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)
If you don't need an exact answer, 9 is the answer, as she'd walk around it 54 times.
The answer is C. 154, because you have a right angle so 90-26= 64 on one side and then add the 90 from the other side.
Answer:
here's the website ans the answers
Step-by-step explanation:
yes its positive and the slope is well I don't know sorry pls give me credit tho :((
x=3, you can find this by using finding common bases for equation b because 243=3^5. Because of that you can get rid of three on both sides to get the equation 2x-1=5 and then just solve for x to get x=3. For the first equation 5^3=625 so that also proves x=3.
Sorry if my explanation is a little disorganized if you need more help feel free to ask
