HALP Solve: |y + 8| ≥ 3

Mathematics

2 answers:

NemiM [27]3 years ago

8 0

The answers gonna be (-17=y)

creativ13 [48]3 years ago

8 0

For
|x|≥y, solve for x such that
x≤-y or x≥y

|y+8|≥3
solve for
y+8≤-3 or y+8≥3
minus 8 from both sides
y≤-11 or y≥-5

the solution set is y≤-11 or y≥-5
basically, all real numbers except z, such that -11<z<-5

the solution set is y≤-11 or y≥-5

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Bezzdna [24]

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A+2/a-5÷a+1/a2-8a+15

krok68 [10]

For $A+\frac{2}{a}-\frac{5}{a}+\frac{1}{a^2}-8a+15$

Combine the fractions $\frac{2}{a} - \frac{5}{a} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{2-5}{a}$

$\mathrm{Subtract\:the\:numbers:}\:2-5=-3 \ \textgreater \ \frac{-3}{a}$

$\mathrm{Apply\:the\:fraction\:rule}:\ \frac{-a}{b}=-\frac{a}{b} \ \textgreater \ -\frac{3}{a}$

Therefore our final answer is $A-\frac{3}{a}+\frac{1}{a^2}-8a+15$

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3 years ago

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Sidana [21]

Answer: 7.2%

Step-by-step explanation:

First find the difference between the amounts.

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Multiply 0.07246 by 100 to convert it to a percent.

0.07246 = 7.2463..%