In the 1st pair of fours the 4 on the left side is 10 times greater than the 4 on the right. The same thing with the 2nd pair of fours. The 4 on the left side is 10 greater than the 4 on the right. Hope this helps!
Tan(-x)csc(-x)sec -1 (-x)cot(-x) i am not sure, I hope you get a good grade, this is what i got.. ',: (
Answer:
21n² - 28n + 7
Step-by-step explanation:
Each term in the second factor is multiplied by each term in the first factor, that is
3n(7n - 7) - 1(7n - 7) ← distribute both parenthesis
= 21n² - 21n - 7n + 7 ← collect like terms
= 21n² - 28n + 7