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Degger [83]
3 years ago
13

PLEASE HELP ASAP 5TH GRADE

Mathematics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

The answer would be B.

Step-by-step explanation:

Hope this helps.

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Which scenario represents a proportional relationship?
kirza4 [7]

Answer:

The answer should be B

Step-by-step explanation:

I hope this helps! Let me know if I am wrong.

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Find the area of the figure shown. Round your answer to the hundredths place. Use 3.14 for pi.
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There is no figure.
5 0
2 years ago
Find a parametric representation for the part of the cylinder y2 + z2 = 49 that lies between the planes x = 0 and x = 1. x = u y
sweet-ann [11.9K]

Answer:

The equation for z for the parametric representation is  z = 7 \sin (v) and the interval for u is 0\le u\le 1.

Step-by-step explanation:

You have the full question but due lack of spacing it looks incomplete, thus the full question with spacing is:

Find a parametric representation for the part of the cylinder y^2+z^2 = 49, that lies between the places x = 0 and x = 1.

x=u\\ y= 7 \cos(v)\\z=? \\ 0\le v\le 2\pi \\ ?\le u\le ?

Thus the goal of the exercise is to complete the parameterization and find the equation for z and complete the interval for u

Interval for u

Since x goes from 0 to 1, and if x = u, we can write the interval as

0\le u\le 1

Equation for z.

Replacing the given equation for the parameterization y = 7 \cos(v) on the given equation for the cylinder give us

(7 \cos(v))^2 +z^2 = 49 \\ 49 \cos^2 (v)+z^2 = 49

Solving for z, by moving 49 \cos^2 (v) to the other side

z^2 = 49-49 \cos^2 (v)

Factoring

z^2 = 49(1- \cos^2 (v))

So then we can apply Pythagorean Theorem:

\sin^2(v)+\cos^2(v) =1

And solving for sine from the theorem.

\sin^2(v) = 1-\cos^2(v)

Thus replacing on the exercise we get

z^2 = 49\sin^2 (v)

So we can take the square root of both sides and we get

z = 7 \sin (v)

4 0
2 years ago
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