You have a lot of questions here, try breaking them up into multiple posts and you may get more responses.
I will help with some.
<span>(x+3)=0
The solution is just -3.
-----------------------------------------------------------------
3m(m-4)=0
The two solutions are 0 and 4.
-----------------------------------------------------------------
(r-3)(r+2)=0
The two solutions are and 3 and -2.
</span>
I hope this gets you started in the right direction.
(9x^4-13x^3-x-7)+(7x^3-2x+1)=
9x^4-13x^3-x-7+7x^3-2x+1=
9x^4+(-13+7)x^3+(-1-2)x-7+1=
9x^4+(-6)x^3+(-3)x-6=
9x^4-6x^3-3x-6
Answer: Option <span>D.)9x^4−6x^3−3x−6</span>
i think it is D because the patten is non linear
The Law of Cosine states
cos

so plugging in the numbers gives us
cos

We now have cosJ so plug that into your calculator and find arccos (arccos(cos(J)) = J)
arccos

rounding your answer will give you 34°