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Unofficialllyy
 one year ago
Will fan and medal!!
Unofficialllyy
 one year ago
Will fan and medal!!

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Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 2x + y2 + 4y = 20

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+2x+y^2+4y=20\]

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0I know I have to complete the square but I have no clue how to do it.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0convert it to the form (x  a)^2 + (y  b)^2 = r^2 complete the square x^2 + 2x and y^2 + 4y

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0We use (hk)^2+(yk)^2=r^2 but it's the same thing as the form you gave me.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0x^2 + 2x + y^2 + 4y = 20 (x + 1)^2  1 + (y + 2)^2  4 = 20 to complete the square you take half of the coefficient of x then you have to subtract this value squared (x + 1)^2 = x^2 + 2x + 1  so you have to subtract 1^2 from this to make x^2 + 2x

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0(x + 1)^2 + (y + 2)^2 = 20+1 + 4 = 25 so if you compare this with the general equation you'll see that center = (1,2) and radius = sqrt25 = 5

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much. I have one more question if you dont mind?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0well i must go in 10 minutes so we'll see how much i can do in that time

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0Prove that the two circles shown below are similar.

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0I know that all circles are similar, but how do I prove it?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0oh must go now I'm sure jhannybean can help

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0Okay thank you!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2 + 2x + y^2 + 4y = 20\]\[(x^2+2x) + (y^2+4y)=20\]\[(x^2+2x+\color{red}{1}) +(y^2+4y+\color{red}{4})=20+\color{red}{1}+\color{red}{4}\]\[(x+1)^2+(y+2)^2=25~~~\checkmark \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha you dont have to @welshfella

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0@Jhannybean Could you help with the circle question? It's my last one. :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hint: try creating triangles within the circles and then using their ratios in comparison.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440188772500:dw

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0So once i have the ratios what do i do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we compare al the sides in a ratio, we would see it has a proportional relationship

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First find the hypotenuse of both triangles, the smaller triangle having a hypotenuse of x. and the bigger one having hypotenuse of y. \[c_1 : x^2= 2^2+2^2 = 8 \iff x=\sqrt{8}\]\[c_2 : y^2=5^2+5^2 = 50 \iff y=\sqrt{50}\]

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0What about dilations? Couldn't we see if they are similar that way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then we can say by the SSS theorem, \[\frac{2}{5} = \frac{\sqrt{8}}{\sqrt{50}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm..i'm not really familiar with dilations! sorry.

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0Okay well thank you!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@cwrw238 could you help me understand dilations? :)

Unofficialllyy
 one year ago
Best ResponseYou've already chosen the best response.0I understand the way you did it, and it makes perfect sense, so no worries. :) thank you for your help!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh cool! no problem.
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