Find the equation of the circle whose center and radius are given. center (0, 8), radius = 8
The equation for a circle is: (x-h)^2+(y-k)^2=r^2 Where (h,k) is the center and r^2 is the radius.
You just plug in the numbers you have for (0,8) and radius of 8
ik but the answer as to be like this __+(___)_______
(x-0)^2+(y-8)^2=8^2 x^2+(y-8)^2=64. When you have a 0 coordinate, it's just x^2 or y^2.
o thank u sooo much i have one more can u help?
Yeah sure!
Find the equation of the circle whose center and radius are given. center ( -2, -5), radius = 1
This would be the same kind of thing. Plug it in: (x-(-2)^2+(y-(-5)^2=1^2
What happens when you have this -(-)? What does it become?
a positive?
Good, so what would the equation be?
(x+2)^2+(y+5)^2=1
Yup!
omg thank u so much!!!!!! r u good with diagnols?
Hm it depends on the question lol
A parallelogram whose vertices have coordinates R(1, -1), S(6, 1), T(8, 5), and U(3, 3) has a shorter diagonal of ___ .
i have 3 answer to chose from
5 square root(13) square root(97)
Well first thing I would do is graph the parallelogram
|dw:1403556069465:dw| So I drew it and drew the diagonals. So to find the length of the diagonals we will use the distance formula with points
Use the formula with: (3,3)(6,1) and (1,-1)(8,5)
Do you know what the distance formula is/how to use it?
yup! ill do it right now dont go nowhere though
I won't haha I'll do it too
haha kk
so the first one i got square root 13 and the seond square root 185! i dont know how to reduce squareroots
Well it's asking for the shorter diagonal anyways. And for the first one I got sqrt(5)
really!!? weird
And for the second I got sqrt(85)
Obviously the smaller one is sqrt(5)
haha ya thank u can u stick around im stuck on alot more!!!!! and u explain really good?
Oops my 1st one is wrong I'm sorry I wrote down (3,3) and it looked like (5,3) lol
ooo hahaha so u gunna redo it?
It is sqrt(13) lol
Which would be the answer because sqrt(13) is smaller than sqrt(85)
hahah thanks knew it heheh!!! ur a big help!!! will u help with more?
Sure I can! Open a new question though, this one is getting lengthy lol
hahaha ok!
Just tag me in your question :)
Join our real-time social learning platform and learn together with your friends!