Answer:
r³ + 15r² - 10r - 4
Step-by-step explanation:
First, note you can remove the brackets, in this case they contribute nothing at all.
Then, group the like terms and add them, e.g., we have 3r³ and -2r², they add up to r³.
Finally, order them from greatest power of r to smallest power of r.
That's all!
Answer:
A
Step-by-step explanation:
Find AB using the Law of Sine:

Thus:

Multiply both sides by sin(40)


AB = 31.0056916 ≈ 31.0 cm (nearest tenth)
By dividing it by a number that = to it
Answer:
It is an identity, proved below.
Step-by-step explanation:
I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.
First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

Another identity is:

Therefore:

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

Another identity:

Therefore:

Hence proved, this is proof by using identity helping to find the specific identity.