The value of t that makes the factor e^(.032t) have the value of 2 can be found using logarithms.
2 = e^(0.032t)
ln(2) = ln(e^(0.032t)) = 0.032t
t = ln(2)/0.032 ≈ 21.66
It would take 21.66 years for the cost to double.
Step-by-step explanation:
f(x) = y= -x²-x+6
When y is max, f'(x) = 0
f'(x) = -2x-1
Equating to 0,
-2x-1= 0
x= -1/2
Put the value of x in f(x)
y = -(-1/2)² -(-1/2) +6
= -1/4+1/2+6
= 6+1/2
y= 13/2
7x^2 + 3
7(4)^2 + 3
7(16) + 3
112 + 3
115.
If n is the first integer, then n+2 is the second one. The equation for the sum can be written as
n + (n+2) = 24 . . . . . . this equation can be used to solve the problem
2n = 22 . . . . . . . . . . . simplify, subtract 2
n = 11
The integers are 11 and 13.
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I like to solve problems like this by looking at the average of the integers. Here, it is 24/2 = 12. The consecutive odd integers whose average is 12 are 11 and 13.
