SOLUTION: Please help me with the study guide below : http://ariggs.buchananschools.com/uploads/9/0/5/7/9057541/chp_6_test_a.pdf

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Question 1018054: Please help me with the study guide below :

http://ariggs.buchananschools.com/uploads/9/0/5/7/9057541/chp_6_test_a.pdf
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
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1) (2^(-3)x^0) / y^(-4) = (y^4 / 8)
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2) 3^0 = 1
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3) 5^(-6) * 5^3 = 5^3 / 5^6 = 1 / 5^3 = (1 / 125)
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4) (-4x^2)^3= (-4)^3 * (x^2)^3 = -64x^6
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5) (1/3a^3)^-4 = (1 / 3^(-4)a^(-12)) = 81a^12
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6) 12(c^-7)(d^6) = (12d^6 / c^7)
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7) -(216)^(1/3) = -6
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8) (64)^(5/3) = 1073741824^(1/3) = 1024
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9) (-8)^(2/3) = (64)^(1/3) = 4
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10) y = -2(3x)
+graph%28+300%2C+200%2C+-4%2C+4%2C+-16%2C+1%2C+-2%2A%283%5Ex%29%29
domain is all real numbers and range is f(x) < 0
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11) y = 3(0.5)^x
+graph%28+300%2C+200%2C+-4%2C+4%2C+-1%2C+16%2C+3%2A%280.5%29%5Ex%29
domain is all real numbers and range is f(x) > 0
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12) 3^x = 1 / 81
(3^x) * 81 = 1
x = -4 since 3^-4 = 1 / 3^4 = 1 / 81
**************************************************
13) 25^(2x-3) = 125^(x+1)
(5^2)^(2x-3) = (5^3)^(x+1)
5^(4x-6) = 5^(3x+3)
4x-6 = 3x+3
x = 9
:
25^(15) = 125^(10)
5^30 = 5^30
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14) A = P(1 + (r/n))^nt
for our problem n=1, that is number of times compounded yearly, then
A = P(1 +r)^t
A is the balance after t years, P is the principal, r is the rate as decimal
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15) A = 500(1 + 0.07)^2 = $572.45
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16) Car costs $6599 and decreases 12% per year
a) V = P(1 - r)^t where V is the depreciated valye, P is principal, r is rate and t is years
b) V = 6599(1 - 0.12)^(2.5) = 4793.854061425 approx $4793.85
c) V = 6599(1 - 0.12)^20 = 511.836875218 approx $511.84
d) 0 = 6599(1 - 0.12)^t
t = 110 years
This will get you started - rest to follow