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siniylev [52]
3 years ago
5

A software designer is mapping the streets for a new racing game. All of the streets are depicted as either perpendicular or par

allel lines. The equation of the lane passing through A and B is -7x + 3y = -21.5. What is the equation of the central street PQ?
A.
-3x + 4y = 3

B.
-1.5x − 3.5y = -31.5

C.
2x + y = 20

D.
-2.25x + y = -9.75

This is a PLATO math question, will give 15 pts to the best answer

Mathematics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

B.   -1.5x − 3.5y = -31.5

Step-by-step explanation:

You forgot to provide the reference image which is essential to answer the question, but I managed to find it... and attach it to my answer.

In the given equation for AB, if we place the y term on the left and x term on the right, we see the slope of that line is 7/3 (y = (7x - 21.5)/3 ==> 7/3x).

We see on the image that the line PQ is perpendicular to AB.  That means that its slope is -3/7.

If we quickly check the slopes of each of the possible answers...

A. -3x + 4y = 3 ----> 4y = 3x - 3 ==> y = (3x-3)/4 => slope = 3/4

Not what we're looking for.

B. -1.5x - 3.5y = -31.5 ==> 3.5y = -1.5x + 31 ===> y = (-1.5x +31)/3.5

that gives us a slope of -1.5/3.5... We can simplify it... -3(0.5)/7(0.5) = -3/7

Exactly as predicted.

Since we have the point P (7,6), we can enter it in the equation to verify:

-1.5x - 3.5y = -31.5

-1.5 (7) - 3.5 (6) = -10.5 - 21 = -31.5  --- Verified

C. 2x + y = 20 ==> y = 20 - 2x ===> slope is -2, not what we want.

D. -2.25x + y = -9.75 ==> y = 2.25x - 9.75 ==> slop is 2.25 cannot be it.

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Step-by-step explanation:

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Anna71 [15]

Just like the Vertex Formula you do this:

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This is where -H gives you the OPPOSITE terms of what they really are, and K gives you the NORMAL term, so just apply that to here. Whenever you are doing a phase shift, you put it in parentheses using the Vertex Formula, but giving the opposite term of what it really is:

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Now here is the graph. You see how I shifted it 4 units away from the origin?

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I am joyous to assist you anytime.

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Evgesh-ka [11]

To find f^{-1}, you can switch the "x" and "f(x) or y" in the equation.

f(x) = \sqrt[3]{x-2} +8

y = \sqrt[3]{x-2}+ 8

x = \sqrt[3]{y-2}+8

Now you need to isolate the "y"

x = \sqrt[3]{y-2}+8   Subtract 8 on both sides

x - 8 = \sqrt[3]{y-2}  Cube ( ³ ) each side to get rid of the ∛

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(x-8)^{3}+2 = y


f^{-1} = (x-8)^{3} + 2


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