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professor190 [17]
4 years ago
11

NEED HELP ASAP!!! These math problem like this are not my strongest when dealing with math. Three boxes are stacked one on top o

f the other. One box is 5 feet 4 inches tall, one is 2 feet 7 inches tall, and one is 3 feet 10 inches tall. How high is the stack? Write your answer in feet and inches. Use a number less than 12 for inches.
Mathematics
1 answer:
Norma-Jean [14]4 years ago
6 0
5' 4" = 64"  (5x12)+4
2' 7" = 31"  (2x12)+7
3' 10" = 46"  (3x12)+10

64+31+46= 141"
141/12= 11.75= 11 3/4
12" per foot / 4 denominator from above= 3
3 numerator from above x 3 quotient from above= 9
11' 9"
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Answer:

The option \frac{(x+5)(x+2)}{x^3-9x} is correct

The difference of the given expression is

\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}

Step-by-step explanation:

Given expression is \frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})

To find the difference of the given expression as below :

\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})

=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-9)})-({\frac{x+1}{x^2-9})

=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-3^2)})-({\frac{x+1}{x^2-3^2})

=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x-3)(x+3)})-({\frac{x+1}{(x-3)(x+3)})  

( using the formula a^2-b^2=(a+b)(a-b) )

=\frac{2x+5(x+3)-(3x+5)-x(x+1)}{x(x-3)(x+3)}

=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x-3)(x+3)} (adding the like terms)

=\frac{x^2+7x+10}{x(x^2-9)} ( by factoring the quadratic polynomial )

=\frac{(x+5)(x+2)}{x^3-9x}

Therefore \frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}

Therefore the difference of the given expression is

\frac{(x+5)(x+2)}{x^3-9x}

Therefore option \frac{(x+5)(x+2)}{x^3-9x} is correct

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