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onegirl Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix i ddi it but can u check my answer?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
sure :O what are you suppose to do? Just find the derivative then plug \(\large a\) in? Or does this have something to do with approximation?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay my teacher said there is an equation, so here is what i did f(x) = cos (x) f'(x) = sin(x) x = a = pi/2 so f'(pi/2) = sin (pi/2) = 1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Your teacher said there is an equation? :) That's not much detail lolol Your derivative looks correct at least.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
yea she said try again there is an equation
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
let me show you how i wrote it first okay?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
i mean the answer i wrote first
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
d/dx ( cos (x)) = sin(x) sin (pi/2) = 1, so the slope is 1, y = mx + b > y = =1x + b cos (pi/2) = 0 so the coordinates pare are (pi/2, 0) so y = 0x + b, 0(pi/2) + b equals to 0, 0 = y = 0x + 0 = y = 0 for my final answer.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Oh so like the previous problems, you're looking for an equation of the tangent line?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
yes sorry i made a mistake
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
yes yes the equation of the tangent line
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large y=x+b\] From here you plugged in your point \(\large \left(\pi/2,\;0\right)\). \[\large 0=\pi/2+b \qquad \qquad \rightarrow \qquad \qquad b=\pi/2\] I think you plugged your \(\large 0\) into the wrong spot.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so instead of 0 = 0(pi/2) + b what would i write
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})\] Plug them into here, match up the colors.\[\huge \color{orangered}{y}=(1)\color{royalblue}{x}+b\] I'm not sure why you're plugging a \(\large 0\) in for your 1....
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so 0 = (1)(pi/2) + b ?
 one year ago
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