Question

Find the center and radius of a circle.

Answer 1

\[16x^2+16y^2+8x+32y+1=0\]

Answer 2

FIGURED IT OUT

Answer 3

THANKS

Answer 4

divide everything by 16
\[\frac{16}{16}x^2+\frac{16}{16}y^2+\frac{8}{16}x+\frac{32}{16}y+\frac{1}{16}=\frac{0}{16}\]
\[x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0\]
\[x^2+\frac{1}{2}x+y^2+2y=-\frac{1}{16}\]
\[x^2+\frac{1}{2}x+(\frac{1}{2*2})^2+y^2+2y+(\frac{2}{2})^2=-\frac{1}{16}+(\frac{1}{2*2})^2+(\frac{2}{2})^2\]

Answer 5

ok nvm then

Answer 6

wow u wrote that out quick

Answer 7

did you get:

center (-1/4 , -1)

Answer 8

radius = 1

Answer 9

\[(x+1/4)^2+(y+1)^2=1\]

Answer 10

so yes! gj!

Answer 11

thanks

Answer 12

Can you help me with one other question and then I am complete my homework?

Answer 13

ok

Answer 14

Solve the inequality in terms of intervals and illustrate the solution set on the real number line.
\[4x <2x + 1\le 3x +2\]

Answer 15

\[4x-4x \le 2x-4x+1\le 3x-4x+2\]
\[0\le-2x+1\le-x+2\]
\[0+x\le-2x+1+x\le-x+2+x\]
\[x\le-x+1\le2\]

......

Answer 16

\[-x+1 \le 2\]
\[-x \le 2-1\]
\[-x \le 1\]
\[x \ge -1\]

and

\[x \le -x+1\]
\[x+x \le -x+x+1\]
\[2x \le 1\]
\[x \le \frac{1}{2}\]

*~~~~*
----|-----|----
-1 1/2

interval notation [-1,1/2]

Answer 17

THANKS

Answer 18

hey i wrote the question down differently

Answer 19

also i went about this a long way lol

Answer 20

its ok i understood..Thanks for your help

Answer 21

\[4x<2x+1\] \[2x+1 \le 3x+2\]

Answer 22

the first one gives 2x<1 which means x<1/2
the second one gives -1<=x which means x>=-1

[-1,1/2)

Answer 23

*~~~()
____|_____|____
-1 1/2

Answer 24

now im happy