Log(2)/log(1.064) ≈ 11.17 . . . . hours
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The population can be given by
p(n) = p₀×1.064ⁿ . . . . where n is the number of hours
You want to find n whe p(n) = 2*p₀.
2p₀ = p₀×1.064ⁿ . . . . . . . . . . . . substitute the given information
2 = 1.064ⁿ . . . . . . . . . . . . . . . . . divide by p₀
log(2) = n×log(1.064) . . . . . . . . take logs to make it a linear equation
log(2)/log(1.064) = n . . . . . . . . divide by the coefficient of n
10x + 2y - 6 = 0
Multiply 5x + y - 3 = 0 by 2
2(5x + y - 3 = 0) = 10x + 2y - 6 = 0
Answer:The answer is 5, 414.52
Step-by-step explanation:
Since number of years isn't imputed, it will be assumed it's for 1 year
Hence using Compound interest formula
A = P( 1 + r/n)^not
Inputting values
A = 5000( 1 + 0.08/9) ^ 9*1
A = 5, 414.52
Hope this helps
-16.9x + 16 is the correct answer. You have to remove the parentheses, collect the like terms, and then simplify.
Answer:
3(x + 4) = 3(x) + 3(4)
3(x + 4) = 3x + 12
3x + 12 = 3x + 12
Subtract 12 from both sides
3x + 12 - 12 = 3x + 12 - 12
3x = 3x
3x - 3x = 3x - 3x
= 0
