F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
Graph A since it crosses all 4 regions of the graph
Answer:
Yes, and if you need extra help try yay math
Step-by-step explanation:
Answer:
D. 8
Step-by-step explanation:
2p=16
divide both sides by 2 to get p alone
p=8
The first step of factoring is to try to factor out a common factor.
The terms x^2 and -9x have the factor x in common.
Factor out x from both terms.
x^2 - 9x = 0
x(x - 9) = 0
Now you have a product of fully factored terms equaling zero, so you can apply the zero product property to solve.
x = 0 or x - 9 = 0
x = 0 or x = 9
Answer: x = 0 or x = 9
