Let that be

Two vertical asymptotes at -1 and 0

If we simply

- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is

Graph attached
4/8 is less full because it's only half compared to 4/6 which is 1.5
a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

Then by linearity, we have

Answer: B: 24q+48r
Step-by-step explanation: 6(4q+8r)
=(6)(4q+8r)
=(6)(4q)+(6)(8r)
=24q+48r
Answer:
B.
Step-by-step explanation:
The problem is saying 2/5 squared by -2 times -3 multiplied by itself 4 times.
2/5 multiplied by itself -2 times looks like this: -2/5 times -2/5.
-3 multiplied by itself 4 times looks like this: -3 times -3 times -3 times -3.
Put the two together and it looks like B.