now, let's recall the rational root test, check your textbook on it.
so p = 18 and q = 1
so all possible roots will come from the factors of ±p/q
now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.
recall that, a root will give us a remainder of 0.
let us use +3.
![\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)](https://tex.z-dn.net/?f=%5Cbf%20x%5E4-7x%5E3%2B13x%5E2%2B3x-18%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Br%7Crrrrr%7D%203%261%26-7%2613%263%2618%5C%5C%20%26%263%26-12%263%2618%5C%5C%20%5Ccline%7B1-6%7D%20%261%26-4%261%266%260%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%28x-3%29%28x%5E3-4x%5E2%2Bx%2B6%29)
well, that one worked... now, using the rational root test, our p = 6, q = 1.
so the factors from ±p/q are ±3, ±2, ±1
let's use 3 again

and of course, we can factor x²-x-2 to (x-2)(x+1).
(x-3)(x-3)(x-2)(x+1).
The answer is the first chart which would be A
Well A seems like the most correct answer, the actual purpose of an afterburner is to inject fuel directly into the exhaust stream to increase thrust by as much as 50%. It’s usually used to go supersonic
Hey there!
1.) let's remember that a acute angel is very tiny, it is less then 90 degrees. So we would want to find one of your options that would be very small.
46 would seem to be the answer.
x=46
Hope this helps you.