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3 years ago

A B C D PLZ HELP ASAP

Mathematics

1 answer:

marishachu [46]3 years ago

8 0

Answer:

B. 4, 4, 4√2

Step-by-step explanation:

45° - 45° - 90° is the measure of special right triangle

the side lengths that sees these angles can be shown as a, a, and a√2 the values that meets these conditions are shown in option b

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$2x - 7x + 2 + 1 - 2$

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2 years ago

Rachel and Sierra are selling boxes of fruit as a fundraiser. Rachel has sold 9/10 boxes of fruit and Sierra has sold 2 has sold

Mandarinka [93]

Rachel sold more because 9/10 is equal to 0.90 and 5/8 is equal to 0.625.

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3 years ago

Find the area of a square whose side is 4xy.

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Step-by-step explanation:

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3 years ago

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What is the sum of 8a<br> 7+ 5b2 – 2a2 and 7b2 + 66_<br> 20a + 3?

barxatty [35]

Answer:

Step-by-step explanation:

7+5b²-2a²+7b²+66-20a-3

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5 0

3 years ago

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago