I don't see a square root sign anywhere, so I'll assume the integral is

First complete the square:

Now in the integral, substitute

so that

Under this change of variables, we have

so that

Under the right conditions, namely that cos(<em>t</em>) > 0, we can further reduce the integrand to


Expand the sine term as

Then


So the integral is

If the answer is wrong then the answer isn't right but it also isn't left hence making it wrong.
Answer:
20
Step-by-step explanation:
-10-5x(-6)
-10+30
20
12 this is because my number for sports is 12 and i just need to answer a ? for point thanks have a good day
Answer:
x = $0.50
y= $0.75
Step-by-step explanation:
1. Multiply the equations to have the same coefficients
5(6x + 6y = 7.5) → 30x + 30y = 37.5
3(10x + 5y = 8.75) → 30x + 15y = 26.25
2. Subtract the equations
30x + 30y = 37.5
<u>- 30x + 15y = 26.25</u>
15y = 11.25
3. Solve for y by dividing both sides by 15
y = 0.75
4. Plug in 0.75 for y into one of the equations
6x + 6(0.75) = 7.5
5. Simplify
6x + 4.5 = 7.5
6. Solve for x
6x = 3
x = 0.5