Ask question

Ask question

All categories

3 years ago

5

I’m a right triangle the length of the hypotenuse is 12and the length of one leg is 8 . What is the length of the other leg?

1 answer:

denis23 [38]3 years ago

3 0

Answer:

other leg is 8.94

Step-by-step explanation:

Pythagorean Theorem is a² + b² = c²

a and b are legs

c is the hypotenuse

8² + b² = 12²

64 + b² = 144

-64 -64

b² = 80

√b² = √80

b ≈ 8.94

You might be interested in

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bs%7D%7B2%7D" id="TexFormula1" title="\frac{s}{2}" alt="\frac{s}{2}" align="absmiddle

Olegator [25]

S = 92

Explanation: I just know

4 0

3 years ago

Read 2 more answers

solve & show ur steps PLEASE Laura jogs 1/5 kilometer in 1/6 hour. What is Lara's rate in kilometers per hour? write your an

velikii [3]

The picture is how I solved it and it will be correct

4 0

3 years ago

Use the Distributive Property to<br> REWRITE 6x(2y - 4x + 1)

BartSMP [9]

Answer:

−24x2+12xy+6x

Step-by-step explanation:

6x(2y−4x+1)

=(6x)(2y+−4x+1)

=(6x)(2y)+(6x)(−4x)+(6x)(1)

=12xy−24x2+6x

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

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

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

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

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

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

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

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

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

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

as

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

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

so

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

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

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

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

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

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

A classroom contains an equal number of boys and girls. if 10 girls leave, twice as many boys as girls remain. what was the orig

PolarNik [594]

the answer of this question is 38 and i hope it s correct

6 0

1 year ago