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3 years ago

How do i solve #13? (linear equations)

Mathematics

1 answer:

Neporo4naja [7]3 years ago

8 0

#12:

Assuming that I have those points correct: (1, 5) and (-4, -10), the slope of this line is

5-(-10)

m = --------------- = 15/5 = 3

1-(-4)

The line passes thru the origin, so the y-intercept is 0.

So the equation of this line is y = 3x.

I see no marks (points) in the illustration for #13.

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The answer is 252.5 ft

y = –0.004x² + x + 2.5
Let's take y = 0 and we'll get the quadratic equation:

–0.004x² + x + 2.5 = 0
The general formula for quadratic equation is:
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a = -0.004
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$x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac} }{2a} =\frac{-1+/- \sqrt{ 1^{2}-4*(-0.004)*2.5} }{2*(-0.004)} = \frac{-1+/- \sqrt{1+0.04} }{-0.008} = \\ 
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$x_1=\frac{-1+1.02}{-0.008} = \frac{0.02}{-0.008} = -2.5 \\ 
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Since distance cannot be negative (x1), the correct answer is x2 = 252.5 ft

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alukav5142 [94]

Hello from MrBillDoesMath!

Answer:

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Discussion:

f(x) = 5 x^3

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(f•g)(x) =

f(g(x)) =

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5 * (x+1)^3 =

5 x^3 + 15 x^2 + 15 x + 5

which is none of the provided answers.

Thank you,

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Answer:

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