The equation v(x)=32500(0.92)^x how much more is the truck worth after two years than after three years

Mathematics

1 answer:

Lyrx [107]3 years ago

6 0

Answer:

the difference: $2,192.64

Step-by-step explanation:

After 1 year, the truck has lost 8% of its value: v(x) = $32,500(0.92)^1, or

$29,900.

After 2 years, v(x) = $32,500(0.92)^2, or $27,500. Finally, after 3 years,

v(x) = $32,500(0.92)^3 = $25,307.36.

Subtracting $25,307.36 (the value after 3 years) from $27,500 (the value after 2 years) yields the difference: $2,192.64

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What is an equation of the line that passes through the points (6,2) and

S_A_V [24]

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6 0

2 years ago

Help Would Be Greatly Appreciated :)

aliina [53]

Answer:

n is $y=\frac{1}{2}x+b$ , p is $y=\frac{1}{2}x-5$ , and r has infinitely many different lines that have this property.

Step-by-step explanation:

If two lines have no solutions, that means that they are parallel. The slopes of two lines are the same if they are parallel. So n's slope is $\frac{1}{2}$ . The y-intercept can be anything except -5, or +b. If two lines have infinitely many solutions, those two lines are exactly the same. So p is just m. If two lines have one and only one solution, that just means the slopes and y-intercepts are different. Since we don't know anything else about line r, we an't model an equation for it.