need help with math homework, faggots. I'll post my nudes if someone gets it right ;P
Need help with math homework, faggots. I'll post my nudes if someone gets it right ;P
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1 = 1
Math done
integral-calculator.com
do it yourself
=404
Look here, nigger. I need to learn this shit.
y=-2
You wrote positive
...
9/6 = 3/2, you ignorant faggot.
Approximation: 8.606609071677304
top kek
thanks babe. no nudes. OP=fat dude
and if you dont understand it, just graph the function
when expressed in terms of x you only get half of the squareroot function, and thus only one intersection
You found the intersection of y not x
This. So much this.
Same thing you're probably just used to the other way around.
I suppose only one rule holds eternally true: OP is a faggot.
Try in respect to dy or solve for x intersctions
>y^2 - 3y = 10
>y = -2
>kek
I am winrar now deliver op
I would like to direct your attention here and here
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Correct answer is:
343/18=19.0556
Op we know where you live
Deliver!
i live at the Breezehome in Whiterun, faggot. Come find me bitch.
What the fuck did you do, you gaylord?!?! You dun goofed!
Solve as a function of y, find the points of intersection, then find the difference between the integrals of both functions between those points?
Im fucking right do integral -2 to 5 (3+y+1/3 (1-y^2)dy = 343/18
The error in your math is you found the points for y not x and you did the integral respect to x dx
Ok if you want
as promised
This guy and skipping most of the fucking function so he can use his derpmath calculator
Lern2Calculus
...
I did it for maths
also..yes..error with da boundaries of integration..as pointed out. forgivable.
Determine should be {5, -2} not {5, 2}
False!
im not posting my nudes. for one, im a guy. for two, my bother is on this thread
i think u r right
you right, bruh
Well
fuck
>my bother
>bother
There you are you gaylord. Now nudes!
>Didn't express you're values as rational numbers
Who's the gaylord now?
bounds of integration are wrong.
go from -2 to 5.
Op underage fag this is for 15-16 year olds, 17 max if op is ameritard
Hes right, but stupid. Went full retard and solved for x. And beyond that, there is only 1 intersection; the first function is a square root of a variable which is non-contunuous between negative and positive infinity. It starts at x=-1/3. The lower bound then is necessarily that value. The upper bound is the intersection at (8,5). So integrate from -1/3 to 8 in terms of x.
Solving for x first you must integrate from 0 to 5 in terms of y. Can be done. But why bother with making y the dependant variable?