<span>measure of ∠EGF = 1/2( 180 - 50)
= 1/2(130)
= 65
</span><span>the measure of ∠CGF = 180 - 65
= 115</span>
Answer:
Yes?
Step-by-step explanation:
Addition is needed because of the subtraction sign.
Answer:
![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
Just take the larger exponent and subtract the smaller exponent and it becomes -->
= ![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)
Hello, please consider the following.
When
and
are two roots, we can factorise as
![ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3Da%28x-x_1%29%28x-x_2%29%3Da%28x%5E2-%28x_1%2Bx_2%29x%2Bx_1x_2%29%3D0)
So for the first equation, we can say that the sum of the zeros is
![\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Ba%5E2%2Bb%5E2%7D%7B2%7D%3D%5Cdfrac%7Ba%5E2%7D%7B2%7D%2B%5Cdfrac%7Bb%5E2%7D%7B2%7D)
and the product is
![\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Ba%5E2b%5E2%7D%7B4%7D%3D%5Cdfrac%7Ba%5E2%7D%7B2%7D%5Cdfrac%7Bb%5E2%7D%7B2%7D)
So we can factorise as below.
![4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0](https://tex.z-dn.net/?f=4x%5E2-2%28a%5E2%2Bb%5E2%29x%2Ba%5E2b%5E2%3D%282x-a%5E2%29%282x-b%5E2%29%3D0)
And the solutions are
![\boxed{\sf \n\bf \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20%5Cn%5Cbf%20%20%5C%20%5Cdfrac%7Ba%5E2%7D%7B2%7D%20%5C%20%5C%20and%20%5C%20%5C%20%5Cdfrac%7Bb%5E2%7D%7B2%7D%7D)
For the second equation, we will complete the square and put the constant on the right side and take the root.
Let's do it!
![9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}](https://tex.z-dn.net/?f=9x%5E2-9%28a%2Bb%29x%2B2a%5E2%2B5ab%2B2b%5E2%3D0%5C%5C%5C%5C9%28x-%5Cdfrac%7Ba%2Bb%7D%7B2%7D%29%5E2-9%5Cdfrac%7B%28a%2Bb%29%5E2%7D%7B4%7D%2B2a%5E2%2B5ab%2B2b%5E2%3D0%5C%5C%5C%5C9%28x-%5Cdfrac%7Ba%2Bb%7D%7B2%7D%29%5E2%3D%5Cdfrac%7B9%28a%2Bb%29%5E2-8a%5E2-20ab-8b%5E2%7D%7B4%7D%5C%5C%5C%5C9%28x-%5Cdfrac%7Ba%2Bb%7D%7B2%7D%29%5E2%3D%5Cdfrac%7B9a%5E2%2B18ab%2B9b%5E2-8a%5E2-20ab-8b%5E2%7D%7B4%7D%5C%5C%5C%5C9%28x-%5Cdfrac%7Ba%2Bb%7D%7B2%7D%29%5E2%3D%5Cdfrac%7Ba%5E2%2Bb%5E2-2ab%7D%7B4%7D%3D%5Cdfrac%7B%28a-b%29%5E2%7D%7B4%7D%5C%5C%5C%5C%28x-%5Cdfrac%7Ba%2Bb%7D%7B2%7D%29%5E2%3D%5Cdfrac%7B%28a-b%29%5E2%7D%7B2%5E2%5Ccdot%203%5E2%7D)
We take the root, and we find the two solutions
![\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dx_1%26%3D%5Cdfrac%7Ba%2Bb%7D%7B2%7D-%5Cdfrac%7Ba-b%7D%7B6%7D%5C%5C%5C%5C%26%3D%5Cdfrac%7B3a%2B3b-a%2Bb%7D%7B6%7D%5C%5C%5C%5C%26%3D%5Cdfrac%7B2a%2B4b%7D%7B6%7D%5C%5C%5C%5C%26%5Cboxed%7B%3D%5Cdfrac%7Ba%2B2b%7D%7B3%7D%7D%5Cend%7Baligned%7D)
![\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dx_1%26%3D%5Cdfrac%7Ba%2Bb%7D%7B2%7D%2B%5Cdfrac%7Ba-b%7D%7B6%7D%5C%5C%5C%5C%26%3D%5Cdfrac%7B3a%2B3b%2Ba-b%7D%7B6%7D%5C%5C%5C%5C%26%3D%5Cdfrac%7B4a%2B2b%7D%7B6%7D%5C%5C%5C%5C%26%5Cboxed%7B%3D%5Cdfrac%7B2a%2Bb%7D%7B3%7D%7D%5Cend%7Baligned%7D)
Thank you.