Answer:
125/6(In(x-25)) - 5/6(In(x+5))+C
Step-by-step explanation:
∫x2/x1−20x2−125dx
Should be
∫x²/(x²−20x−125)dx
First of all let's factorize the denominator.
x²−20x−125= x²+5x-25x-125
x²−20x−125= x(x+5) -25(x+5)
x²−20x−125= (x-25)(x+5)
∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx
x²/(x²−20x−125) =x²/((x-25)(x+5))
x²/((x-25)(x+5))= a/(x-25) +b/(x+5)
x²/= a(x+5) + b(x-25)
Let x=25
625 = a30
a= 625/30
a= 125/6
Let x= -5
25 = -30b
b= 25/-30
b= -5/6
x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)
∫x²/(x²−20x−125)dx
=∫125/6(x-25) -∫5/6(x+5) Dx
= 125/6(In(x-25)) - 5/6(In(x+5))+C
Answer: The answer is 29, because that 29 would be the outlier of the numbers.
Step-by-step explanation:
X=1.3
Hope this helps I not I’m truly sorry
Answer:
Yes, it passes the vertical line test
Step-by-step explanation:
This graph has a one to one correspondence and will pass the vertical line test, there fore it is a function
