<u>Answer-</u>

<u>Solution-</u>
Rational Root Theorem-

All the potential rational roots are,

The given polynomial is,

Here,

The potential rational roots are,


From, the given options only
satisfies.
12(3+1)
12(4)
The number 12 times the number 4 is 48
Answer:
x= 110°
Step-by-step explanation:
if you continue drawing the top purple line you will have a triangle with three angles
70° ; is given
180-140= 40°; is the same angle as the supplement angle of 140° (complementary angles are congruent)
180-70-40 = 70°; because sum of angles in a triangle is 180°
So angle x= 180 -70 = 110 because are a linear pair

solve for "k", to find k or the "constant of variation"
then plug k's value back to

now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
Answer: No solution
Step-by-step explanation: -12x-14=-62-12x
You would have to simplify the -12x but you can't so this would be a no solution.