Answer:
$40.60
Step-by-step explanation:
Answer:
g(t) = 10000(0.938)^t
Step-by-step explanation:
Given data:
car worth is $10,000 in 2012
car worth is $8000 in 2014
let linear function is given as
P(t) = at + b
which denote the value of car in year t
take t =0 for year 2012
at t =0, 10,000 = 0 + b
we get b = 10,000
take t =2 for year 2014
at t =2, P(2) = 2a + b
8800 = 2a + 10,000
a = - 600
Thus the price of car at year t after 2012 is given as p(t) = -600t + 10000
let the exponential function 
where t denote t = 0 at 2012
putting t = 0 P(0) = 10,000 we get 10,000 = ab^0
a = 10,000
putting t = 2 p = 8800
b = 0.938
g(t) = 10000(0.938)^t
Answer: -62
Step-by-step explanation:
9( a + 2b) + c
Substitute correct values for all a, b and c.
9( -3 + 2(-2) ) + 1
9( -3 - 4 ) + 1
9(-7) + 1
-63 + 1
-62.
f(x) = 5x is linear. Just a straight line with a slope of +5. So if the intervals are both a difference of 1, then the average rate of change will be the same.
f(x2) - f(x1) over x2 - x1. That's the formula for average rate of change.
So for Section A:
f(x) = 5x, (0,1)
[f(1) - f(0)]/(1-0)
= [5(1) - 5(0)]/1
=(5)/1
=5
Do the same for section B and you'll get 5 as well.
I hope this helps you because I have no clue if my answer is right
<span>If you are adding a constant, then the graph is either raised or lowered k units.
For example.... If you have the graph of y = x^2
and now you add 3, so your new graph of x^2 + 3, will be the same graph as x^2 but raised vertically 3 units.
The graph of x^2 - 7 will be the graph of x^2 lowered vertically 7 units.
I hope my answer has come to your help. God bless and have a nice day ahead!</span>
