Answer:
y=0x+6
Step-by-step explanation:
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:
c would be the answer you are looking for and please mark brainliest
Step-by-step explanation:
Step-by-step explanation:
Set-builder notation is a notation for describing a set by indicating the properties that its members must satisfy.
C={2,4,6,8}
C={2×1,2×2,2×3,2×4}
thus set builder form is:
{x:x=2n,n∈N} and this set is also call even number set in description mathod
Sqrt(486) - sqrt(24) + sqrt(6)
find the factors of 486 that we can remove from under the square root sign
2 * 243
2 * 3 * 81
2 * 3 * 9 * 9 (we have 2 nines, we can move a 9 outside the sqrt sign)
sqrt(486) = 9 sqrt(6)
Repeating for sqrt(24)
2 * 12
2 * 2 * 6
2 * 2 * 2 * 3 (we can move a 2 outside the sqrt
sqrt(24) = 2 sqrt(6)
Finally, add all 3 terms together
9 sqrt(6) - 2 sqrt(6) + sqrt(6)
8 sqrt(6)
8 times square root of 6 is the final answer
