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).
Answer:
Don
Step-by-step explanation:
1. We make the fractions have common denominators so it is easier to compare them. We can do this buy multiplying 2/3 by a factor of 16, so it becomes 32/48. For 11/16, we multiply by a factor of 3 so it becomes 33/48. It is now apparent that Don has the longer pipe.
- About 127.3(square root of 16,200, or 90√2) Look at this problem like a right triangle. Each leg is 90 feet, so the hypotenuse is the square root of 90^2 + 90^2
- About 52.3(square root of 2735.64) Another right triangle problem! Once again, with Pythagorean theorem (a^2 + b^2 = c^2) You can deduce that 60^2 = 29.4^2 + the width of the TV^2.
- About 11.6(Square root of 134.75)Another right triangle problem, you can deduce that 9.5^2 + The pool length^2 = 15^2
Hope it helps <3
(If it does, please give brainliest, only need one more for rank up :) )
So since the vertex falls onto the axis of symmetry, we can just solve for that to get the x-coordinate of both equations. The equation for the axis of symmetry is 
, with b = x coefficient and a = x^2 coefficient. Our equations can be solved as such:
y = 2x^2 − 4x + 12: 
y = 4x^2 + 8x + 3: 
In short, the vertex x-coordinate's of y = 2x^2 − 4x + 12 is 1 while the vertex's x-coordinate of y = 4x^2 + 8x + 3 is -1.
