The width of his office on the floor plan is 2.5 in.
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}](https://tex.z-dn.net/?f=-7r-4%5Cge%20%5C%3A%5C%3A4r%2B2%5Cquad%20%5C%3A%3A%5Cquad%20%5C%3A%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ar%5Cle%20%5C%3A%5C%3A-%5Cfrac%7B6%7D%7B11%7D%5C%3A%5C%5C%20%5C%3A%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ar%5Cle%20%5C%3A%5C%3A-0.54545%5Cdots%20%5C%3A%5C%5C%20%5C%3A%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%5C%3A%2C%5C%3A-%5Cfrac%7B6%7D%7B11%7D%5D%5Cend%7Bbmatrix%7D)
Please check the attached line graph below.
Step-by-step explanation:
Given the expression

Add 4 to both sides

Simplify

Subtract 4r from both sides

Simplify

Multiply both sides by -1 (reverses the inequality)

Simplify

Divide both sides by 11

Simplify

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}](https://tex.z-dn.net/?f=-7r-4%5Cge%20%5C%3A%5C%3A4r%2B2%5Cquad%20%5C%3A%3A%5Cquad%20%5C%3A%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ar%5Cle%20%5C%3A%5C%3A-%5Cfrac%7B6%7D%7B11%7D%5C%3A%5C%5C%20%5C%3A%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ar%5Cle%20%5C%3A%5C%3A-0.54545%5Cdots%20%5C%3A%5C%5C%20%5C%3A%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%5C%3A%2C%5C%3A-%5Cfrac%7B6%7D%7B11%7D%5D%5Cend%7Bbmatrix%7D)
Please check the attached line graph below.
Answer:
75 degrees, we know that already so look below lol.
Step-by-step explanation:
This triangle is an isosceles triangle, LP is equivalent to LQ, and angles LQP and LPQ are also equivalent. We understand that PLQ is equal to 30 degrees, since its complementary to triangle LQM.
180 - 30 = 150
150/2 = 75.
<em>Have a nice day, fam. Spread The Love. Thanks for the opportunity.</em>
1/3 x 14
2/3 x 7
4/6 x 7
2/6 x 14
1/3 x 28/2
1/3 x 42/3
Answer: A=351.68 cm²
Step-by-step explanation:
To find the area of the shaded region, we would subtract the area of the circle by the area of the inner circle.
Area of Circle



Area of inner (white) circle



Now that we have the area to the circle and inner circle, we would subtract to find the area of the shaded region.


