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Likurg_2 [28]
3 years ago
7

Rewrite the equation using Ax+By=C form

Mathematics
2 answers:
cupoosta [38]3 years ago
6 0

Answer:

6x - y = - 6

Step-by-step explanation:

Given

y + 6 = 6(x + 2) ← distribute parenthesis

y + 6 = 6x + 12 ( subtract y from both sides )

6 = 6x - y + 12 ( subtract 12 from both sides )

- 6 = 6x - y OR 6x - y = - 6 ← in the form Ax +By = C

VLD [36.1K]3 years ago
4 0

Step-by-step explanation:

y + 6 = 6(x + 2) \\  \\  \therefore \: y + 6 = 6x + 12 \\  \\  \therefore \: y = 6x + 12 - 6 \\  \\ \therefore \: y = 6x + 6 \\  \\  \therefore \: - 6x +  y = 6 \\  \\ or \:   \:  \:  \:  6x - y =  - 6  \\ which \: is \: in \: the \: form \: of \: \\  Ax+By = C \\

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Step-by-step explanation:

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The zookeeper starts with 100.75 pounds of food and tracks how much the otter eats and any food deliveries. What is the food sup
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Answer:

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Step-by-step explanation:

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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
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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)

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