Solution
Given the quadratic equation
we need to find the zeros of the equation
To do that, we use the completing the square method
Step 1. Add 38 to both sides
Step 2: add the square of half of the coefficient of x to both sides
That is;
Step 3: Simplify the above expression;
![\begin{gathered} \Rightarrow x-1=\pm\sqrt[]{39} \\ \\ \Rightarrow x=1\pm\sqrt[]{39} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20x-1%3D%5Cpm%5Csqrt%5B%5D%7B39%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20x%3D1%5Cpm%5Csqrt%5B%5D%7B39%7D%20%5Cend%7Bgathered%7D)
Answer:
(13y + 15) -9y
-117y - 135
Step-by-step explanation:
Distribute -9y to the inside numbers in the parentheses -117y and -135 are not compatible to each other ( because the -135 doesn't have a y ) so that would give you your final answer of:
-117y -135
Answer: 10
Step-by-step explanation:
X/4 = 12
⇒ x= 12*4
⇒ x= 48
The final answer is x= 48~
21/99 or 7/33 I'm pretty Shure because the numbers repeat, it's probably divided by 9 or some multipule of 9
if 2 numbers repeat, it's probably divided by 99
like 0.abababababab=ab/99
