P(r)=r^2-0.1
144
What is the approximate value of P(3)?
A.0.20
B.0.034
C.0.062
D.0.041
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
Directrix (directrix):
What is the 144 in the posted problem?
OpenStudy (ambermarie151):
It's supposed to be divide by 2-0.1/144
Directrix (directrix):
@ambermarie151
Would you click on the blue DRAW tab and write the equation as it appears in your text? Thanks.
jimthompson5910 (jim_thompson5910):
The equation is probably
\[\Large P(r) = \frac{r^2-0.1}{144}\]
but that's just an educated guess
OpenStudy (ambermarie151):
Yes that's it
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
ok so you just replace every copy of r with 3
\[\Large P(r) = \frac{r^2-0.1}{144}\]
\[\Large P({\color{red}{r}}) = \frac{({\color{red}{r}})^2-0.1}{144}\]
\[\Large P({\color{red}{3}}) = \frac{({\color{red}{3}})^2-0.1}{144}\]
then you evaluate the right side
jimthompson5910 (jim_thompson5910):
hopefully that makes sense @ambermarie151
OpenStudy (ambermarie151):
So would you do 3•3 -0.1 then divide by 144 ?
jimthompson5910 (jim_thompson5910):
correct since 3^2 means 3*3
OpenStudy (ambermarie151):
I am still confused
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (ambermarie151):
I get a decimal
jimthompson5910 (jim_thompson5910):
what decimal value did you get?
jimthompson5910 (jim_thompson5910):
it's perfectly normal in this case to get a decimal value