbers $(z,w)= (a+i b, c+i d)$. As such they are elements of ${\bf R}^2\times {\bf R}^2$.
(3) Quaternions are ordered pairs consisting of a real number $a$ and a vector $x i+y j+z k$ in 3-space that is they are elements of ${\bf R}\times {\bf R}^3$.
In the problems and articles on this site we mainly use the third representation.
While real numbers and complex numbers form fields, the arithmetic and algebra of quaternions is the same as a field in all respects except that multiplication is not commutative. For this reason the structure for the algebra of quaternions is called a skew field.
To read about number systems, where quaternions fit in, why there are no three dimensional numbers and numbers in higher dimensions, see the NRICH article What Are Numbers?
Notes here
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# Thread: 2 Linear algebra questions
1. ## 2 Linear algebra questions
1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.
I can do it in the complex case by proving it is equal to it's conjugate but it doesn't work in the real case.
2) if A and B are similar, that is if AP=PB for invertible P, is there an easy way to find powers of A?
2. ## Re: 2 Linear algebra questions
Originally Posted by boromir
1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.
I can do it in the complex case by proving it is equal to it's conjugate but it doesn't work in the real case.
2) if A and B are similar, that is if AP=PB for invertible P, is there an easy way to find powers of A?
2) $\displaystyle AP=PB$
$\displaystyle A=PBP^{-1}$
$\displaystyle A^2=PBP^{-1}PBP^{-1}=PB^2P^{-1}$
By induction, we get $\displaystyle A^n=PB^nP^{-1}$
3. ## Re: 2 Linear algebra questions
1) If $\displaystyle A$ is a real matrix then $\displaystyle A$ is an hermitian matrix if and only if $\displaystyle A$ is a symmetric matrix. You should know all the eigenvalues of an hermitian matrix are real (do you know why?). So if $\displaystyle A$ is symmetric and real then $\displaystyle A$ is an hermitian matrix thus the eigenvalues of $\displaystyle A$ are real.
4. ## Re: 2 Linear algebra questions
Originally Posted by boromir
1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.
I suppose you mean $\displaystyle A=(a_{ij})\in M_{n\times n}(\mathbb{R})$ , otherwise the result is not true. Another way (without using the concept of hermitian matrix): let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$ (a priori complex) and $\displaystyle 0\neq X=(x_i)\in\mathbb{C}^n$ such that $\displaystyle AX=\lambda X$ . Then, $\displaystyle \bar{X}^tAX=\lambda \bar{X}^tX$ . But $\displaystyle \bar{X}X=\sum \bar{x_i}x_i=\sum |x_i|^2\in\mathbb{R}-\{0\}$ . On the other hand $\displaystyle \bar{X}^tAX=\sum a_{ij}x_i\bar{x_j}\in\mathbb{R}$ because its conjugate is the same: $\displaystyle \sum a_{ij}\bar{x_i}x_j\in\mathbb{R}$ ($\displaystyle A$ is symmetric). So, $\displaystyle \lambda=\frac{\bar{X}^tAX}{\bar{X}^tX}\in\mathbb{R }$ .
5. ## Re: 2 Linear algebra questions
Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on $\displaystyle {C}^n$. Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,
t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.
Is this the correct reasoning?
6. ## Re: 2 Linear algebra questions
Originally Posted by boromir
Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on $\displaystyle {C}^n$. Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,
t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.
Is this the correct reasoning?
I think that is correct. I have a slightly different proof:
Let $\displaystyle A$ be an hermitian $\displaystyle (n,n)$ matrix and supposte $\displaystyle x\in \mathhb{C}^n$ is an eigenvector of $\displaystyle A$ and suppose $\displaystyle c$ is the corresponding eigenvalue. Then we have $\displaystyle A\cdot x=c\cdot x$ and thus $\displaystyle ^H x\cdot A\cdot x=c\cdot ||x||^2$
$\displaystyle \Rightarrow \overline{c}\cdot ||x||^2=^H(^H x\cdot A\cdot x)=^H x \cdot A \cdot x=c\cdot ||x||^2$
Therefore $\displaystyle \overline{c}=c$ and thus $\displaystyle c$ is real.
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www.reference.com Math Numbers
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1. ## Financial Mathematics
This problem is in the sequences and series review section of my book, so could someone help me with the problem using sequences and series concepts?
Michele invested 1500 francs at an annual rate of interest of 5.25 per cent.
a) Find the value of Michele's investment after 3 years. Give your answer to the nearest franc.
b) How m any complete years will it take for Michele's investment to double in value?
c) What should the interest rate be if Michele's initial investment were to double in value in 10 years?
What I did was...
a) 1500 x 1.0525^2 = $1662 (which is incorrect, the book has it as:$1749
b) I got the correct answer: 14 complete years
c) 3000 = 1500r^9
r=1.08
Therefore, the interest rate is 8% (which is incorrect, the book has it as: 7.18%)
What did I do wrong, and what are the correct steps to the solutions? Thanks!
2. Originally Posted by bhuang
This problem is in the sequences and series review section of my book, so could someone help me with the problem using sequences and series concepts?
Michele invested 1500 francs at an annual rate of interest of 5.25 per cent.
a) Find the value of Michele's investment after 3 years. Give your answer to the nearest franc.
b) How m any complete years will it take for Michele's investment to double in value?
c) What should the interest rate be if Michele's initial investment were to double in value in 10 years?
What I did was...
a) 1500 x 1.0525^2 = $1662 (which is incorrect, the book has it as:$1749
b) I got the correct answer: 14 complete years
c) 3000 = 1500r^9
r=1.08
Therefore, the interest rate is 8% (which is incorrect, the book has it as: 7.18%)
What did I do wrong, and what are the correct steps to the solutions? Thanks!
In both a and c you have used the wrong time. In (a) you should use 3 not 2, in (c) use 10 not 9.
If you are using the a x r^(n-1) formula, you need to be careful with what n is. Draw a timeline.
3. For part c, you need to find the following:
$\displaystyle 1500*r^{10} = 3000$
A bit of simple arithmetic gives us:
$\displaystyle r^{10}=2$
taking the log of each side:
$\displaystyle 10log(r) = log 2$
which simplifies to
$\displaystyle log(r) = \frac{log 2}{10}$
take the inverse log of this gives you the answer 1.07177: 7.18% interest.
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# Ordering Decimals On A Number Line Worksheet
Ordering Decimals On A Number Line Worksheet – Fantastic, exciting and totally free math worksheets must be able to provide a numerical issue in different ways. Math is after all simply a numeric manifestation of some of life’s easiest questions: How much cash do you have left should i buy a soda? In the end of each week, how much of my everyday allowance will I be able to conserve if I don’t?
Each time a youngster learns to relate math to daily questions, he is going to be great at it from your simplest addition all the way to trigonometry. To transform percentages, decimals and fractions is thus one essential skill. The amount of an apple inc cake continues to be consumed? The solution to this query could be conveyed in percentages, 50%; or perhaps in decimals, .5; or perhaps in fraction, ½. Quite simply, [%half of|50 % of|one half of%] mom’s tasty apple inc cake is gone. The number of youngsters in school have performed their groundwork? Once more this can be clarified in several techniques: in rates, 70%; or even in percentage, 7:10; Both these indicate out of 15 kids in class there are 7 good types who did and a few not-so-excellent kinds who didn’t. The end result is that kids learn math significantly better when it makes sense.
Thus, the Ordering Decimals On A Number Line Worksheet that you get to your youngsters ought to include intriguing term problems that help them to with all the useful use of the lessons they find out. It must also provide the identical difficulty in many different methods to ensure that a child’s understanding of any topic is further and thorough.
There are numerous standard workout routines which train individuals to convert percentages, decimals and fractions. Changing percentage to decimals for example is really as basic as shifting the decimal point two places on the left and losing the [%percent|%|percentage%] indication “%.” Hence 89% is the same as .89. Expressed in fraction, that would be 89/100. Whenever you drill youngsters to get this done usually enough, they learn to do conversion process nearly instinctively.
Ratios and dimensions are likewise great math training with lots of fascinating practical programs. If 3 pans of pizza, one particular kilo of pasta, two buckets of chicken can effectively nourish 20 starving close friends, then just how much pizza, pasta and poultry does mommy must prepare for birthday party with 30 children?
## Start Teaching your kids with Ordering Decimals On A Number Line Worksheet
Wouldn’t it be fantastic if your kid figured out how you can mathematically physique this out? To help teach him, begin by offering him plenty of totally free Ordering Decimals On A Number Line Worksheet!
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# How many grams are in 12 moles of carbon?
Contents
## How much is a mole of carbon-12?
Exactly 12 grams of pure carbon-12 powder is known as one mole. The number of atoms of carbon-12 present in this one mole sample is 6.022 136 7 x 1023. This number is known as Avogadro’s number.
1.992×10−23 gm.
## How many grams is 5 moles of carbon?
Carbon-12 refers to the isotope of carbon that has an atomic mass of 12.0000 g. Thus, the mass of one mole of carbon-12 is 12.0000 g. Therefore, the mass of 5 moles of C-12 is 5 × 12.0000 = 60.0000 g.
## Why is a mole 12 grams of carbon-12?
as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.022 X 1023). So the mole is the title used for the amount 6.022 x 1023 much the same way the word “dozen” is used for the amount 12. … This means that the atomic mass or atomic weight (12 grams) of carbon is equal to exactly 1 mole of carbon.
## What is the 12 in carbon?
The 12 in carbon-12 is the mass number of the isotope. This means that all atoms of carbon-12 have a mass of 12 atomic mass units.
## Why do we use carbon-12 as the basis for the mole?
Carbon-12 is the basis for the mole because the atomic mass of 12 grams of Carbon-12 is exactly the Avogadro’s number which defines a mole.
THIS IS AMAZING: Frequent question: Can Aztec clay remove pimples?
## Why is carbon 12 the standard?
Carbon-12 is of particular importance in its use as the standard from which atomic masses of all nuclides are measured, thus, its atomic mass is exactly 12 daltons by definition. Carbon-12 is composed of 6 protons, 6 neutrons, and 6 electrons.
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# What does P Y mean on a financial calculator?
44
payments per year
In this regard, what is PY and CY on financial calculator?
I/Y – nominal annual rate of interest per year (entered as a %; NOT a decimal) C/Y – # of interest compounding periods per year P/Y – # of payment periods per year PV – present value (the amount of money at the beginning of the transaction.)
Subsequently, question is, how do you calculate compound interest on a financial calculator? Formulas where n = 1 (compounded once per period or unit t)
1. Calculate Accrued Amount (Principal + Interest) A = P(1 + r)t
2. Calculate Principal Amount, solve for P. P = A / (1 + r)t
3. Calculate rate of interest in decimal, solve for r. r = (A/P)1/t - 1.
4. Calculate rate of interest in percent.
5. Calculate time, solve for t.
One may also ask, what does PMT mean on a financial calculator?
Payment (PMT) This is the payment per period. To calculate a payment the number of periods (N), interest rate per period (i%) and present value (PV) are used.
What does f01 mean on financial calculator?
C01 is cash flow at time period 1. ? F01 = frequency of C01, and so on.
Pundit
## How is interest rate calculated?
Divide your interest rate by the number of payments you'll make in the year (interest rates are expressed annually). So, for example, if you're making monthly payments, divide by 12. 2. Multiply it by the balance of your loan, which for the first payment, will be your whole principal amount.
Pundit
## What does N mean on a financial calculator?
PV (Present Value) \$27,360.09. N (Number of Periods) 10.000. I/Y (Interest Rate)
Pundit
## What is the present value formula?
Present Value Formula
PV = Present value, also known as present discounted value, is the value on a given date of a payment. r = the periodic rate of return, interest or inflation rate, also known as the discounting rate.
Pundit
## What is P Y and C Y on TI 84?
P/Y stands for "payments per year." If you set this value to, say, 12 then the calculator will assume monthly compounding and adjust the interest rate appropriately. For example, if you have quarterly payments but the interest rate is compounded monthly, then you would set P/Y to 4 and C/Y to 12.
Pundit
## How do I calculate future value?
The Future Value Formula
PV is the present value and INT is the interest rate. You can read the formula, "the future value (FVi) at the end of one year equals the present value (\$100) plus the value of the interest at the specified interest rate (5% of \$100, or \$5)."
Teacher
## What is a TVM Solver?
You can use the TVM Solver on the TI-83 graphing calculator to find the future and the present value of money. In the context of a savings or investment account, the future value of your money is the amount of money in the account after a specified time period.
Teacher
## What does PMT stand for?
PMT is an abbreviation for premenstrual tension.
Teacher
## What does C Y stand for in finance?
What does CY stand for?
Rank Abbr. Meaning
CY Current Year
CY Container Yard
CY Current Yield (finance)
Teacher
## What is P Y and C Y on TI 83?
P/Y stands for "payments per year." If you set this value to, say, 12 then the calculator will assume monthly compounding and adjust the interest rate appropriately. For example, if you have quarterly payments but the interest rate is compounded monthly, then you would set P/Y to 4 and C/Y to 12.
Reviewer
## How do you solve PMT?
PMT formula examples
To calculate a loan payment amount, given an interest rate, the loan term, and the loan amount, you can use the PMT function. In the example shown, the formula in C10 is: =PMT(C6/12,C7,-C5) How this formula works To solve for an annuity payment, you can use the PMT function.
Reviewer
## What is the annuity formula?
The annuity payment formula is used to calculate the periodic payment on an annuity. An annuity is a series of periodic payments that are received at a future date. The present value portion of the formula is the initial payout, with an example being the original payout on an amortized loan.
Reviewer
## What is the best financial calculator?
So here is my list of The 10 Best Financial Calculators.
• HP 12CP Financial Calculator.
• Calculated Industries 3405 RE Financial Calculator.
• Texas Instruments BAII Plus Financial Calculator.
• HP 10bII+ Financial Calculator.
• HP 17BII+ Financial Calculator.
• Casio FC-200V Financial Calculator.
• SwissMicros DM15L Financial Calculator.
Beginner
## What is the principal in compound interest?
P = principal amount (the initial amount you borrow or deposit) r = annual rate of interest (as a decimal) t = number of years the amount is deposited or borrowed for. A = amount of money accumulated after n years, including interest.
Beginner
## What does compounded quarterly mean?
Compounded quarterly means, you do it for every three months. So after every three months, your interest will be added to principal and the total sum becomes the principal for next quarter. But, if you use simple interest, then after two quarters, the interest would be \$60 and the principal amount would never change.
Co-Authored By:
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Other
# How do you find unit rate?
## How do you find unit rate?
To find the unit rate, divide the numerator and denominator of the given rate by the denominator of the given rate. So in this case, divide the numerator and denominator of 70/5 by 5, to get 14/1, or 14 students per class, which is the unit rate.
### How do you find the cost per gallon?
Take the mileage of the total distance of your trip and divide it by your miles per gallon to get the number of gallons of gas you will need on your trip. Then multiply that figure by the current price of gas, and the result is the estimated cost of gas for your road trip.
#### What is the unit rate for a car that can travel 256 miles on 8 gallons of gas?
8 goes into 256 32 times, so the unit rate is 32/1. 32 miles per 1 gallon. So divide 256 by 32 and you’ll get 8.
Which rate is equivalent to 15 miles per gallon?
The equivalent unit rate for 15 miles per gallon is 60 miles per 4 gallons ( ) .
What is a unit rate give an example?
A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate.
## What is the difference between a rate and a unit rate?
A. The difference between a rate and a unit rate is that a rate is the ratio between two different units of measure, while a unit rate is the ratio of between two different units of measure for a single thing.
### What is the unit rate of 297 words in 5.5 minutes?
Answer: The unit rate of 297 words for 5.5 minutes is, 54 words per minutes.
#### How to calculate how much gas you use per mile?
Record the number of gallons or liters required to fill the tank once again. This is the total number of gallons or liters you used for the trip (or the time period). Record the trip ending odometer reading (this might also be the starting reading for your next trip). What is it costing you in gas per mile? Enter the price per gallon.
What’s the difference between miles per gallon and miles per liter?
When you calculate miles/gallon (mpg) or kilometers/liter (km/l) you are calculating fuel economy in terms of distance per unit volume or distance/volume. The following outline is generally applicable to both calculations. When you calculate liters per 100 kilometers (l/100km) you are calculating volume per 100 units of distance.
How to calculate nautical miles per gallon ( mpg )?
1 nmi/gal [US] = 0.9582212144591 MPG [US]. 1 x 0.9582212144591 MPG [US] = 0.9582212144591 Miles Per Gallon US. Always check the results; rounding errors may occur.
## How many miles can you go with 30 gallons of gas?
But you aren’t going to be able use every last drop of gas before you refill so say you use 30 gallons and refill when you have about 1/4 of a tank left. That means you can go about 300 miles before you have to refill. 30 gal x 10 mpg = 300 miles.
### Which is correct mpg or miles per gallon?
Always remember mpg means miles per gallon and “per” always means “divided by”. It always helps to sort out the units so you can find out the formula. MPG could be written as m/g. So the units would be as follows: miles / (miles/gallon) = gallons.
#### Do you need to enter number of miles to calculate gas?
For a successful calculation you need to enter at least the number of miles to travel on your trip as well as the mpg of you car or vehicle. If you also enter the gas price you expect to pay we can also give you an estimate of what the gas would cost for your trip.
What is the formula to calculate miles per gallon?
The formula for calculating Miles Per Gallon is simply miles driven divided by gallons used. So to calculate the miles driven you take the miles at the end of the tank and subtract the Miles at the start of the tank, and then you divide that by the total gallons of gas used.
But you aren’t going to be able use every last drop of gas before you refill so say you use 30 gallons and refill when you have about 1/4 of a tank left. That means you can go about 300 miles before you have to refill. 30 gal x 10 mpg = 300 miles.
Ruth Doyle
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# Difference between revisions of "1986 AIME Problems/Problem 7"
## Problem
The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.
## Solutions
### Solution 1
Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$, so in binary form we get $1100100$. However, we must change it back to base 10 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}$.
### Solution 2
Notice that the first term of the sequence is $1$, the second is $3$, the fourth is $9$, and so on. Thus the $64th$ term of the sequence is $729$. Now out of $64$ terms which are of the form $729$ + $'''S'''$, $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$, i.e. $972$, is greater than the largest term which does not, or $854$. So the $96$th term will be $972$, then $973$, then $975$, then $976$, and finally $\boxed{981}$
### Solution 3
After the $n$th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$th power of 3 is equal to the number of terms before the $n$th power, because those terms after the $n$th power are just the $n$th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of $3$ and the terms that come after them, we see that the $100$th term is after $729$, which is the $64$th term. Also, note that the $k$th term after the $n$th power of 3 is equal to the power plus the $k$th term in the entire sequence. Thus, the $100$th term is $729$ plus the $36$th term. Using the same logic, the $36$th term is $243$ plus the $4$th term, $9$. We now have $729+243+9=\boxed{981}$
### Solution 4
Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$. From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ = $729$ and the 128th term is $3^7$ = $2187$. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the ($2^n$+$2^{n+1}$)/2 term is equal to $3^n$ + $3^{n-1}$. From here, we know that the 96th term is $3^6$ + $3^5$ = $972$. From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is $972 + 1 = 973$, the 98th term is $972 + 3 = 975$, the 99th term is $972 + 3 + 1 = 976$, and finally the 100th term is $972 + 9 = \boxed{981}$
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# How to Find the Surface Area of a Triangular Prism?
To help visualize a triangular prism, imagine a classic camping tent. Prisms are three-dimensional shapes, with two identical polygon ends. These polygon ends dictate the prism's overall shape sinc... Read More »
http://www.ehow.com/how_8124824_surface-area-triangular-prism.html
Top Q&A For: How to Find the Surface Area of a Triangular Prism
## How to Find the Area of a Triangular Prism?
A prism is defined as a solid figure with a uniform cross section. There are many different types of prisms, from rectangular to circular to triangular. You can find the surface area of any type of... Read More »
http://www.ehow.com/how_8165114_area-triangular-prism.html
## What is the formula to find the volume of a triangular prism?
The formula for the volume of a triangular prism is 1/2 x B x H x L where B and H represent the base and height of the triangle and L represents the length of the triangular prism.Source:Maths Teac... Read More »
## What is the surface area&volume of a triangular pyramid?
The surface area for a triangular pyramid is determined by the area of the base plus ½ times the perimeter x the length of the side (surface area = base area + ½ perimeter x side length). The vol... Read More »
## How to Calculate the Surface Area of a Rectangular Prism?
A rectangular prism is a fancy name for a 6-sided object that is very familiar to everybody—the box. Think of the common brick, or a shoebox, and you know exactly what a rectangular prism is. Thi... Read More »
http://www.wikihow.com/Calculate-the-Surface-Area-of-a-Rectangular-Prism
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JEE Advanced Level Test: Vector - JEE MCQ
# JEE Advanced Level Test: Vector - JEE MCQ
Test Description
## 30 Questions MCQ Test Mathematics (Maths) Class 12 - JEE Advanced Level Test: Vector
JEE Advanced Level Test: Vector for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The JEE Advanced Level Test: Vector questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Vector MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Vector below.
Solutions of JEE Advanced Level Test: Vector questions in English are available as part of our Mathematics (Maths) Class 12 for JEE & JEE Advanced Level Test: Vector solutions in Hindi for Mathematics (Maths) Class 12 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Advanced Level Test: Vector | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) Class 12 for JEE Exam | Download free PDF with solutions
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JEE Advanced Level Test: Vector - Question 1
### If the vector is collinear with the vector (2√2, -14) and = 10, then
Detailed Solution for JEE Advanced Level Test: Vector - Question 1
a = 2(2)½ i - j + 4k
|b| = 10
[(2(2λ)2 + (1λ)2 + (4λ)2]½
=[ 8λ2 + λ2 + 16λ2]½ = 10
= 25λ2 = (10)2
25λ2 = 100
λ = +-2
b = +-(4(2)½ i - 2j + 8k)
Therefore, 2a +- b
JEE Advanced Level Test: Vector - Question 2
### The vertices of a triangle are A(1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the internal bisector of the angle A is
Detailed Solution for JEE Advanced Level Test: Vector - Question 2
Let AD is the bisector of ∠A.Then,
BD/DC = AB/ACeq(1)
Given the vertices of triangle,
AB = √32 + 22 + 12 = √14
AC = √12 + 22 + 32 = √14
As AB = BC, So, BD = DC(from eq(1))
It means D is middle point of BC. So, vertices of D will be (3,3,3).
So, vector AD will be 2iˆ+ 2jˆ+ kˆ.
JEE Advanced Level Test: Vector - Question 3
### Let . The point of intersection of lines
JEE Advanced Level Test: Vector - Question 4
If and then is equal to
JEE Advanced Level Test: Vector - Question 5
Angle between diagonals of a parallelogram whose side are represented by
Detailed Solution for JEE Advanced Level Test: Vector - Question 5
D1 = a+b
D2 = a-b
D1 = 3i + 0j + 0k
D2 = i + 2j + 2k
|D1| = 3
D1.D2 = |D1| . |D2| . cos θ
3 + 0 + 0 = (3) . (3) cos θ
3 = 9 cosθ
cos-1 = (⅓)
JEE Advanced Level Test: Vector - Question 6
Vector make an angle θ = 2π/3. if , then is equal to
Detailed Solution for JEE Advanced Level Test: Vector - Question 6
JEE Advanced Level Test: Vector - Question 7
Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices
A, B, C is
Detailed Solution for JEE Advanced Level Test: Vector - Question 7
Δ = 1/2(a * b) = 1/2(b*c) = 1/2(c*a)
2Δ = a*b = b*c = c*a
unit vector = 1/2Δ[a*b + b*c + c*a]
JEE Advanced Level Test: Vector - Question 8
The value of is equal to the box product
Detailed Solution for JEE Advanced Level Test: Vector - Question 8
Matrix {(1,2,-1) (1,-1,0) (1,-1,-1)} [a b c]
= {1(1) -2(-1) -1(0)} [a b c]
= 3[a b c]
JEE Advanced Level Test: Vector - Question 9
If are two non-collinear vectors such that , then is equal to
JEE Advanced Level Test: Vector - Question 10
Vector of length 3 unit which is perpendicular to and lies in the plane of and
Detailed Solution for JEE Advanced Level Test: Vector - Question 10
JEE Advanced Level Test: Vector - Question 11
Vector satisfying the relation
Detailed Solution for JEE Advanced Level Test: Vector - Question 11
A * X = B
(A * X)*A = B * A
=> -A(x.A) + x(A.A) = B * A
=> -Ac + x|A|2 = B * A
x|A|2 = B * A + Ac
x = [B * A + Ac]/|A|2
JEE Advanced Level Test: Vector - Question 12
If a ,b,c are linearly independent vectors, then which one of the following set of vectors is linearly dependent ?
Detailed Solution for JEE Advanced Level Test: Vector - Question 12
xa + yb + zc = 0
We have to prove x = y = z = 0
a,b,c are non planner
x(a-b) +y(b-c) +c(c-a) = 0
Let x = 1, y = 1, z = 1
So, we get a - b + b - c + c - a = 0
JEE Advanced Level Test: Vector - Question 13
Let be vectors of length 3,4,5 respectively. Let be perpendicular to , and . then
Detailed Solution for JEE Advanced Level Test: Vector - Question 13
|A| = 3, |B| = 4, |C| = 5
Since A.(B + C) = B.(C+A) = C(A+B) = 0...........(1)
|A+B+C|2 = |A| + |B|2 + |C|2 + 2(A.B + B+C + C.A)
= 9+16+25+0
from eq(1) {A.B + B+C + C.A = 0}
therefore, |A+B+C|2 = 50
=> |A+B+C| = 5(2)1/2
JEE Advanced Level Test: Vector - Question 14
Given the vertices A (2, 3, 1), B(4, 1, –2), C(6, 3, 7) & D(–5, –4, 8) of a tetrahedron. The length of the altitude drawn from the vertex D is
JEE Advanced Level Test: Vector - Question 15
for a non zero vector If the equations hold simultaneously, then
JEE Advanced Level Test: Vector - Question 16
The volume of the parallelopiped constructed on the diagonals of the faces of the given rectangular parallelopiped is m times the volume of the given parallelopiped. Then m is equal to
Detailed Solution for JEE Advanced Level Test: Vector - Question 16
Vi=[a→,b→,c→]
(a→ + b→)(b→ + c→)(a→ + c→)
Vf=[(a→ + b→)(b→ + c→)(a→ + c→)]
=(a→ + b→)⋅[(b→ + c→)⋅(a→ + c→)]
=(a→ +b→)[b→⋅a→ + c→⋅a→ + b→⋅c→ + c→⋅c→]
=[b→ c→ a→]+[a→ b→ c→]
Vf=2[a→ b→ c→]
Vf=2Vi.
JEE Advanced Level Test: Vector - Question 17
If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for
JEE Advanced Level Test: Vector - Question 18
The value of a, for which the points A,B,C with position vectors and respectively are the vertices of a right angled triangle with C = π/2 are
Detailed Solution for JEE Advanced Level Test: Vector - Question 18
JEE Advanced Level Test: Vector - Question 19
The distance between the line and the plane is
JEE Advanced Level Test: Vector - Question 20
A particle is acted upon by constant forces which displace ot from a point to the point . The workdone in standard units by the force is given by
JEE Advanced Level Test: Vector - Question 21
If are non-coplaner vectors and λ is a real number, then the vectors are non-coplaner for
JEE Advanced Level Test: Vector - Question 22
Let be non zero vectors such that , If θ is the acute angle between the vectors , then sin θ equals is
Detailed Solution for JEE Advanced Level Test: Vector - Question 22
(a→*b→)*c→ = (1/3)|b→||c→||a→|
⇒ − c→*(a→*b→) = (1/3)|b→||c→||a→|
⇒(c→.a→)b→ −(c→.b→)a→ =(1/3)|b→||c→|a→|
Now, as all of them are non-collinear, (c→⋅a→) can be 0 that means,
(c→.b→) = (−1/3)|b→||c→|
⇒|b→|.|c→|cosθ = (−1/3)|b→||c→|
⇒ cosθ = −1/3
sinθ = √1−(1/3)^2
= √8/9
= (2√2)/√3
JEE Advanced Level Test: Vector - Question 23
are three vectors, such that then is equal to
JEE Advanced Level Test: Vector - Question 24
If are three non-coplaner vectors, then equals
JEE Advanced Level Test: Vector - Question 25
JEE Advanced Level Test: Vector - Question 26
The vectors are the sides of a triangle ABC. The length of the median through A is
Detailed Solution for JEE Advanced Level Test: Vector - Question 26
The length of median through A = vector(AB + BC)/2
= (3i + 4k + 5i - 2j + 4k)/2
= (8i - 2j + 8k)/2
= 4i - j + 4k
Length = √(16 + 1 + 16) = √33
JEE Advanced Level Test: Vector - Question 27
JEE Advanced Level Test: Vector - Question 28
If is equal to
Detailed Solution for JEE Advanced Level Test: Vector - Question 28
∣u × v∣=∣(a − b) × (a + b)∣
=2∣a × b∣ (∵a × a = b × b = 0)
and ∣a × b∣2 + (a ⋅ b)2
=(ab sinθ)+ (abcosθ)2
=a2b2
⇒∣a×b∣ = a2b2−(a ⋅ b)2
So, ∣u × v∣ = 2∣a × b∣
=2[a ^ 2b− (a ⋅ b)2]1/2
=2[(2)2(2)2−(a⋅b)2]1/2
=2(16−(a⋅b)2)1/2
∴∣a∣=∣b∣=2
JEE Advanced Level Test: Vector - Question 29
Let and be three non-zero vectors such that is a unit vector perpendicular to both . if the angle between is π/6, then is equal to
Detailed Solution for JEE Advanced Level Test: Vector - Question 29
According to the given conditions,
(c1)2+c(2)2+(c3)22 =1,a⋅c=0,b⋅c=0
and cosπ/6 = [(3)1/2]/2
(a1b1+a2b2+a3b3]/[(a1)2+(a2)2 + (a3)^2)]^1/2 [(b1)^2 + (b2)2 + (b3)2]1/2
Thus a1c1+a2c2+a3c3=0, b1c1+b2c2+b3c3=0
and [(3)^1/2]/2[(a1)2+(a2)2 + (a3)2)1/2 ((b1)2 + (b2)2 +(b3)2]1/2
=a1b1+a2b2 +a3b3
[(a1)2+(a2)2 + (a3)2] -(a1b1 + a2b2 + a3b3)2
[(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]
= -3/4 [(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]
= 1/4 [(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]
JEE Advanced Level Test: Vector - Question 30
A point taken on each median of a triangle divides the median in the ratio 1 : 3, reckoning from the vertex.
Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is
Detailed Solution for JEE Advanced Level Test: Vector - Question 30
Let A (0,0); B (4m , 0) and C(4p , 4q)
M1 (2m + 2p, 2q)
M2 (2p , 2q) and M3 (2m , 0)
Let E , F and G be the point on the median.
E = (2m + 2p) / 4, 2q / 4) = ((m + p) / 2, q / 2)
F = ((2p + 12m) / 4, (2q + 0) / 4) = ((p + 6m) / 2, q / 2)
G = ((2m + 12p) / 4, (0 + 12 q) / 4) = ((m + 6p ) / 2, 3q)
Area of traingle ABC = 1/2
Area of traingle ABC = 1/2 {(0,0,0) (4m,0,1) (4p,4q,1)}
=1/2(16) = 8 unit
Area of triangle EFG = 1/2 {((m + p) / 2, q / 2, 1)) ((p + 6m) / 2, q / 2, 1) ((m + 6p)/2, 3q, 1)}
= 25/8 unit
ar (EFG) /ar (ABC) = {25 / 8} / 8
= 25/64
## Mathematics (Maths) Class 12
204 videos|288 docs|139 tests
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# polynomial and facor Create a polynomial and factor it using trial factors
## Expert Answers
Since the problem does not specify the polynomial, I suggest you to consider the next polynomial: x^2-7x+6.
You need to form a set comprising all divisors of constant term 6 such that:
D:{-1,-2,-3,-6,1,2,3,6}
You need to find two of these divisors that multipliesd yield 6 such that:
1*6; 2*3,(-1)*(-6),(-2)(-3)
You need to find two of these divisors that added yield 7 such that:
1+6=7
This is the only pair that yields 7.
Hence, since the divisors 1 and 6 respect the simultaneous conditions, the trial factor method lays stress on these two divisors, thus 1 and 6 are the roots of polynomial.
You may write the factored form of polynomial such that:
x^2-7x+6=(x-1)(x-6)
Approved by eNotes Editorial Team
This is a very vague question. A polynomial has to be created and its factors determined.
• Let the polynomial be f(x) = x^2 - 4
The factors of this polynomial are (x - 2) and (x + 4)
• Consider the polynomial f(x) = x^2 + 2x + 1
It can be seen that x^2 + 2x + 1 = (x + 1)^2
The factor of f(x)=x^2 + 2x + 1 is (x + 1)
Approved by eNotes Editorial Team
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# 155615 (number)
155,615 (one hundred fifty-five thousand six hundred fifteen) is an odd six-digits composite number following 155614 and preceding 155616. In scientific notation, it is written as 1.55615 × 105. The sum of its digits is 23. It has a total of 2 prime factors and 4 positive divisors. There are 124,488 positive integers (up to 155615) that are relatively prime to 155615.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 23
• Digital Root 5
## Name
Short name 155 thousand 615 one hundred fifty-five thousand six hundred fifteen
## Notation
Scientific notation 1.55615 × 105 155.615 × 103
## Prime Factorization of 155615
Prime Factorization 5 × 31123
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 155615 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 155,615 is 5 × 31123. Since it has a total of 2 prime factors, 155,615 is a composite number.
## Divisors of 155615
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 186744 Sum of all the positive divisors of n s(n) 31129 Sum of the proper positive divisors of n A(n) 46686 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 394.481 Returns the nth root of the product of n divisors H(n) 3.33323 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 155,615 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 155,615) is 186,744, the average is 46,686.
## Other Arithmetic Functions (n = 155615)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 124488 Total number of positive integers not greater than n that are coprime to n λ(n) 62244 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 14277 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 124,488 positive integers (less than 155,615) that are coprime with 155,615. And there are approximately 14,277 prime numbers less than or equal to 155,615.
## Divisibility of 155615
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 5 7 5
The number 155,615 is divisible by 5.
## Classification of 155615
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (155615)
Base System Value
2 Binary 100101111111011111
3 Ternary 21220110112
4 Quaternary 211333133
5 Quinary 14434430
6 Senary 3200235
8 Octal 457737
10 Decimal 155615
12 Duodecimal 7607b
20 Vigesimal j90f
36 Base36 3c2n
## Basic calculations (n = 155615)
### Multiplication
n×y
n×2 311230 466845 622460 778075
### Division
n÷y
n÷2 77807.5 51871.7 38903.8 31123
### Exponentiation
ny
n2 24216028225 3768377232233375 586416022993996650625 91255129418210788787009375
### Nth Root
y√n
2√n 394.481 53.7878 19.8615 10.9247
## 155615 as geometric shapes
### Circle
Diameter 311230 977758 7.60769e+10
### Sphere
Volume 1.57849e+16 3.04308e+11 977758
### Square
Length = n
Perimeter 622460 2.4216e+10 220073
### Cube
Length = n
Surface area 1.45296e+11 3.76838e+15 269533
### Equilateral Triangle
Length = n
Perimeter 466845 1.04858e+10 134767
### Triangular Pyramid
Length = n
Surface area 4.19434e+10 4.44108e+14 127059
## Cryptographic Hash Functions
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Hi,
I heard about this awesome language from a friend in college. And it has been a joy to use, albeit a bit weird to get the syntax. Basically, instead of making a function call like foo(bar, baz, qux), you put the parenthesis out front and the function name first, with its arguments separated by white space. Here is how the aforementioned function call would look in real Lisp code:
(foo bar baz qux)
This means take foo, and apply it to the arguments bar baz and qux. Now how about a real example?
Suppose that we want to implement the classic n! problem in Lisp. In the languages we know and love, this would look somewhat like this:
def factorial(n):
""" Return the factorial of n."""
if n <= 0: return 1
else: return n*factorial(n-1)
Now, let's take this definition and make it Lispy!
(defun factorial (n)
"Return the factorial of n"
(if (<= n 0)
1
(* n (factorial (- n 1)))))
Now, I guess I've got some 'splaining to do.
We'll take this chunk by chunk:
1. The defun macro basically tells your Lisp interpreter "Okay, interpreter, the next thing is going to be called a function called factorial. It will take as an argument a number n."
2. The "Returns the factorial of n" bit is a docstring, akin to what is in Python.
3. Now, we get to the bit of code that does the heavy lifting. This (if (<= n 0) part is a call to the if function. Its first argument is a call to the <= function, to see if n is <= 0. Note, however, that I did not close off the if list; if I did this, the forms that follow would not evaluate.
4. If the first condition, n <= 0 yields true, we simply return the atom 1 (atoms are essentially basic values in Lisp). This is the base case for our recursive call; 0! = 1 and we need to stop there, else we will recursively loop to infinity.
5. Now, we get to the last bit of code.
a. We'll start out by starting a call to the * function. Its first argument is the value we passed in originally, n.
b. Now, we make a call to the factorial function.
c. Then, we call the - function of n-1. Since n! is defined as n*(n-1)!, this is our recursive call.
6. And now, we close off the lists of function calls; in Lisp, it seems to be customary to put all the closing parenthesis on the last line.
And there you have it, an example of Lisp code. Not exactly revolutionary, but I think it might start an interesting discussion. To test the code and prove it, you'll need, naturally, a Lisp interpreter. As I am on Linux, I use SBCL, which I belive has been ported to Windows and Mac.
Happy hacking!
+1
Lisp is too complicated for me, but It's certainly interesting to study it.
For anyone who wants to learn lisp (or, more accurately, the Scheme derivative), you can check out Andrew Plotkin's Lists and Lists, which is a Scheme interpreter written in Inform.
“Can we be casual in the work of God — casual when the house is on fire, and people are in danger of being burned?” — Duncan Campbell
“There are four things that we ought to do with the Word of God – admit it as the Word of God, commit it to our hearts and minds, submit to it, and transmit it to the world.” — William Wilberforce
+1
This is exactly why I dislike LISP and it's derrivitives. The syntax is entirely comprised of lists, which makes it too confusing for most people to grasp. Not everyone can keep track of a thousand parentheses... at least languages like Python have some common-language elements to them.
"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question." — Charles Babbage.
I have to agree with you Ethin. It seems just too complicated.
“Can we be casual in the work of God — casual when the house is on fire, and people are in danger of being burned?” — Duncan Campbell
“There are four things that we ought to do with the Word of God – admit it as the Word of God, commit it to our hearts and minds, submit to it, and transmit it to the world.” — William Wilberforce
I used to think the same thing. It is much easier when you use an interface like SLIME for Emacs; it will keep track of your parenthesis, give you documentation for function calls, and a whole host of other things.
@king gamer222, that doesn't matter. We shouldn't have to rely on other tools to keep track of something we should be able to keep track of ourselves. LISP is simply no longer a language that very many use -- C++ is far, far more preferable.
"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question." — Charles Babbage.
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You are on page 1of 19
# 1
Matrices
Elemento: aij
a11 a12 a1n
Tamao: m n a21 a22 a2 n
Matriz cuadrada: n n
(orden n)
am1 am 2 amn
Elementos de la diagonal: aii
a1
Vector columna a2
(matriz n x 1)
an
Vector fila
(a1 a2 an )
(matriz 1 x n)
2
2 1 3 4 7 8
Suma:
A 0 4 6 , B 9 3 5
6 10 5 1 1
2
24 1 7 3 (8) 6 6 5
AB 09 43 65 9 7 11
6 1 10 (1) 5 2 5 3
9
## Multiplicacin por un escalar:
ka11 ka12 ka1n
ka21 ka22 ka2 n
kA ( k aij ) mn
kam1 kam 2 kamn
3
Si A, B, C son matrices mn, k1 y k2 son escalares:
(i) A+ B =B +A
(ii) A + (B + C) = (A + B) + C
(iii) (k1k2) A = k1(k2A)
(iv) 1 A = A
(v) k1(A + B) = k1A + k1B
(vi) (k1 + k2) A = k1A + k2A
4
Multiplicacin:
4 7 9 2
(a) A ,B
3 5 6 8
4 * 9 7 * 6 4 * (2) 7 * 8 78 48
AB
3 * 9 5 * 6 3 * (2) 5 * 8 57 34
5 8
(b) 4 3
A 1 0 , B
2 7 2 0
5 * (4) 8 * 2 5 * (3) 8 * 0 4 15
AB 1 * (4) 0 * 2 1 * (3) 0 * 0 4 3
2 * (4) 7 * 2 2 * (3) 7 * 0 6 6
Nota: En general, AB BA
5
Transpuesta de una matriz A:
a11 a21 am1
a12 a22 am 2
A
T
a1n a2 n amn
(i) (AT)T = A
(ii) (A + B)T = AT + BT
(iii) (AB)T = BTAT
(iv) (kA)T = kAT
Nota: (A + B + C)T = AT + BT + CT
(ABC)T = CTBTAT
6
Matriz cero 0 0
0 0 0
0 ,0 , 0 0 0
0 0 0 0 0
A+0=A
A + (A) = 0
Matrices triangulares
2 0 0 0 0
1 2 3 4
1 6 0 0 0
0 5 6 7 8 9 3 0 0
0 0 8 9
1 1 1 2 0
0 0 0 1
15 2 3 4 1
Triangular superior Triangular inferior
7
Matriz diagonal:
Matriz cuadrada n n, i j, aij = 0
7 0 0
5 0 1 0
0 1 0 5
0 5 0 1
2
0 1
0
Matriz identidad:
1 0 0 0
A: m n, entonces 0 1 0 0
Im A = A In = A
0 0 0 1
8
Una matiz An n es simtrica si AT = A.
1 2 7
A 2 5 6
7 6 4
1 2 7
A 2 5 6 A
T
7 6 4
9
Resolucin de sistemas de ecuaciones lineales:
a11x1 a12 x2 a1n xn b1
a21x1 a22 x2 a2 n xn b2
am1x1 am 2 x2 amn xn bm
Matriz aumentada
asociada, para resolver a11 a12 a1n b1
el sistema de ecuaciones
lineales. a21 a22 a2 n b2
am1 am 2 amn bm
10
2x1 + 6x2 + x3 = 7 2 6 1 7 1 2 1 1
R12
x1 + 2x2 x3 = 1 1 2 1 1 2 6 1 7
5x1 + 7x2 4x3 = 9 5 7 4 9 5 7 4 9
2 R1 R2 1 2 1 1 1 2 1 1
5 R1 R3
1
2 R2
0 2 3 9 0 1 3
2
9
2
0 3 0 3
1 14 1 14
1 2 1 1 1 2 1 1
3 R2 R3
2 R
11 3
0 1
3
2
9
2 0 1
3
2
9
2
0 0 11 55 0 0
2 2 1 5
x1 2 x2 x3 1
x3 = 5, x2 = 3, x1 = 10 3
x2 x3
9
2 2
x3 5 11
Resolver mediante el mtodo de Gauss-Jordan
x1 + 3x2 2x3 = 7
4x1 + x2 + 3x3 = 5
2x1 5x2 + 7x3 = 19
1 3 2 7 4 R1 R2 1 3 2 7
2 R1 R3
4 1 3 5 0 11 11 33
2 5 7 19 0 11 11 33
111R2
111R3
1 3 2 7 3 R2 R1 1 0 1 2
R2 R3
0 1 1 3 0 1 1 3
0 1 1 3 0 0
0 0
Entonces: x2 x3 = 3
x1 + x3 = 2
Haciendo x3 = t, tenemos x2 = 3 + t, x1 = 2 t.
12
Resolver: x1 + x2 = 1
4x1 x2 = 6
2x1 3x2 = 8
1 1 1 1 0 1
4 1 6 0 1 2
2 3 0 0 16
8
0 + 0 = 16 !! No tiene soluciones.
13
Determinantes
a11 a12
det A a11a22 a12 a21
a21 a22
a11 a12 a13
det A a21 a22 a23
a31 a32 a33
a11a22a33 a12a23a31 a13a21a32 a13a22a31
a11a23a32 a12a21a33 .
## a22 a23 a21 a23 a21 a22
det A a11 a12 a13
a32 a33 a31 a33 a31 a32
Expansin por cofactores a lo largo de la primera fila.
14
a11 a12 a13 El cofactor de aij es
det A a21 a22 a23 Cij = (1)i+ j Mij
a31 a32 a33 donde Mij se llama menor.
## a22 a23 a21 a23 a21 a22
C11 C12 C13
a32 a33 a31 a33 a31 a32
## ... O por la tercera fila:
det A = a31C31 + a32C32 + a33C33
Podemos expandir por filas o columnas.
15
det AT = det A
5 7 5 3
det A 41 det A
T
41
3 4 7 4
## Si dos filas (columnas) de una matriz A de n n
son idnticas, entonces det A = 0.
6 2 2 6 2 2
A 4 2 2 det A 4 2 2 0
9 2 2
9 2 2
16
Si todos los elementos de una fila (columna) de una
matriz A de n n son cero, entonces det A = 0.
## Si B es la matriz obtenida por intercambio de
dos filas (columnas) de una matriz An n,
entonces:
det B = det A
2 1 3 4 1 9
det B 6 0 7 6 0 7 det A
4 1 9 2 1 3
17
Si B se obtiene de una matriz An n multiplicando
una fila (columna) por un nmero real k, entonces:
det B = k det A
det B kai1Ci1 kai 2Ci 2 kainCin
k (ai1Ci1 ai 2Ci 2 ainCin ) k det A
expansin de det A por cofactores a lo largo de la i -sima fila
5 8 1 8 1 1
5 58
20 16 4 16 4 2
1 1
582 80(1 2) 80
2 1
18
Si A y B son matrices n n, entonces
det AB = det A det B.
2 6 3 4
A , B
1 1 3 5
12 22
AB
6 9
## det AB = 24, det A = 8, det B = 3,
det AB = det A det B.
19
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https://www.shaalaa.com/question-bank-solutions/if-20-of-a-b-then-b-of-20-is-the-same-as-percentage-discount-and-partnership-entrance-exam_108035
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# If 20% of a = B, Then B% of 20 is the Same as - Mathematics
MCQ
If 20% of a = b, then b% of 20 is the same as
#### Options
• 4% of a
• 5% of a
• 20% of a
• None of these
#### Solution
4% of a
Explanation :
20% of a = b
⇒ 20/100a = b
∴ b% of 20 = 20/100axx1/100xx20
4/100a=4% of a
Concept: Percentage, Discount and Partnership (Entrance Exam)
Is there an error in this question or solution?
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https://socratic.org/questions/how-do-you-solve-x-2-8x-10-0
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# How do you solve x^2 + 8x + 10 = 0?
Solve $y = {x}^{2} + 8 x + 10 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 64 - 40 = 24$ -->$d = \pm 2 \sqrt{6}$
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http://geometricfunctions.org/foundations/technology_foundation.shtml
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Technologically Embodied
Geometric Functions
With Technologically Embodied Geometric Functions, students develop conceptual metaphors that directly relate computer-based sensory motor experiences of abstract function concepts. This approach relies on four foundations:
Use the Materials, Support the Work
We’re working hard to create the sketches, worksheets, and support materials for these Web Sketchpad activities. Ours is a volunteer effort on the part of everyone involved, both curriculum developers and field testers, and we make these activities freely available for you to download and use with your own students.
As we develop, revise and expand the activities, we need your help. Email your commentary (What worked well? What didn’t?) and your suggested improvements to the webmaster.
Technology Foundation of Geometric Functions
Existing Technology
Several forms of technology have been used, to greater or lesser extents, in students’ study of function concepts:
• graphing calculators,
• computer algebra systems (CAS systems),
• computer-based rangers (CBRs) (Bazzini, 2001; Borba, 2004), and
• dynagraphs (Goldenberg, Lewis & O’Keefe, 1992).
These are all useful supports as students investigate various kinds of functions. However, it’s worth noting that these technologies are based for the most part on functions defined by algebraic equations, and that the two most common (graphing calculators and CAS systems) provide little opportunity for students to directly vary an independent variable and observe the effect on the dependent variable.
Geometric Functions Technology
The Geometric Functions activities are based on a geometric approach to function concepts, and provide a particularly close connection between students’ sensory-motor activities and the corresponding abstract function concepts. For instance, students drag a physical finger or mouse in two dimensions (on a touch screen or mouse pad) to vary a geometric point (the independent variable) in two dimensions (on the Euclidean plane). As they drag, they observe (and can trace) the motion of a transformed point (the dependent variable) on the two-dimensional screen. This experience provides an extraordinarily close connection between the concrete physical reality and the corresponding abstract concepts: The two-dimensional physical motion corresponds to 2D mathematical variation, the two points (the dragged point and its observed image) correspond to the independent and dependent variables.
In cognitive science terms, we can use the term conceptual metaphor to refer to the connection between the manipulated and observed concrete objects on one hand (the dragged point and its observed transformed image), and the abstract mathematical concepts on the other hand (an independent and a dependent variable related by a function). This connection seems so close and so direct that we might venture to say that it shows a remarkably short metaphorical distance—that is, the concrete reality is extraordinarily similar to the abstract function concepts to which the metaphor refers.
The activities in their current form use The Geometer’s Sketchpad (Jackiw, 2009), due to its user friendliness and its powerful function capabilities. The affordances of Sketchpad allow the activities to address aspects of function (such as function notation, restricted domain, and composition), to incorporate real-time experience of covariation and rate of change, and to exploit powerful function-embodying technology such as locus constructions and custom transformations of arbitrary objects, including pictures. (See the November 2012 Mathematics Teacher article on “Multiple Representations of Composition of Functions.”)
Update History:
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https://rememberingsomer.com/13-60-as-a-percentage/
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It"s really common as soon as learning about fractions to want to know how transform a fraction like 13/60 right into a percentage. In this step-by-step guide, we"ll present you exactly how to turn any fraction into a percent really easily. Let"s take it a look!
Want to easily learn or present students just how to convert 13/60 come a percentage? play this an extremely quick and fun video clip now!
Before we acquire started in the fraction to percentage conversion, let"s walk over some very quick portion basics. Remember the a numerator is the number above the portion line, and also the denominator is the number listed below the portion line. We"ll use this later in the tutorial.
You are watching: 13/60 as a percentage
When we space using percentages, what we room really speak is the the percentage is a fraction of 100. "Percent" means per hundred, and so 50% is the exact same as saying 50/100 or 5/10 in fraction form.
So, because our denominator in 13/60 is 60, we could change the portion to do the denominator 100. To execute that, we division 100 by the denominator:
100 ÷ 60 = 1.6666666666667
Once we have that, we have the right to multiple both the numerator and denominator by this multiple:
13 x 1.6666666666667/60 x 1.6666666666667=21.666666666667/100
Now we have the right to see the our fraction is 21.666666666667/100, which method that 13/60 together a portion is 21.6667%.
We can likewise work this out in a simpler way by an initial converting the portion 13/60 come a decimal. To execute that, we merely divide the molecule by the denominator:
13/60 = 0.21666666666667
Once we have the answer to the division, we deserve to multiply the price by 100 to do it a percentage:
0.21666666666667 x 100 = 21.6667%
And there you have actually it! Two different ways to transform 13/60 come a percentage. Both space pretty straightforward and also easy to do, however I personally prefer the convert to decimal technique as it takes much less steps.
I"ve checked out a most students get confused whenever a question comes up about converting a portion to a percentage, however if you monitor the procedures laid out right here it must be simple. The said, you might still need a calculator for more facility fractions (and you can always use ours calculator in the type below).
If you desire to practice, grab yourself a pen, a pad, and a calculator and shot to transform a few fractions to a portion yourself.
Hopefully this tutorial has actually helped you to understand exactly how to convert a fraction to a percentage. You can now go forth and convert fountain to percentages as much as your tiny heart desires!
If you discovered this content valuable in her research, please do us a an excellent favor and use the tool below to make sure you properly reference us wherever you usage it. We really evaluate your support!
"What is 13/60 as a percentage?". rememberingsomer.com. Accessed on November 12, 2021. Https://rememberingsomer.com/calculator/fraction-as-percentage/what-is-13-60-as-a-percentage/.
"What is 13/60 as a percentage?". rememberingsomer.com, https://rememberingsomer.com/calculator/fraction-as-percentage/what-is-13-60-as-a-percentage/. Accessed 12 November, 2021.
See more: What Is The Gcf Of 96 And 84 And 96? Gcf Of 84 And 96
What is 13/60 as a percentage?. rememberingsomer.com. Retrieved native https://rememberingsomer.com/calculator/fraction-as-percentage/what-is-13-60-as-a-percentage/.
## Fraction as Percentage
Enter a numerator and denominator
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Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is : Kaysons Education
# Let ABCD be A Quadrilateral With Area 18, With Side AB parallel To The Side CD and AB = 2CD. Let AD be Perpendicular To AB and CD. If A Circle Is Drawn Inside The Quadrilateral ABCD touching All The Sides, Then Its Radius Is
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## Question
### Solution
Correct option is
2
Let CD = α so that AB = 2α be two parallel lines. Taking A as origin the co-ordinate are A (0, 0), B(2α, 0), D(0, 2r) and C(α, 2r). Since the circle is touching the axes of co-ordinates it is of form
The above line (2) is a tangent to circle (1). Apply the condition of tangency i.e., p = r we have
Area of quadrilateral i.e. trapezium ABCD is
#### SIMILAR QUESTIONS
Q1
The lines joining the origin to the points of intersection of the line 4x + 3y = 24 with the circle (x – 3)2 + (y – 4)2 = 25 are
Q2
If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1= 0 intersect in two distinct points P and Q then the line
5x + by – a = 0 passes through P and Q for:
Q3
The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Equation of the circle on AB as diameter is:
Q4
A square is formed by following two pairs of straight lines y2 – 14y + 45 = 0 and x2 – 8x + 12 = 0. A circle is inscribed in it. The centre of the circle is
Q5
Let PQ and RS be tangents at the extremities the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals
Q6
A diameter of x2 + y2 – 2x – 6y + 6 = 0 is a chord to circle (2, 1), then radius of the circle is
Q7
If α, β, γ, δ be four angles of a cyclic quadrilateral taken in clockwise direction then the value of will be:
Q8
A square is inscribed in the circle x2 + y2 – 2x + 4y + 3 = 0. Its sides are parallel to the co-ordinate axes. Then one vertex of the square is
Q9
A circle passes through the point (–1, 7) and touches the line y = x at (1, 1). Its diameter is
Q10
Let a circle be given by 2x(x – a) + y(2y – b) = 0, (a 0, b ≠ 0). Find the condition on a and b, if two chords each bisected by the x – axis can be drawn to the circle from
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# Cut and Paste Logic Puzzles for Preschoolers
My oldest were thrilled with this set of pattern block puzzles that I shared recently (get the entire ebook of puzzles here!) and so I decided to make a simpler version for my younger kiddos. These cut and paste logic puzzles are simpler (they are only 3×3 squares instead of 4×4) but still a fun way to get kids using their logical reasoning skills, and learning shapes and colors at the same time!
## Materials Needed for the Cut and Paste Logic Puzzles:
• Cut and Paste Puzzles (available free in my shop)
• Scissors
• Glue
## How to Use the Cut and Paste Logic Puzzles:
First have your kids cut out the shape cards at the bottom of the puzzle. Depending on the age of your kids, they may need some help with this.
You can also take some time to review the shapes and have a discussion about where they see these shapes around them. For instance, ask them to point out a square in the room or name something that is a circle.
Then, explain that each row and each column must have one of each shape, without repeating.
While it may be tempting to also point out the row or column that already has two of the shapes and lead them to figure out which one must be missing, step back and let them try first!
The point of these puzzles is to help kids learn to think and problem solve and use logic. So give them the space to think through it on their own.
If they complete the puzzle and have repeating shapes in a row, simply show them WHY that’s not right (“I see there are two circles in this row. You need to have one of each shape. How can you fix it?”) rather than explaining how to solve it.
I guarantee they will catch on quickly and be able to solve the first and the rest of the puzzles independently.
Once they have correctly solved the puzzle, allow them to glue the pieces down in the correct squares. Try to be careful that they solve the puzzle first, before glueing, so that they aren’t trying to pull up and move around pieces that have already been glued.
And once they’ve got the hang of these puzzles, try these pattern block logic puzzles! These fun, hands on challenges will keep kids busy and get their brains thinking!
### {Click HERE to go to my shop and grab the Cut and Paste Logic Puzzles!}
LOVE this idea? Grab my ebook packed with fun, hands on puzzles for kids ages 4-9!
Want more fun puzzles for kids? Try one of these:
### Never Run Out of Fun Math Ideas
If you enjoyed this post, you will love being a part of the Math Geek Mama community! Each week I send an email with fun and engaging math ideas, free resources and special offers. Join 163,000+ readers as we help every child succeed and thrive in math! PLUS, receive my FREE ebook, 5 Math Games You Can Play TODAY, as my gift to you!
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There are three basic formulae, involving the ratio of the lengths of the sides. Given a right angled triangle, we first label the sides for the angle we have:
If we have the angle shown, we label the sides for this angle. Opposite =is the side opposite the angle x Hypotenuse =is the longest side, always opposite the right angle. Adjacent=a is the remaining side/ We are to calculateWe use one of the formulae We choose which formula to use depending on which side we gave got and which side we need to find. In this case we haveand need to findso we use the formula Example: Now we haveand want to findWe use the formula Example: Now we haveand want to findWe use the formula Example: We have o and want to find h. We use the formula
It is important to realise the difference between the formulae
and
For the first we writebut for the second we actually work out
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https://math.stackexchange.com/questions/690118/derivative-with-a-square-root-in-denominator
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# Derivative with a Square root in Denominator
$f(x) = \dfrac{-3}{\sqrt{3x^2 + 3}}$
I can't seem to figure this problem out. I think you would make the bottom(3x^2+3)^(1/2) and then use the chain rule on bottom and then use the quotient rule. This is the only question I cant seem to figure out on my homework so if you could give step by step detailed instructions i'd be forever grateful. Thanks for your time.
• You could do that. But the Quotient rule is what you'd use first. – David Mitra Feb 25 '14 at 17:47
One route to go is to write your function $$f(x) = -3(3x^2 + 3)^{-1/2}$$ and then use the chain rule!
$$f'(x) = -(1/2)(-3) (3x^2 + 3)^{-3/2}\cdot(6x)= 9x(3x^2 + 3)^{-3/2}$$
$$= \dfrac{9x}{(3x^2 + 3)^{3/2}}$$
• Have a wonderful time my friend. I'll try to get back soon. One of my article is going to be published at Korean Mathematical Society. :-) – mrs Feb 26 '14 at 9:27
• I'm so glad to hear from you, AND so glad that you're going to be published! – Namaste Feb 26 '14 at 13:36
Exponent rules! Remember that $\dfrac{1}{\sqrt{x}} = x^{-1/2}$. Then use the power and chain rule.
Just use the chain rule, first bringing up
$f(x) = -3(3x^2 + 3)^\frac{-1}{2}$
Then, taking the derivative of what we have raised to the (-1/2) power is just the use of the chain rule, and we will have,
$$f'(x) = \frac{-1}{2} \frac{-3}{(3x^2 + 3)^\frac{3}{2}}(6x) = \frac{9x}{(3x^2 + 3)^\frac{3}{2}}$$
An alternative approach: Quotient Rule tells you that $$f'(x)=\cfrac{\sqrt{3x^2+3}\cdot\frac{d}{dx}[-3]-(-3)\cdot\frac{d}{dx}\left[\sqrt{3x^2+3}\right]}{\left(\sqrt{3x^2+3}\right)^2}.$$
Since $\frac{d}{du}\left[\sqrt u\right]=\frac1{2\sqrt u},$ then by Chain Rule, $$\frac{d}{dx}\left[\sqrt{3x^2+3}\right]=\frac1{2\sqrt{3x^2+3}}\cdot\frac{d}{dx}[3x^2+3].$$
Can you take it from there?
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https://asvabtestpro.com/quiz/all-of-the-following-are-examples-of-composite-numbers-except-which-one-6278d7121cb4089b0e7a6b6f/
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All of the following are examples of composite numbers except which one?
2
Explanation
A composite number is a whole number that can be divided by itself, 1, and one or more other whole numbers. 6 is a composite number because it can be divided by 1, 2, 3, and 6; 9 is a composite number because it can be divided by 1, 3, and 9; and 12 is a composite number because it can be divided by 1, 2, 3, 4, 6, and 12.
A prime number is a whole number that can only be divided by itself and 1. 2 is a prime number because its factors are only 1 and 2. A prime number is a whole number that can only be divided by itself and 1. 2 is a prime number because its factors are only 1 and 2.
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https://www.inchcalculator.com/convert/coulomb-second-to-kiloampere/
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# Coulombs per Second to Kiloamperes Converter
Enter the electric current in coulombs per second below to get the value converted to kiloamperes.
## Result in Kiloamperes:
1 C/s = 0.001 kA
Do you want to convert kiloamperes to coulombs per second?
## How to Convert Coulombs per Second to Kiloamperes
To convert a measurement in coulombs per second to a measurement in kiloamperes, divide the electric current by the following conversion ratio: 1,000 coulombs per second/kiloampere.
Since one kiloampere is equal to 1,000 coulombs per second, you can use this simple formula to convert:
kiloamperes = coulombs per second ÷ 1,000
The electric current in kiloamperes is equal to the electric current in coulombs per second divided by 1,000.
For example, here's how to convert 500 coulombs per second to kiloamperes using the formula above.
kiloamperes = (500 C/s ÷ 1,000) = 0.5 kA
Coulombs per second and kiloamperes are both units used to measure electric current. Keep reading to learn more about each unit of measure.
## What Are Coulombs per Second?
One coulomb per second is equal to one coulomb of charge over one second.
Coulombs per second can be abbreviated as C/s; for example, 1 coulomb per second can be written as 1 C/s.
In the expressions of units, the slash, or solidus (/), is used to express a change in one or more units relative to a change in one or more other units.[1] For example, C/s is expressing a change in electric charge relative to a change in time.
## What Is a Kiloampere?
One kiloampere is equal to 1,000 amperes, which are the electrical current equal to the flow of one coulomb per second.
The kiloampere is a multiple of the ampere, which is the SI base unit for electric current. In the metric system, "kilo" is the prefix for thousands, or 103. A kiloampere is sometimes also referred to as a kiloamp. Kiloamperes can be abbreviated as kA; for example, 1 kiloampere can be written as 1 kA.
## Coulomb per Second to Kiloampere Conversion Table
Table showing various coulomb per second measurements converted to kiloamperes.
Coulombs Per Second Kiloamperes
1 C/s 0.001 kA
2 C/s 0.002 kA
3 C/s 0.003 kA
4 C/s 0.004 kA
5 C/s 0.005 kA
6 C/s 0.006 kA
7 C/s 0.007 kA
8 C/s 0.008 kA
9 C/s 0.009 kA
10 C/s 0.01 kA
20 C/s 0.02 kA
30 C/s 0.03 kA
40 C/s 0.04 kA
50 C/s 0.05 kA
60 C/s 0.06 kA
70 C/s 0.07 kA
80 C/s 0.08 kA
90 C/s 0.09 kA
100 C/s 0.1 kA
200 C/s 0.2 kA
300 C/s 0.3 kA
400 C/s 0.4 kA
500 C/s 0.5 kA
600 C/s 0.6 kA
700 C/s 0.7 kA
800 C/s 0.8 kA
900 C/s 0.9 kA
1,000 C/s 1 kA
## References
1. National Institute of Standards and Technology, NIST Guide to the SI, Chapter 6: Rules and Style Conventions for Printing and Using Units, https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-6-rules-and-style-conventions-printing-and-using
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How to show $\sin x \geq \frac{2x}{\pi}, x \in [0, \frac{\pi}{2}]$? [duplicate]
I have tried the following: $$f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0$$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
$\sin(x)$ is concave on that range and the line pass through $(0,0)$ and $(\pi/2,1)$ is $y=\frac{2}{\pi}x$
Let $$f(x) = \frac{\sin{x}}{x} \quad\implies f'(x) = \frac{g(x)}{x^2}~~ \text{with} \quad g(x) = x\cos{x} -\sin{x}$$ and $$g'(x) = -x\sin{x}\le 0$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, then $g$ is decreasing whereas $g(0) = 0$.
Thus $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, hence $f$ is decreasing on $[0,\frac{\pi}{2}]$. That is $$\frac{\sin{x}}{x} =f(x) \ge f(\pi/2) =\frac{2}{\pi}~~for ~~~0\le x\le \pi/2$$
Note that $f$ first increases from $0$ to some value and then decreases again to $0$, so it must be greater than $0$ all this time. The precise value is where $f'(x)=0$ and the graph of $f(x)$ will roughly look like a parabola.BTW, i.e. $f(\arccos(2/\pi))=\sin(\arccos(2/\pi))-(2/\pi)\arccos(2/pi)=\sqrt{\pi^2-4}/\pi-(2/\pi)\arccos(2/pi)$.
I would say, you have sort of completed this question. What you want to show is that f(x) >= 0 . You can say f' has critical point at around 0.881 (solve f'(x) =0 ). Then you plug 0, 0.881, pi/2 into f(x). Then you show the statement sin(x) >= 2x/pi is true. It would be easier if you plot the graph and think about it. For plotting, I recommend, https://www.desmos.com/calculator
you can type, f(x) = sin(x) -2pi/x g(x) =d/dx f(x) and so on, to create 2nd, 3rd,,, derivative.
• Let $f(x) = \left\{\begin{array}{ll}\frac{\sin(x)}{x} & \mbox{if$x \in \left(0,\frac{\pi}{2}\right)$.} \\ 1 & \mbox{ if$x=0$.} \end{array}\right.$
• By Mean value theorem applied to the function $g(x)=\sin(x)$ we get a $t\in (0,x)$ such that $\cos(t) =\frac{\sin(x)}{x}.$
• Note for $t \in (0,x)$ we have $$f'(x)=\frac{x\cos(x)-\sin(x)}{x^{2}} =\frac{\cos(x)-\frac{\sin(x)}{x}}{x}=\frac{\cos(x)-\cos(t)}{x} < 0$$ for all $x \in (0,\frac{\pi}{2}]$.
• I don't understand what you want to prove... Mar 27 '17 at 14:29
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# Help with a formula please.
=?Utf-8?B?Um9yeQ==?=
Guest
Posts: n/a
24th Apr 2006
Can anyone write a formula to do the following,
Add 53 % to a cell with a figure in it then round of the pence to 95 ie
101.95 also if possible if i drag down a coloum of figures to apply the
formula to, ignore any cells in the coloum with have a nil value.
--
Rory Moore
=?Utf-8?B?VG9wcGVycw==?=
Guest
Posts: n/a
24th Apr 2006
Do you always round up or down?
127*1.53=194.31 ==> 193.95 or 194.95?
137*1.53=209.61 ==> 208.95 or 209.95?
If it up, take integer value and add 0.95
A1=Value
B1=%
=IF(A1<>0,INT(A1*(1+B1))+0.95,0)
Round down
=IF(A1<>0,INT(A1*(1+B1)-1)+0.95,0)
HTH
"Rory" wrote:
> Can anyone write a formula to do the following,
> Add 53 % to a cell with a figure in it then round of the pence to 95 ie
> 101.95 also if possible if i drag down a coloum of figures to apply the
> formula to, ignore any cells in the coloum with have a nil value.
> --
> Rory Moore
=?Utf-8?B?VG9wcGVycw==?=
Guest
Posts: n/a
24th Apr 2006
.... change to
=IF(A1<>0,INT(A1*(1+B1))+0.95,"") if you nil values to return blank
"Toppers" wrote:
> Do you always round up or down?
>
> 127*1.53=194.31 ==> 193.95 or 194.95?
> 137*1.53=209.61 ==> 208.95 or 209.95?
>
> If it up, take integer value and add 0.95
>
> A1=Value
> B1=%
>
> =IF(A1<>0,INT(A1*(1+B1))+0.95,0)
>
> Round down
>
> =IF(A1<>0,INT(A1*(1+B1)-1)+0.95,0)
>
> HTH
>
> "Rory" wrote:
>
> > Can anyone write a formula to do the following,
> > Add 53 % to a cell with a figure in it then round of the pence to 95 ie
> > 101.95 also if possible if i drag down a coloum of figures to apply the
> > formula to, ignore any cells in the coloum with have a nil value.
> > --
> > Rory Moore
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# How to calculate annualized rate of return over multiple transactions
The annual rate of return on an investment can be calculated as
And the top response to this question suggests that the return on a portfolio of such purchases is simply the weighted average of returns on individual transactions, where the weighting coefficient is the number of shares bought:
However, by this definition, it seems like the calculated portfolio return can be positive, despite losing money.
Consider the following example: I purchase two shares of stock, one for \$100 and one for \$10, and sell both for \$50, the first held for 1 year and the second held for 0.5 years.
So I've spent \$110 and received \$100, but the calculated return is +1175% [ (-50% + 2400%)/2 ].
What am I doing wrong?
1st equation
``````R = (Pt/Po)^(1/t) - 1
``````
annualised returns for shares 1 & 2 are
``````r1 = (50/100) - 1 = -50 %
r2 = (50/10)^(1/0.5) - 1 = 2400 %
``````
2nd equation
``````R = Sum[mi*Ri]/n
``````
Calculating half-year returns corresponding to the holding periods:
"weighted average of returns on individual transactions" for each half-year
``````hr1 = (100*(1 + r1)^(1/2) + 10*(1 + r2)^(1/2))/110 - 1 = 9.73698 %
hr2 = (1 + r1)^(1/2) - 1 = -29.2893 %
``````
giving the return over the year as
``````(1 + hr1)*(1 + hr2) - 1 = -22.4042 %
``````
• Got it, thanks. For a collection of purchases with different holding periods, it looks like this calculation could become quite complex. Is internal rate of return (IRR) the best approach in this case? – Michael Boles May 6 '20 at 23:22
• Calculation of CAGR is an IRR problem, e.g. `0 = 10 - 50/(1 + r)^0.5``r = -50%` but if you mean put all the cash flows in one IRR like so: `0 = 110 - 50/(1 + r)^0.5 - 50/(1 + r)``r = -11.875%` you get a quite different result because you are omitting certain information. The closest to reality is a time-weighted return, which you could use if you knew the value of share 1 when you sold share 2. Let's say it was \$72. Then `hr1 = (72 + 50)/(100 + 10) - 1 = 10.9%` and `hr2 = 50/72 - 1 = -30.56%` so `r = (1 + hr1)*(1 + hr2) - 1 = -22.98%`. So the approach in the answer is closer to reality. – Chris Degnen May 7 '20 at 9:07
• However, if you have numerous transactions IRR (or money-weighted return) is usually a fair enough approximation. It gets less accurate compared to time-weighted return when there are large mid-period cash flows. – Chris Degnen May 7 '20 at 9:17
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http://math.stackexchange.com/questions/880081/the-sum-of-two-irrational-square-roots
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# The sum of two irrational square roots
This is very similar to this question, but I was wondering if there was a simpler proof. In particular, a proof that would prove that $\sqrt{x}+\sqrt{y}$ is an irrational number if both $\sqrt{x}$ and $\sqrt{y}$ are irrational.
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Hint: If $xy$ is not a perfect square, show that $(\sqrt x + \sqrt y )^2 = x + y + 2\sqrt{ xy}$ is irrational, and use the fact that and if $u^2$ is irrational then so is $u$.
If $xy = n^2$ then show that $\sqrt x + \sqrt y = \sqrt x\left(1+ \frac n {x}\right)$is irrational.
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This gets about 80% of the way... but what if, e.g., $x=2$ and $y=8$? – Steven Stadnicki Jul 28 '14 at 1:04
Should get the whole way now :) – Mathmo123 Jul 28 '14 at 1:19
With the tiny caveat that $x$ itself can't be a rational square, of course, but this looks good! – Steven Stadnicki Jul 28 '14 at 1:25
That was implicit in the question I believe - otherwise of course the theorem is false. – Mathmo123 Jul 28 '14 at 1:27
@just1question Well, you know that $1+\frac nx$ is rational, and $\sqrt{x}$ is irrational, so... – Steven Stadnicki Jul 28 '14 at 7:31
If $x$ and $y$ are integers then so is $x - y$ and further if $\sqrt x + \sqrt y$ is rational then $\sqrt x - \sqrt y = \frac{x - y}{\sqrt{x} + \sqrt{y}}$ is also rational. Then by sums and differences so are both $\sqrt{x}$ and $\sqrt{y}$. Therefore if we assume that at least one of $\sqrt{x}$ and $\sqrt{y}$ are irrational than so is their sum.
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http://mathhelpforum.com/advanced-statistics/120358-obtain-maximum-likelihood-estmate-average-arrival-rate.html
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# Thread: Obtain the maximum likelihood estmate of the average arrival rate
1. ## Obtain the maximum likelihood estmate of the average arrival rate
Question : It is assumed that the arrival of number of calls $X$ per hour follows Poisson distribution with parameter $\lambda$ . A random sample $X_1 = x_1, X_2 = x_2,.....X_n = x_n$ is taken. Obtain the maximum likelihood estimate of the average arrival rate
2. Originally Posted by zorro
Question : It is assumed that the arrival of number of calls $X$ per hour follows Poisson distribution with parameter $\lambda$ . A random sample $X_1 = x_1, X_2 = x_2,.....X_n = x_n$ is taken. Obtain the maximum likelihood estimate of the average arrival rate
Do you know what sort of distribution the waiting time follows? That is your starting point .... Go back and look it up in your class notes.
3. ## This is what u mean
Originally Posted by mr fantastic
Do you know what sort of distribution the waiting time follows? That is your starting point .... Go back and look it up in your class notes.
Do u mean exponential distribution????
4. Originally Posted by zorro
Do u mean exponential distribution????
Yes, but I realise now that I misread the question - this distribution is not relevant (which is not to say that it's not relevant to you in the broader context of developing and consolidating your general understanding).
Here is a Christmas present:
The likelihood function is $L(x_1, x_2, .... x_n) = \frac{e^{-\lambda} \lambda^{x_1}}{x_1!} \cdot \frac{e^{-\lambda} \lambda^{x_2}}{x_2!} \cdot .... \frac{e^{-\lambda} \lambda^{x_n}}{x_n!} = e^{-n \lambda} \left( \frac{\lambda^{x_1 + x_2 + .... + x_n}}{x_1! x_2! .... x_n!} \right)$.
Therefore $\ln L = -n \lambda + (x_1 + x_2 + .... + x_n) \ln \lambda - \ln (x_1! x_2! .... x_n!)$.
Your job is to solve $\frac{d \ln L}{d \lambda} = 0$ for $\lambda$. The answer is not surprising.
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https://artofproblemsolving.com/wiki/index.php?title=1987_AJHSME_Problems/Problem_8&oldid=97884
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# 1987 AJHSME Problems/Problem 8
## Problem
If $\text{A}$ and $\text{B}$ are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is
$\begin{tabular}[t]{cccc} &9 & 8 & 7 & 6 \\ && A & 3 & 2 \\ +& & B & 1 \\ \hline \end{tabular}$ (Error compiling LaTeX. ! Extra alignment tab has been changed to \cr.)
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ \text{depends on the values of A and B}$
## Solution
The minimum possible value of this sum is when $A=B=1$, which is $$9876+132+11=10019$$
The largest possible value of the sum is when $A=B=9$, making the sum $$9876+932+91=10899$$
Since all the possible sums are between $10019$ and $10899$, they must have $5$ digits.
$\boxed{\text{B}}$
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Solving equations using iteration – Higher tier
Approximate solutions to more complex equations can be found using a process called iteration. Iteration means repeatedly carrying out a process. To solve an equation using iteration, start with an intial value and substitute this into the equation to obtain a new value, then use the new value for the next substitution, and so on.
Example
Find the solution to the equation using the initial value , giving the answer to 3 decimal places.
First, rearrange the equation to leave on its own on one side of the equation.
One way to do this is:
To solve the equation, use the iterative formula
We are given the initial value
Substituting this into the iterative formula gives
Substituting iteratively gives:
(3 dp)
(3 dp)
(3 dp)
(3 dp)
(3 dp)
(3 dp)
Since and give the same value to 3 decimal places, the iteration stops. The solution to the equation is 2.113 to 3 decimal places.
If your calculator has an ANS button, use it to keep the value from one iteration to substitute into the next iteration. To solve the equation on a calculator with an ANS, type 2 =, then type to find the first iteration, then just press = for the next iteration.
Example
Show that the equation has a solution that lies between 1 and 2.
Use the iterative equation to find the solution to the equation to 3 decimal places.
Substituting into the left hand side of the equation gives
Substituting into the left hand side of the equation gives
The value on the right hand side of the equation, 7, lies between these two values, 3 and 12, so the solution to the equation must be between 1 and 2.
Using the initial value and substituting this iteratively gives:
(3dp)
(3 dp)
(3 dp)
(3 dp)
(3 dp)
(3 dp)
(3 dp)
So the solution to the equation is 1.569 to 3 decimal places.
Move on to Test
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https://www.slideshare.net/abhijitgaikwad/4-recursion-details
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# 4 recursion details
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### 4 recursion details
1. 1. More on Recursion More techniques 1
2. 2. Binary search algorithm• Binary searching for a key in an array is similar to looking for a word in dictionary• Take the midpoint of the dictionary and see if the key is in the lower half or upper half• Pick the half the contains the key and take the midpoint of the half and see which quarter the key is in• Repeat the above step until the key is matched or the dictionary cannot be halved any more 2
3. 3. Use helper method• Given int[] list and int key• You are asked to return – the position of key in the array if matched or – -1 if not matched• What additional information do you need to apply the algorithm above? 3
4. 4. /** Use binary search to find the key in the list */ public static int recursiveBinarySearch(int[] list, int key) { int low = 0; int high = list.length - 1; return recursiveBinarySearch(list, key, low, high); // use helper method }// recursiveBinarySearch method/** Use binary search to find the key in the list between list[low] and list[high] */ public static int recursiveBinarySearch(int[] list, int key, int low, int high) { if (low > high) // The list has been exhausted without a match return -1; int mid = (low + high) / 2; if (key < list[mid]) return recursiveBinarySearch(list, key, low, mid - 1); else if (key == list[mid]) return mid; else return recursiveBinarySearch(list, key, mid + 1, high); }// recursiveBinarySearch method 4
5. 5. What is the difference between:public static int recursiveBinarySearch( int[] list, int key){ . . . }andpublic static int recursiveBinarySearch( int[] list, int key, int low, int high) { . . . }The first method calls the second method.Is it recursive call? 5
6. 6. Recursion versus iterationIteration Recursion• Uses for or while loop for • Recursion achieves repetition but no recursion repetition without a loop • Price paid: memory overhead and run time (Every time a method is called, some space in memory must be reserved or allocated to store the method’s local• Why use recursion? variables and formal parameters, if any) 6
7. 7. public class NestedCalls { public static void m1() { int m1_x = 1; int m1_y = 2; m2(); // m1 calls m2 } // m1 method public static void m2() { int m2_ = 3; int z = 4; z = m3(); //m2 calls m3 }// m2 method public static int m3() { int m3_x =5; int m3_y = 6; return 1; }//m3 method}//NestedCalls class 7
8. 8. Run NestedCalls.java with Bluej Debugger Observe the stack memory and calling sequences 8
9. 9. View of stack at various points of execution ofNestedCalls.java Stack frame for m3: m3_x: 5 m3_y: 6 Stack frame for m2: Stack frame for m2: m2_x: 3 m2_x: 3 m2_z: 4 m2_z: 4Stack frame for m1: Stack frame for m1: Stack frame for m1:m1_x: 1 m1_x: 1 m1_x: 1m1_y: 2 m1_y: 2 m1_y: 2Just before Just before Just before m3calling m2 calling m3 returns 9
10. 10. Stack of Binary Search just before returning from thelast recursive call 10
11. 11. Why use recursion?In some problems, iterative solutions are hard tofind.But recursion is easy - See the Towers of Hanoiproblem in textbook. Animation by Michael Iverson 11
12. 12. Other problems that are neatly solved by recursion:Fractals:• The Koch snowflake (demo)• Sierpinski triangle (demo) 12
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https://www.jiskha.com/questions/1009326/how-would-you-solve-each-system-by-the-substitution-method-1-y-6x-5-y-x-9
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# Math
How would you solve each system by the substitution method?
1. Y=6x-5
Y=-x+9
1. 👍 0
2. 👎 0
3. 👁 109
1. well, we know that both are y so
6x-5 = -x+9
7 x = 14
x = 2
then y = -x+9 = -2+9 = 7
1. 👍 0
2. 👎 0
posted by Damon
2. I still don't get it.
1. 👍 0
2. 👎 0
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