For all the math lovers =)

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LadiiKay

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  1. Medical Student
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take out those graphing calculators, set your minimum and maximum for x and y to 2 (or -2 depending on the side) and graph this:
x^2 + (y - (x^2)^(1/3))^2 = 1

using wolframalpha for help with this one automatically makes you a *itch*ss 😀
it's stupid but I guarantee a smile when you graph it.
I'll be posting the simpler version already in y equal's format tomorrow morning for those of you too lazy to do it yourselves. it took me a little while to get it, but it was a great study break for me and so with that... enjoy m'dears.
:luck::luck::luck:
 
aww thats so sweet of you, fitting for valentines day
 
👍

Happy Valentines Day to you too! 🙂
 
y^2 - 2yx^(2/3) = 1 - x^(4/3) - x^2
 
Do I need to use the quadratic formula? How do you solve for y?
 
Totally inappropriate.

This will make more people graph it guaranteed
 
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wiingy you gave it away! but as promised here's the simpler version:
y1 = (x^2)^(1/3) - (1 - x^2)^(1/2)
y2 = (1 - x^2)^(1/2) + (x^2)^(1/3)
😍

Sorry, changed my post so it's clear I was cheating. 🙄
 
Pretty cool!
If you're willing to go through the trouble...

Turn axes off, set scales
xmin = -1.88
xmax= 1.88
ymin = -1.2
ymax = 1.7

y1= abs(x) + sqrt(1-x^2)
y2 = abs(x) - sqrt(1-x^2)

2nd, Draw, Shade, Vars, Y-Vars, Function, Y1 (insert comma here), Vars, Y-Vars, Function, Y2. ENTER

voila~
 
Then the answer is probably C, and I'm not going to attempt to dig up a graphing calculator.

a rough plot by hand resulted in what appears to be a heart shaped curve 😕
 
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