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nalin [4]
2 years ago
12

The credit remaining on a phone card (in dollors) is a linear function of the total calling trime made with the card ( in minute

s). The remaining credit after 22 minutes is $36.70, and the remaining credit after 52 mintues is $32.20. What is the remaining credit after 85 mintues of calls?
Mathematics
1 answer:
KATRIN_1 [288]2 years ago
7 0

After 85 minutes of calls, there are $27.25 left on the card.

<h3>What is the remaining credit after 85 minutes of calls?</h3>

A linear equation in slope-intercept form is:

y = a*x + b

Where a is the slope.

If the line passes through two points (x₁, y₁) and (x₂, y₂) the slope is:

a = \frac{y_2 - y_1}{x_2 - x_1}

In this case, we know that the line passes through the points (22, 36.7) and (52, 32,20)

(points of the form (time, dollars)).

So the slope is:

a = \frac{32.2 - 36.7}{52 -22} = -0.15

The linear equation is then:

y = -0.15*x + b

To find the value of b, we use the point (22. 36.7)

36.7 = -0.15*22 + b

36.7 + 0.15*22 = b = 40

Then the linear equation is:

y = -0.15*x +40

The amount remaining in the credit card after 85 minutes is given by evaluating the above equation in x = 85.

y = -0.15*85 + 40 = 27.25

This means that after 85 minutes of calls, there are $27.25 left on the card.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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A triangular pyramid is formed from three right triangles as shown below.
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7 0
3 years ago
Over the three evenings,Karen received a total of 137 phone calls at the call center. The third evening, she received 4 times as
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Answer:

1st: 22, 2nd: 27, 3rd: 88

Step-by-step explanation:

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Hope this helps!

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3 0
3 years ago
Read 2 more answers
Line segment AB has a slope of 4/3 and contains point A (6,-5). What is the y-coordinate of point Q(2, y) if QA is perpendicular
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\bf A(\stackrel{x_1}{6}~,~\stackrel{y_1}{-5})\qquad Q(\stackrel{x_2}{2}~,~\stackrel{y_2}{y}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{y-(-5)}{2-6}=\stackrel{\textit{QA's slope}}{-\cfrac{3}{4}}\implies \cfrac{y+5}{-4}=\cfrac{-3}{4} \\\\\\ 4y+20=12\implies 4y=-8\implies y=\cfrac{-8}{4}\implies \boxed{y=-2}

8 0
3 years ago
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