Answer: 
<u>Step-by-step explanation:</u>
Convert everything to "sin" and "cos" and then cancel out the common factors.
![\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)](https://tex.z-dn.net/?f=%5Cdfrac%7Bcot%28x%29%2Bcsc%28x%29%7D%7Bsin%28x%29%2Btan%28x%29%7D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%5B%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%5D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29cos%28x%29%7D%7Bcos%28x%29%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29)
Answer:
m∠ TRS = 60° , m∠ SRW = 120°
Step-by-step explanation:
First, find x
∠TRS = ∠VRW (vertically opposite angles are equal)
x + 40° = 3x
x - 3x = -40
-2x = -40
x = -40/-2
x = 20
m∠ TRS = 60° [x + 40 = 20+40 = 60]
m∠ SRW + m∠ TRS = 180° (linear pair)
m∠ SRW + 60° = 180°
m∠ SRW = 180° - 60°
m∠ SRW = 120°
hope this helps you
You are given Kayla's walking distance along the edge of the river 100 ft and marks point C, then she walks 100 ft further and marks point D and she turns 90° and walks until her location, point A, and point C are collinear. She marks point E at this location. ABC and EDC are congruent.
1, B
2, C
3, C
4, C
5, a
6, b
7, b
8, C
9, a I think.
Hoped this helps.
I will create a set of arbitrary constants (x1,y1) (x2,y2)
slope = y2-y1/x2-x1
y = (y2-y1/x2-x1)x + b
y2 = (y2-y1/x2-x1)x2 + b
b = y2 - (y2-y1/x2-x1)x2
y = (y2-y1/x2-x1)x + [y2 - (y2-y1/x2-x1)]
Choose any points and just
Plug the values and you have a linear function.
NOT SURE IF THAT'S WHAT THE QUESTION WANTS.
