
Integrating gives

To compute the integral, substitute
, so that
. Then

Since
for all
, we can drop the absolute value, so we end up with

Given that
, we have

so that

do 12.5/5 then you will get ur answer
For the first quarter you would use the equation of $2000+(2000 times 0.08)=the money for the first quarter or $2160. the second quarters equation is $2000+(2000 times 0.08)2= the money for the second quarter or $2320. the thirds quarters equation is $2000+(2000 times 0.08)3= the money for the third quarter or $2480. The fourth quarters equation is $2000+(2000 times 0.08)4=the money for the fourth quarter or $2640.
Answer:
The best place to jack off is by her window my dude
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
It looks like figure A shows lower scores since it doesn't start at 0