For the given equation;
We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.
Let us begin by expanding the parenthesis;
Now that we have expanded the left side of the equation, we would have;
We shall now solve the resulting quadratic equation using the quadratic formula as follows;
![\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20%5Ctext%7BWhere%3B%7D%20%5C%5C%20a%3D9%2Cb%3D-30%2Cc%3D150%20%5C%5C%20x%3D%5Cfrac%7B-%28-30%29%5Cpm%5Csqrt%5B%5D%7B%28-30%29%5E2-4%289%29%28150%29%7D%7D%7B2%289%29%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B900-5400%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-4500%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%5Ctimes5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%7D%5Ctimes%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BTherefore%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B30%2B30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%2Cx%3D%5Cfrac%7B30-30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BDivide%20all%20through%20by%206%2C%20and%20we%27ll%20have%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
ANSWER:
![x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D)
Answer:
7/33 = 21 repeating and I'm pretty sure the question wants you to write it as an improper fraction. so your answer would be 40/33
Step-by-step explanation:
If triangles AMN and ABC are similar, then
AM/AB = AN/AC
or
AM/(AM + MB) = AN/(AN + NC)
Check if this is true:
AM/AB = 21/(21 + 9) = 21/30 = 7/10
AN/AC = 14/(14 + 6) = 14/20 = 7/10
The angle at vertex A is common to both of the triangles.
Then by the side-angle-side (SAS) similarity theorem, the triangles are indeed similar.
Given the compound statement <span>(p∨q)∧r
where: p: 5 < -3
q : All vertical angles are congruent.
r: 4x = 36, then x = 9.
Recall the in logic, '</span>∨' symbolises "or" while '∧' symbolises "and".
Therefore, the compound statement <span>(p∨q)∧r can be written as follows:
Either 5 < -3 or all vertical angles are congruent, and if 4x = 36, then x = 9.
</span>
5 Yellow = 3 Blue
2 Yellow = 3/5 *2 = 6/5 Blue, 1.2 cans of Blue .
<u>Award brainliest if helped!</u>
