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3 years ago

PLEASE HELP ASAP 15 POINTS

Mathematics

1 answer:

stepan [7]3 years ago

5 0

Answer: Its C

Step-by-step explanation: Hope this Helped!!!

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How do you tell if a line is parallel perpendicular or neither?

cluponka [151]

You can tell that a line is parallel if the lines can go on forever without crossing or intersecting. A line is perpendicular when the lines intersect at a right angle. If the slopes are either equal or negative reciprocals, they cannot be parallel or perpendicular.

Hope this helps you!

3 0

3 years ago

What are the vertical asymptotes of the function f(x) = the quantity of 2 x plus 8, all over x squared plus 5 x plus 6? x = −3 a

lbvjy [14]

For this question the answer would be x=-3 and x=-2

3 0

3 years ago

6 − x = −12 please answer

lorasvet [3.4K]

Answer:

x=18

Step-by-step explanation:

6-x=-12

Subtracting 6 from both sides

-x=-12-6

-x= -18

Dividing both sides by -1

x=18

6 0

3 years ago

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Amia went to a flower farm and picked f flowers. When she got home, she put the flowers in 4 vases, with 22 flowers in each vase

Whitepunk [10]

One equation that represents this situation is f/4 = 22.

This shows the total number of flowers (f) being broken into 4 groups (4 vases) with the result being 22 flowers in each vase.

7 0

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