Answer:
f(x) = 26500 * (0.925)^x
It will take 7 years
Step-by-step explanation:
A car with an initial cost of $26,500 depreciates at a rate of 7.5% per year. Write the function that models this situation. Then use your formula to determine when the value of the car will be $15,000 to the nearest year.
To find the formula we will use this formula: f(x) = a * b^x. A is our initial value which in this case is $26500. B is how much the value is increasing or decreasing. In this case it is decreasing by 7.5% per year. Since the car value is decreasing we will subtract 0.075 from 1. This will result in the formula being f(x) = 26500 * (0.925)^x. Now to find the value of the car to the nearest year of when the car will be 15000 we plug 15000 into f(x). 15000 = 26500 * (0.925)^x. First we divide both side by 26500 which will make the equation: 0.56603773584=(0.925)^x. Then we will root 0.56603773584 by 0.925. This will result in x being 7.29968 which is approximately 7 years.
Answer:
A
Step-by-step explanation:
how long is the ball in the air ?
that is the same as asking : after how many seconds will the ball hit the ground (= reach the height of 0) ?
so, that means we need to find the zero solution of h(t).
at what t is h(t) = 0 ?
when at least one of the factors is 0 :
2(-2 - 4t)(2t - 5)
we have 3 factors
2 : can never be 0.
(-2 -4t) : can only be 0 for negative t, which does not make sense in our scenario (we cannot go back in time, only forward).
(2t - 5) : is 0 when 2t = 5 or t = 2.5
so, A is the right answer.
FYI : the starting height (on the hill) is given by t = 0 :
2(-2 - 0)(0 - 5) = 2×-2×-5 = 20 ft
Answer:
A. length = 23 cm; width = 5 cm
Step-by-step explanation:
So if 4 times the width would be 
and 3 more than would be 
, the equation would be 
. If the perimeter is 56, then the new equation, 
can be formed. This simplifies to 
, then to 
, then to 
, and finally, 
. So the width is 5, and now plug it into the first equation.
, becomes 
, becomes 
.
So the dimensions are: length = 23 cm, width = 5 cm.
*And just to be sure, 
which is correct.
