Solve for r.
You want to get r by itself on one side on the equal sign.
bh + hr = 25
Subtract bh from both sides.
hr = 25 - bh
Divide h on both sides.
r = 25 - bh / h
The two h's cancel each other out.
r = 25 - b
Hope this helps!
12. the change in temperature is -1 C. -3 C + -1 C= -4 C
13. The net gain. Lin gained 14 yards, then loss 7 yards, and lastly gained 9 yards
14 + (-7) + 9= 16 yards
14. Pythagoras was born 582 BC. Newton 1643 AD
1643-582= 1,061 years
15. Sonny has $75 but he needs to buy something costs $93
he needs to borrow $18.
93-75= $18
The formula for the perimeter of a rectangle is P = 2L + 2W, where L is the length and W is the width. Because we don't know either the length or the width we can't solve the problem...too many unknowns. BUT we do have some information that will help with this problem. We are told that the length is 2 feet longer than the width, so we can use that: L = W+2. Now we can make the substitution into the formula along with the value for the perimeter that was given to us: 36=2(W+2) + 2W, and 36 = 2W + 4 + 2W; 36 = 4W + 4; 32 = 4W and W = 8. Now go back to where you said that the length is 2 feet longer than the width. If the width is 8, then 8+2 = 10 for the length.
Answer:
N = 920(1+0.03)^4t
Step-by-step explanation:
According to the given statement a car repair center services 920 cars in 2012. The number of cars serviced increases quarterly at a rate of 12% per year after 2012.
Rate is 12 % annually
rate in quarterly = 12/4= 3%
We will apply the compound interest equation:
N=P( 1+r/n)^nt
N= ending number of cars serviced.
P= the number of cars serviced in 2012,
r = interest rate
n = the number of compoundings per year
t= total number of years.
Number of compoundings for t years = n*t = 4t
Initial number of cars serviced=920
The quarterly rate of growth = n=4
r = 3%
The growth rate = 1.03
Compound period multiplied by number of years = 920(1.03)^4t
Thus N = 920(1+0.03)^4t
N = number of cars serviced after t years...
Answer:
5
Step-by-step explanation:
k(x)=5 says I'm 5 no matter the value of x...
So therefore
k(-4)=5
k(56)=5
k(66378)=5
k(whatever)=5
k(x) is constantly 5 for whatever input x.
