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2 years ago

Round your answer to the nearest hundredth LA

Mathematics

1 answer:

m_a_m_a [10]2 years ago

8 0

Answer:

Step-by-step explanation:

Since this is a right triangle, we will use a right triangle trig identity to solve our problem. The side opposite the reference angle A is given as 5, and the side adjacent to the reference angle is given as 7. The trig ratio that uses the sides opposite and adjacent is the tangent ratio:

$tanA=\frac{5}{7}$

Hit the 2nd button on your calculator then the tan button which displays:

$tan^{-1}($

Enter 5/7 after that open parenthesis and hit = to get that

A = 35.54°

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Which value of x makes the inequality -* < 8 true?

elena55 [62]

Answer:

D: x = 2

Step-by-step explanation:

Go down the multiple choices answers and plug each value in for x until the inequality is true:

1/4(32) < 8

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8 < 8

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2 years ago

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

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I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, 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

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

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3 years ago