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
Step-by-step explanation:
A = -7.
B = -8 x-(-6) is the same as x + 6
C = 13
D = 2
E = -3
(1/4)^-2 - (5^0 x 2) x 1^-1 =
(4/1)^2 - (1 x 2) x 1 = 16-2 = 14
If you raise something to the power of -2, swap numerator and denominator and remove the minus.
So (1/4)^-2 = 4^2 = 16
Also 1^-1 is just 1, not -1.
Hi ky name is ni ole and i xont know the ansewer bye
|-6 1/2|, |-1|, |0|, |9|, these are simply going from smallest to largest, once you get past 0 on a number line it goes back by one, for example 0,-1,-2,-3 and so forth, don’t think of negatives like whole numbers