How do I solve this using f.o.i.l (Factoring trinomials)

What is the average rate of change of the line whose equation is:

dedylja [7]

Answer:

Rate of change = -12

Step-by-step explanation:

y − 3 = −12(x + 4)

Expanding, we have;

y - 3 = -12x - 48

y = -12x - 48 + 3

y = -12x - 45

Now, formula for rate of change is;

Rate of change = (y(x + h) - y)/h

y(x + h) is; y = -12(x + h) - 45

This gives;

Rate of change = (-12(x + h) - 45 + 12x + 45)/h

Rate of change = (-12x - 12h - 45 + 12x + 45)/h

Rate of change = -12h/h

Rate of change = -12

3 0

3 years ago

To find the quotient Start Fraction 5 Over 7 End Fraction divided by one-third,

posledela

Answer:

its C.

Step-by-step explanation:

i took the test

8 0

3 years ago

10,000 round to the nearest ten thousand

wlad13 [49]

Answer:

i think it would be 10,000

I hope it helped!

5 0

2 years ago

Read 2 more answers

A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s

andre [41]

Answer:

$x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)$

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation:

Let

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio,

Length of the pool (with the patio)
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio)
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0 (1)

Let

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\ = \frac{-2(l + w) \pm \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l + w)^2) + 16lw}}{8} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 6lw + w^2)}}{8}$
$= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}
\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

Since $(l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \ 0$ , $-\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right)$ is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by

$\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

7 0

3 years ago

3 less than a number p is 25

Goryan [66]

Answer:

The number is 28

Step-by-step explanation:

Step 1: Make an expression

$p - 3 = 25$