The depreciating value of a semi-truck can be modeled by y = Ao(0.85)x, where y is the remaining value of the semi, x is the tim
e in years, and it depreciates at 15% per year.
What is the value of the truck initially, Ao, and how would the graph change if the initial value was only $65,000?
$70,000, and the graph would have a y-intercept at 10,000
$75,000, and the graph would have a y-intercept at 65,000
$75,000, and the graph would fall at a slower rate to the right
$75,000, and the graph would fall at a faster rate to the right
What is the value of the truck initially, Ao, and how would the graph change if the initial value was only $65,000?
$70,000, and the graph would have a y-intercept at 10,000
$75,000, and the graph would have a y-intercept at 65,000
$75,000, and the graph would fall at a slower rate to the right
$75,000, and the graph would fall at a faster rate to the right
2 answers:
<span>I've answered this question before, my answer was:
y = Ao(0.85)x
</span><span>B. <span>$75,000, and the graph would have a y-intercept at 65,000
Based on the graph, when x = 0 ; y = 75,000
If the initial value will be 65,000 instead of 75,000, then the y-intercept is 65,000.
The rate of the same all throughout. It is still depreciating at 15% per year.
</span></span>
y = Ao(0.85)x
</span><span>B. <span>$75,000, and the graph would have a y-intercept at 65,000
Based on the graph, when x = 0 ; y = 75,000
If the initial value will be 65,000 instead of 75,000, then the y-intercept is 65,000.
The rate of the same all throughout. It is still depreciating at 15% per year.
</span></span>
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