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2 years ago

13

Quadrilateral ABCD has vertices A(-3, 1) B(0,4) C(4,0) D(1, -3).

1 answer:

Rina8888 [55]2 years ago

7 0

I think the correct answer is D 1,-3

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Step-by-step explanation:

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3 years ago

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2 years ago

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2 years ago

4/5 divided by 1/3 plus 1/5 minus 3/5

r-ruslan [8.4K]

Answer:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

Step-by-step explanation:

Considering the expression

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

Solution Steps:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

as

$\mathrm{Combine\:the\:fractions\:}\frac{1}{5}-\frac{3}{5}:\quad -\frac{2}{5}$

so

$=\frac{\frac{4}{5}}{\frac{1}{3}-\frac{2}{5}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$=\frac{4}{5\left(\frac{1}{3}-\frac{2}{5}\right)}$

join $\frac{1}{3}-\frac{2}{5}:\quad -\frac{1}{15}$

so

$=\frac{4}{5\left(-\frac{1}{15}\right)}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=\frac{4}{-5\cdot \frac{1}{15}}$

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

$=-\frac{4}{5\cdot \frac{1}{15}}$

$\mathrm{Multiply\:}5\cdot \frac{1}{15}\::\quad \frac{1}{3}$

so

$=-\frac{4}{\frac{1}{3}}$

$\mathrm{Simplify}\:\frac{4}{\frac{1}{3}}:\quad \frac{12}{1}$

so

$=-\frac{12}{1}$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$=-12$

Therefore

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

4 0

2 years ago