Ok so this is conic sectuion
first group x's with x's and y's with y's
then complete the squra with x's and y's
2x^2-8x+2y^2+10y+2=0
2(x^2-4x)+2(y^2+5y)+2=0
take 1/2 of linear coeficient and square
-4/2=-2, (-2)^2=4
5/2=2.5, 2.5^2=6.25
add that and negative inside
2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0
factor perfect squares
2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0
distribute
2(x-2)^2-8+2(y+2.5)^2-12.5+2=0
2(x-2)^2+2(y+2.5)^2-18.5=0
add 18.5 both sides
2(x-2)^2+2(y+2.5)^2=18.5
divide both sides by 2
(x-2)^2+(y+2.5)^2=9.25
that is a circle center (2,-2.5) with radius √9.25
Answer:
44
Step-by-step explanation:
the mode of a set of data is the value that occurs most often, in this case it is 44
Answer:
no
Step-by-step explanation:
Answer:
d
Step-by-step explanation:
Answer:
Step-by-step explanation:
[...] if you can write the LHS as a perfect square, or if you can't spot a factorization of it right away, if and only if the discriminant 
(or, if b is an even number, 1/4 of it) is zero.
<u>I see it! I see it!</u>
Stare at it for a while. First term is 
, third term is 
, we are missing a double product, but we can play with k. For the LHS to be 
you just need 
.
<u>I don't see it...</u>
Then number crunching it is. Set the discriminant to 0, solve for k
