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)
The answer for either field being harvested is 6.25 days (acres/(rate of harvest))
The answer is probably seven because, even though a day can be divided into parts, the fields are done the seventh not the sixth. (I'm assuming the problem calls for whole days)
Answer:
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Answer:
4
Step-by-step explanation:
20×25 = 500
20 guest
25 dollars per guest
75s+200= 500
75s = 300
S = 4
They can afford to buy 4 cases of snacks and stay within their budget.
