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anonymous
 one year ago
nick1234567
Area of one leaf of the rose r=3sin(4theta)
anonymous
 one year ago
nick1234567 Area of one leaf of the rose r=3sin(4theta)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I'm very sorry, it is 2:38 a.m. (Italy time zone) from me so I have to go to sleep

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Need answer in next 3 mins please

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1ok any ideas on where to start?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope no clue and it's my last prob due in 3 mins

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can u please save me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Only have 1min left thank u again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Needs suggestion in the ext 39 sec

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1Things like this cant be done in that much time. You need to make sure you have time for it

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1You will need to use \[\huge~\rm~\int\limits_{}^{}1/2r^2d \emptyset\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2if this is an emergency, this is a brief attempt: \(r=3sin(4 \theta)\) \(r = 0, \ 4\theta = 0, \ \pi \implies \theta = 0, \pi/4, ....\) \(A = \frac{1}{2} \int_{0}^{\pi / 4} 9 sin^2 (4 \theta) \ d \theta \) \( = \frac{9}{4} \ \int_{0}^{\pi / 4} 1  cos \ 8 \theta \ d \theta\) \( = \frac{9}{4} \ [ \theta  \frac{1}{8}sin \ 8 \theta \ ]_{0}^{\pi / 4}\) \( = \frac{9}{4} \ [ \frac{\pi}{4} ] = \frac{9 \pi}{16}\)

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1^or that works too ;p

pooja195
 one year ago
Best ResponseYou've already chosen the best response.1You will need to use \[\huge~\rm~\int\limits_{}^{}1/2r^2d \theta\] \[\huge~\rm~\int\limits\limits_{\alpha }^{\beta }1/2r^2d \theta \] \[\huge~\rm~\int\limits\limits\limits_{0 }^{\ \pi/4 }1/2(3\sin(4\theta))^2d \theta =\frac{ 9 }{ 16 }\pi \]

Jack1
 one year ago
Best ResponseYou've already chosen the best response.0lol... why is this q orange?
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