Step-by-step explanation:
angle AOB = x+3
angle AOC = 2x + 11
angle BOC = 4x-7
angle AOC = angle AOB + angle BOC
=> 2x +11 = (x+3) + (4x-7)
2x +11 = 5x - 4
=> 3x = 15
x = 5
subst x = 5 in the given formulas
angle AOB = x +3 =8
angle AOC = 2x + 11 = 21
angle BOC = 4x - 7 = 13
The answer to that question is B.
Germany surrendered on November 11, 1918, and all nations had agreed to stop fighting.
Answer:
x=8.06
Step-by-step explanation:
Since it's a right triangle, we can use Pythagorean Theorem, which is a^2+b^2=c^2. The hypotenuse, or longest side is always C, so in this case we're going to do C^2-a^2=b^2. 9^2=81, 4^2=16, 81-16=65, square root of 65 is 8.06, so x=8.06
let's first off, recall that the factoring of the quadratic, will come from the factoring of the constant, whose FOIL factors added will give the middle term.
now, let's firstly multiply both sides by the LCD of 3, to do away with the denominator.
now, since our constant of -9, can only be factored as 3*3 OR 9*1, so
![\bf 3x^2-vx-9t=0\implies \begin{array}{llll} (3x\pm 3)(x\pm 3)=0\\\\ (3x\pm 1)(x\pm 9)=0 \end{array} \\\\[-0.35em] ~\dotfill\\\\ 3x\pm 3=0\implies 3x=\mp 3\implies x=\mp\cfrac{3}{3}\implies \boxed{x=\mp 1} \\\\[-0.35em] ~\dotfill\\\\ x\pm 3=0\implies \boxed{x=\mp 3} \\\\[-0.35em] ~\dotfill\\\\ 3x\pm 1=0\implies 3x=\mp 1\implies \boxed{x=\mp \cfrac{1}{3}} \\\\[-0.35em] ~\dotfill\\\\ x\pm 9 = 0\implies \boxed{x=\mp 9}](https://tex.z-dn.net/?f=%5Cbf%203x%5E2-vx-9t%3D0%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20%283x%5Cpm%203%29%28x%5Cpm%203%29%3D0%5C%5C%5C%5C%20%283x%5Cpm%201%29%28x%5Cpm%209%29%3D0%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%203x%5Cpm%203%3D0%5Cimplies%203x%3D%5Cmp%203%5Cimplies%20x%3D%5Cmp%5Ccfrac%7B3%7D%7B3%7D%5Cimplies%20%5Cboxed%7Bx%3D%5Cmp%201%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20x%5Cpm%203%3D0%5Cimplies%20%5Cboxed%7Bx%3D%5Cmp%203%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%203x%5Cpm%201%3D0%5Cimplies%203x%3D%5Cmp%201%5Cimplies%20%5Cboxed%7Bx%3D%5Cmp%20%5Ccfrac%7B1%7D%7B3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20x%5Cpm%209%20%3D%200%5Cimplies%20%5Cboxed%7Bx%3D%5Cmp%209%7D)
now, we could have also used the <u>rational root test</u> for that, and get the same variations.
