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miskamm [114]
3 years ago
8

the sum of two numbers is -17 one of the numbers is seven less than the other what are the two numbers?​

Mathematics
1 answer:
dmitriy555 [2]3 years ago
5 0

Answer:

-12 and -5

Step-by-step explanation:

a+b=-17

b-a=7

b=7+a

a+7+a=-17

2a+7=-17

2a=-17-7

2a=-24

a=-12

-12+b=-17

b=-17+12

b=-5

so a and b are -12 and -5 respectively

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1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

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\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

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\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

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\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

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