Answer:

Step-by-step explanation:
![\sf 3a^5-18a^3+6a^2\\\\HCF = 3a^2\\\\Take \ 3a^2 \ common\\\\= 3a^2(a^3-6a+2)\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%203a%5E5-18a%5E3%2B6a%5E2%5C%5C%5C%5CHCF%20%3D%203a%5E2%5C%5C%5C%5CTake%20%5C%203a%5E2%20%5C%20common%5C%5C%5C%5C%3D%203a%5E2%28a%5E3-6a%2B2%29%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
X² + 3x = 10
Convert to standard form.
x² + 3x - 10 = 0
factor x² = x * x
factor -10 = 5 * -2
(x + 5) (x - 2) = 0
x(x -2) + 5(x-2) = 0
x² - 2x + 5x - 10 = 0
x² + 3x - 10 = 0
x + 5 = 0 x² + 3x = 10
x = -5 -5² + 3(-5) = 10
25 - 15 = 10
10 = 10
x - 2 = 0 x² + 3x = 10
x = 2 2² + 3(2) = 10
4 + 6 = 10
10 = 10
An expression to show would be : g/12
First thing to do is to change the radians to degrees so it's easier to determine our angle and where it lies in the coordinate plane.

. If we sweep out a 210 degree angle, we end up in the third quadrant, with a 30 degree angle. In this quadrant, x and y are both negative, but the hypotenuse, no matter where it is, will never ever be negative. So the side across from the 30 degree reference angle is -1, and the hypotenuse is 2, so the sine of this angle, opposite over hypotenuse, is -1/2