Yz =5
Xz= 7
Hope this helps!!!
Use the binomial series to find the maclaurin series for the function. (use (2n)! 2nn! for 1 · 3 · 5 (2n − 1).) f(x) = 1 1 − x5 vfdgfdbfdvbcbvcb f fbv vbbvcbvb
What? Is that two problems? I'll change my answer, give me my answer.
Answer:
x = -1/2 ( 3±sqrt(37))
Step-by-step explanation:
x^2 + 3x − 7 = 0
Add 7 to each side
x^2 + 3x =7
Using complete the square
Taking the coefficient of x
3
Divide by 2
3/2
Square it
(3/2)^2 = 9/4
Add this to each side
x^2 + 3x+ 9/4 = 7+9/4
( x+ 3/2) ^2 = 28/4 + 9/4
( x+ 3/2) ^2 = 37/4
Take the square root of each side
x+3/2 = ±sqrt(37/4)
x+3/2 = ±sqrt(37) / sqrt(4)
x+ 3/2 = ±sqrt(37) / 2
Subtract 3/2 from each side
x = -3/2 ±sqrt(37) / 2
x = -1/2 ( 3±sqrt(37))
If you multiply x over, you get
8x = 3/4 + 1
8x = 7/4
x = 7/4 * 1/8
x = 7/32
