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
Answer:
x=2y-8
We use the substitution of x in the second equation
2*(2y-8)+3y= -2
4y-16+3y=-2
7y-16= -2
7y= -2 +16
7y=14
y=14Ș7
y=2
so x= 2*2 -8
x=4-8
x=-4
the solution is (x,y)=(-4, 2)
Step-by-step explanation:
It should be 185 pounds. Have a GREAT day!!!!! :)
Answer:
-56
Step-by-step explanation:
t=-9 u=7
(-9)(7) + (7) =
-63 + (7) =
-56
Answer:
Step-by-step explanation:
for Which of the functions represents a linear function? Explain your reasoning, then determine the rate of change for that function over the interval 2≤x≤4.
For a linear function, the rate of change is represented by the parameter m in the slope-intercept form for a line: y=mx+b, and is visible in a table or on a graph.
