~~ANSWER THESE FOR A FAN + MEDAL~~ http://prntscr.com/5215jf
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OpenStudy (mathmath333):
\(\large\tt \begin{align} \color{black}{
f(x)=2(x)^2+5 \sqrt{x+2}\\~\\
f(0)=2(0)^2+5 \sqrt{0+2}\\
}\end{align}\)
like this u have to do 9th
OpenStudy (anonymous):
the answer is 5sqrt of 2, but how do i put that
OpenStudy (anonymous):
it said round to the nearest hundredth, i got 7.07 from 5sqrt of 2
OpenStudy (mathmath333):
wait
OpenStudy (mathmath333):
nearest hundredth means having 2 decimal places
\(5\sqrt2=7.07\) is correct
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OpenStudy (anonymous):
ok and 9. is f(19) = -sqrt of 27/7 = -2.00? because rounding from 1.96
OpenStudy (anonymous):
-1.96*
OpenStudy (anonymous):
or is it -2.06
OpenStudy (mathmath333):
first what is the value of f(19) u got?
OpenStudy (anonymous):
- sqrt of 27/7
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OpenStudy (mathmath333):
r us sure u calculated correctly ,cuz i m getting diff value
OpenStudy (anonymous):
i mean just -27/7 sorry
OpenStudy (mathmath333):
ok ,thats correct
OpenStudy (anonymous):
oh i had put the square root on it lol
OpenStudy (anonymous):
that goes to -3.85
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OpenStudy (anonymous):
the answer is -3.90?
OpenStudy (mathmath333):
wait
OpenStudy (anonymous):
-3.85714 , the answer is -3.86? rounded to hundredths
OpenStudy (mathmath333):
well m not sure ,let me call somone
@satellite73 ,@surjithayer
OpenStudy (mathmath333):
@triciaal
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OpenStudy (anonymous):
i got the answers for 8 and 9, what about 10?
OpenStudy (mathmath333):
oh,so whats the answer for 9th
OpenStudy (anonymous):
-3.86
OpenStudy (mathmath333):
the answer for nth would be \(-3.857\)
u did nearest hundredth , but in the question nearest \(thousand \) is given
OpenStudy (anonymous):
ohhhh
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OpenStudy (anonymous):
i read wrong gosh, thanks
OpenStudy (anonymous):
i dont understand how to get 10
OpenStudy (mathmath333):
for the last one the value of x should not be less than 2 other wise
we will get complex numbers so for \(10^{th}\)
\(\underline {greater}~~~ than~~or ~~equal~~to~2\)
OpenStudy (anonymous):
yea cause i put 1 for the x'es in 10 and got 2+5i and thats complex?
OpenStudy (mathmath333):
yes
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OpenStudy (anonymous):
thanks :) i have more questions if you can keep helping
OpenStudy (mathmath333):
well sry i hv to go now but there are lots of people here so just post the question and get helped