Solve for x over the real numbers: by completing the square:
5 x^2 - 30 x = 5
Divide both sides by 5:
x^2 - 6 x = 1
Add 9 to both sides:
x^2 - 6 x + 9 = 10
Write the left hand side as a square:
(x - 3)^2 = 10
Take the square root of both sides:
x - 3 = sqrt(10) or x - 3 = -sqrt(10)
Add 3 to both sides:
x = 3 + sqrt(10) or x - 3 = -sqrt(10)
Add 3 to both sides:
Answer: | x = 3 + sqrt(10) or x = 3 - sqrt(10)
Answer: PART A: The answer is 307
PART B: The prime numbers between 200 and 300 are 211, 223, 227, 229,...., 281, 283, 293 Total Count of Numbers = 16 step 2 Find the sum of prime numbers between 200 and 300.
Step-by-step explanation: Yw and I sure hope this helps u. And mark me as BRAINIEST pls
Answer:
A 4/81
Step-by-step explanation:
(2/9)² is the same as 2/9×2/9, which is 4/81.
Answer:
-16
Step-by-step explanation:
fog(x)=f(gx)
= -5×4+4
= -20+4
= -16
Answer:
x = 8
Step-by-step explanation:
well using the theorem we have
![\frac{6}{9}=\frac{x}{12}\\\\\frac{2}{3}=\frac{x}{12}\\\\12[\frac{2}{3}]=12[\frac{x}{12}]\\\\4\cdot 2=x\\\\x=8](https://tex.z-dn.net/?f=%5Cfrac%7B6%7D%7B9%7D%3D%5Cfrac%7Bx%7D%7B12%7D%5C%5C%5C%5C%5Cfrac%7B2%7D%7B3%7D%3D%5Cfrac%7Bx%7D%7B12%7D%5C%5C%5C%5C12%5B%5Cfrac%7B2%7D%7B3%7D%5D%3D12%5B%5Cfrac%7Bx%7D%7B12%7D%5D%5C%5C%5C%5C4%5Ccdot%202%3Dx%5C%5C%5C%5Cx%3D8)
