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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1sure :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?

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0okay 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

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

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

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

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

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0d/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.

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

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

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

zepdrix
 one year ago
Best 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.

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

zepdrix
 one year ago
Best 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....

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