Its about a 42 percent chance that both marbles will be red.
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:
Abram payed $12.50 for the cab ride in all.
The consecutive integers are i, i+1, i+2, i+3.
3 less than sum of the integers
= i + i +1 + i + 2 + i + 3 - 3 = 4i + 3
So it is asking you to group like term so
x terms can be grouped/added/subtracted to other x terms, but not to x^2 or x^3 terms
x^2 terms to x^2 and so on so
1. 9-3k+5k=
9+(5k-3k)=
9+2k
2. k^2+2k+4k=
k^2+(2k+4k)=
k^2+6k=
