A small business purchases a piece of equipment for $875. After 5 years the equipment will be outdated, having no value. (a) Write a linear equation giving the value of the equipment y in terms of the time (b) Find the value of the equipment when X=2 (c) Estimate (to two-decimal-place accuracy) the time when the value of the equipment is $200.
y=at+875 0=5a+875
where did A come from?
-5a=875 y=-175x+875
y=mx+b I am trying to find m(or slope)
\[f(x)=-175x+875\]
waaah why is it +875?
y(0) = 875 y(5) = 0 equipment loses value at rate of 875/5 = 175 per year a) y(t) = 875 - (875/5)t y(t) = 875 - 175t b) y(2) = 875 - 175(2) = 875-350 = 525 c) y(t) = 200 = 875-175(t) t = (200-875)/(-175) = 675/175 = 3.86 years.
because you start with 875
f(x)=-175(0)+875 =875
the equipment was bought at 875 it went down to 0..
tbh i wil never b good at word problems i cant make sense of the words correctly.
its like you added information thats not there.. nothingg hints to adding 875 to 875/5..
its like you added information thats not there.. nothingg hints to adding 875 to 875/5..
Dude, the thing was bought for $875 at time 0, and costs $0 at time 5. You can write this as two ordered pairs: \[(0, 875)\]\[(5,0)\]You can then find the slope in between those two points: \[m=\frac{0-875}{5-0}=\frac{-875}{5}=-175\]Then you can use the point-slope formula to finish it up: \[y-y_{0}=m(x-x_{0})\]\[y-875=-175(x-0)\]\[y=-175x+875\]Is that clearer?
You then use this formula the way lalaly did in her post above to figure out the answers for b and c.
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