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Alina [70]
3 years ago
15

Pls help with this question! Its due today!

Mathematics
1 answer:
blagie [28]3 years ago
5 0

Answer:

down below.

Step-by-step explanation:

straight lines are always 180.

x+34+2x+2=180\\3x+36=180\\3x+36-36=180-36\\3x=144\\3x/3=144/3\\x=48

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A survey found that 12 out of every 15 people in the united States prefer eating at a restaurant over cooking at home. if 400 pe
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500 people took the survey.

12 out of 15 people is 80% of the population.  12/15 = 0.80
Thus, 80% of the population prefers eating in the restaurant.
 
If 400 represents the people who prefers eating in the restaurant; then, 400 is the 80% of the population. To get the total population or 100%, we must divide 400 by 0.80 or 80%

400 / 0.80 = 500 people.

Out of the 500 people, 400 selected eating in the restaurant while the remaining 20% of the population or 100 people selected cooking at home.
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A line passes through the points (7, -5) and (3, 1). Determine the slope of the line.
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2 years ago
What is the slope of -2x + 4y = 12 ?
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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤
MrMuchimi
The average rate of change of a function f(x) in an interval, a < x < b is given by
\frac{f(b) - f(a)}{b - a}

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by \frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by \frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by \frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by \frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by \frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by \frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0
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3 years ago
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