Answer:
hmmmm.... each question here is relatively straight forward, the issue is that these questions basically represent an entire unit (possible two) of an Algebra II class... I can provide answers, but the explanation (if you don't know the basic ideas might be elusive, But I will try to help)
Step-by-step explanation:
1) A = P (1+ r/n) ^ nt
interest equation ... monthly compounding so n = 12 thus r gets divided by 12 and t get multiplied by 12
A = 10,000(1 + .04/12)^(3*12)
A = 10000(1.00333)^36
A = $11,272.71
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2) [tex]A = pe^{rt}[/tex]
A = 15,000 [tex]e^{.06 t}[/tex]
18,000= 15,000 [tex]e^{.06 t}[/tex]
1.2= [tex]e^{.6 t}[/tex]
ln(1.2) = .06t (ln e)
.1823 / .06 = 3.03 years
Note: ln e = 1
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[tex]3^{5x+1} = 27[/tex]
[tex]3^{5x+1} = 3^{3}[/tex]
thus: 5x+ 1 = 3
5x = 2
x = 2/5
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f(x) = 5x - 12
y = 5x - 12
inverse
x = 5y - 12
5y = x + 12
y = 1/5 x + 12/5
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fog(X) =
f(X) =6x+12
g(x) = x^2 + 3
fog(X) = 6(x^2 + 3)+12
= 6x^2 + 18 + 12
= 8x^2 + 30
Domain (- infinity, + infinity)
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vertex 3x^2 - 6x + 2
verted = [-b/2a, f(-b/2a)]
a=3 , b=-6 , c = 2
-b/2a = 6/6 = 1
f(-b/2a) = -1
vertex = (1,-1)
=======================
30x - .4x^2
-30/-.8 = 37.5
max rev = 30(37.5) - .4(37.5)^2 = $562.5
