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nasty-shy [4]
3 years ago
11

Please help I don't understand 120 - 28a = 78

Mathematics
2 answers:
stiks02 [169]3 years ago
6 0
Simplifying
120 + -28a = 78

Solving
120 + -28a = 78

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-120' to each side of the equation.
120 + -120 + -28a = 78 + -120

Combine like terms: 120 + -120 = 0
0 + -28a = 78 + -120
-28a = 78 + -120

Combine like terms: 78 + -120 = -42
-28a = -42

Divide each side by '-28'.
a = 1.5

Simplifying
a = 1.5
Gemiola [76]3 years ago
4 0
120 - 28a = 78
28a = 120-78
28a = 42
a =  \frac{42}{28}  \\ a =  \frac{3}{2}
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3 years ago
A) what is average rate of change of f(x) over the interval from x=5 to x=9? (table shows values of f(x). graph shows function o
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Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

The average rate of change of a function over the interval [a,b] is equal to \frac{f(b)-f(a)}{b-a}, hence:

\frac{f(b)-f(a)}{b-a}\\\\\frac{f(9)-f(5)}{9-5}\\\\\frac{14-(-4)}{9-5}\\ \\\frac{14+4}{4}\\ \\\frac{18}{4}\\ \\\frac{9}{2}

Therefore, the average rate of change of f(x) over the interval [5,9] is \frac{9}{2}.

<u>Part B</u>

Do the same thing as in Part A:

\frac{f(b)-f(a)}{b-a}\\ \\\frac{f(1)-f(0.25)}{1-0.25}\\ \\\frac{2-5}{0.75}\\ \\\frac{-3}{0.75}\\ \\-4

Therefore, the average rate of change of g(x) over the interval [0.25,1] is -4.

<u>Part C</u>

To interpret our answer from Part B in terms of the real world it represents, we say that between 0.25 seconds and 1 second, the ball falls at a rate of 4 feet per second (since our average rate of change is negative).

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Answer:190

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The correct answer is B. 130,099.

Hope this helps!

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The perpendicular line will have a slope of b/a.

To find this, we first have to solve our equation for slope intercept form.

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So we know the slope of this equation to be -a/b. Since perpendicular lines have opposite and reciprocal slopes, we know we can simply flip the fraction and make it a negative to get the new slope of b/a.

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