WHAT??? HELP ME PLEASE///8TH GRADE MATH

You can use the Pythagorean theorem, a^2+b^2=c^2, to solve this.

b= square-root of (c^2)-(a^2)

b= square-root of (9^2)-(2^2)

b= 8.77.... which in the exact form is square root of 77.

5 0

2 years ago

Find the value of x.

scZoUnD [109]

Step-by-step explanation:

$90 + 90 + 52 + x = 360 \\ x = 360 - 90 - 90 - 52 \\ x = 128$

3 0

2 years ago

What is the answer to this solution. -7r−4≥ 4r+2

tia_tia [17]

Answer:

The solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$

Please check the attached line graph below.

Step-by-step explanation:

Given the expression

$-7r-4\ge \:4r+2$

Add 4 to both sides

$-7r-4+4\ge \:4r+2+4$

Simplify

$-7r\ge \:4r+6$

Subtract 4r from both sides

$-7r-4r\ge \:4r+6-4r$

Simplify

$-11r\ge \:6$

Multiply both sides by -1 (reverses the inequality)

$\left(-11r\right)\left(-1\right)\le \:6\left(-1\right)$

Simplify

$11r\le \:-6$

Divide both sides by 11

$\frac{11r}{11}\le \frac{-6}{11}$

Simplify

$r\le \:-\frac{6}{11}$

Therefore, the solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$

Please check the attached line graph below.

7 0

2 years ago