8/12 and 4/6 and10/15 and20/30 and 40/60
100-99=1+98=99-97=2+96=98-95=3
2-1=1
Answer: The dimensions are: " 1.5 mi. × ³⁄₁₀ mi. " .
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{ length = 1.5 mi. ; width = ³⁄₁₀ mi. } .
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Explanation:
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Area of a rectangle:
A = L * w ;
in which: A = Area = (9/20) mi.² ,
L = Length = ?
w = width = (1/5)*L = (L/5) = ?
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A = L * w ; we want to find the dimensions; that is, the values for
"Length (L)" and "width (w)" ;
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Plug in our given values:
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(9/20) mi.² = L * (L/5) ; in which: "w = L/5" ;
→ (9/20) = (L/1) * (L/5) = (L*L)/(1*5) = L² / 5 ;
↔ L² / 5 = 9/20 ;
→ (L² * ? / 5 * ?) = 9/20 ?
→ 20÷5 = 4 ; so; L² *4 = 9 ;
↔ 4 L² = 9 ;
→ Divide EACH side of the equation by "4" ;
→ (4 L²) / 4 = 9/4 ;
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to get: → L² = 9/4 ;
Take the POSITIVE square root of each side of the equation; to isolate "L" on one side of the equation; and to solve for "L" ;
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→ ⁺√(L²) = ⁺√(9/4) ;
→ L = (√9) / (√4) ;
→ L = 3/2 ;
→ w = L/5 = (3/2) ÷ 5 = 3/2 ÷ (5/1) = (3/2) * (1/5) = (3*1)/(2*5) = 3/10;
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Let us check our answers:
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(3/2 mi.) * (3/10 mi.) =? (9/20) mi.² ??
→ (3/2)mi. * (3/10)mi. = (3*3)/(2*10) mi.² = 9/20 mi.² ! Yes!
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So the dimensions are:
Length = (3/2) mi. ; write as: 1.5 mi.
width = ³⁄₁₀ mi.
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or; write as: " 1.5 mi. × ³⁄₁₀ mi. " .
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1 1/3 is the answer
you have to subtract the numerator of the improper fraction from the denominator and the number you have left is the fraction like for 4/3 4-3=1 so the fraction is 1/3 the amount that is NOT a fraction counts as the whole depending on how many times it can go into the denominator.
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
