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Tomtit [17]
2 years ago
6

A business buys a computer for $3,000. After 4 years the value of the computer is expected to be $250. The value, v, can be rela

ted to the time in years, t, in a linear equation. Which equation models the relationship between v an t?
a.v=250t
b.v=-250t
c.v=687.50t-3,000
d.v=-687.50t+3,000
Mathematics
1 answer:
yuradex [85]2 years ago
6 0

You are given two points in the linear function. At time 0 years, the value is $3000. At time 4 years, the value is $250. This means you have points (0, 3000) and (4, 250). You need to find the equation of the line that passes through those two points.

y = mx + b

m = (y2 - y1)/(x2 - x1) = (3000 - 250)/(0 - 4) = 2750/(-4) = -687.5

Use point (0, 3000).

3000 = -687.5(0) + b

b = 3000

The equation is

y = -687.5x + 3000

Since we are using points (t, v) instead of (x, y), we have:

v = -687.5t + 3000

Answer: d. v = -687.50 t + 3,000

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Find the difference (7a-6b+7) - (8a-2)
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Let's set up our expression:

(7a-6b+7)-(8a-2)

In order to simplify, we can use that subtraction sign and distribute it, using the distributive property. We have:

7a-6b+7-8a+2

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Answer:

<h2>{7}^{3} (as \: a \: single \: power \: of \: 7)</h2>

Step-by-step explanation:

<em><u>Given</u></em><em><u> </u></em><em><u> </u></em>

{7}^{4}  \div 7

<em><u>Since</u></em><em><u>,</u></em>

{a}^{x}  \div  {a}^{y}  =  {a}^{x - y}

<em><u>Hence</u></em><em><u>,</u></em>

=  {7}^{4 - 3}

=  {7}^{3} (ans)(as \: a \: single \: power \: of \: 7)

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