Answer:
just by looking at the graph you can see that the y int is one, so that eliminates two answers already. Then the slope is rise over run. (best thing to remember in math. rise over run) so pick two points. (0, 1) and (4,0) were the first ones i saw. to get from those two points you go from (0,1) over 4 (thats your run) and from there down 1 (thats your rise) to get to (0,4). You would then do rise over run (and because your rise goes down its negative) to get -1/4 as your slope. The answer is slope= -1/4, Y intercept = 1
120
6x5x4=120
hope this helps
X=t-yz and we are asked to solve for y so, subtract t from both sides
x-t=-yz divide both sides by -z
(x-t)/(-z)=y
y=(x-t)/(-z) multiply the numerator and denominator by -1
y=(t-x)/z
Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
