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...
The arithmetic sequence general formula is an = a1 + d *(n-1) where d is the arithmetic difference and n is an integer. In this case, upon derivation, the formula that represents the sum of the series is S = (a1 + an)*(n/2). Substituting, S = (3+75)*(10/2) equal to 390
Answer: X = -5
Step-by-step explanation:
3(x-5)=2(x-10)
3x - 15 = 2x -20
3x = 2x -5
x = -5
Answer:
Step-by-step explanation:
6.1 / 5 ( 12b + 5 )
b = 5/6
by substituting,
6.1 / 5 ( 12 × 5/6 + 5 )
1.22 × 15
= <u><em>18.3</em></u>
Hope this helps
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