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3 years ago

What is missing to make this true? 2x^2+4x prove it with the distributive property

1 answer:

5 0

Take you the common factors. those are 2 and x. So if you divided 2x^2 by 2x you end with only one x. and if you divided 4x by 2x you end with 2. The final answer should look like 2x(x+2)=2x^2+4x

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lozanna [386]

Answer:

552

Step-by-step explanation:

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3 years ago

Find the variance for the given sample data, Round your answer to one more decimal place than the original data 7 725 1 A. 11.8

VladimirAG [237]

Answer:

D. 7.7

Step-by-step explanation:

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3 years ago

What are the first 4 prime numbers

ddd [48]

2, 3, 5 and 7 the the first 4 prime numbers

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3 years ago

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2 Points Which of the following is the product of the rational expressions shown below?

Norma-Jean [14]

Answer:

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3 years ago

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If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago