There are two ways in which you can find a point that lies in the circle. One of them is to do y the subject of the formula, and another one is to determine the center of the circumference, and with the information of the radius, you can sum this value upward or downward.

the general equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$

with center at (h,k)

you have the following equation:

$(x+5)^2+(y-9)^2=8\\\\h=-5\\\\k=9\\\\r=\sqrt{8}$

Then, the center is (-5,9)

if you sum the value of the radius in one of the fourth directions (up, down, left, right), for example upward you have

$(-5,9+r)=(-5,9+\sqrt{8})$

Then, one point that lies in the circle is (-5,9+√8)

3 0

2 years ago

Please help me I tried to do it

anyanavicka [17]

You do 9x27=243 then take 2x31=62 add 243+62=305 then take 305 and subtract it from 28 305-28=277
so if i did it right you should get n=277

8 0

3 years ago

Plz help I'm really bad at these

GREYUIT [131]

Y = mx + b, where m = slope = rise/run and b, the y intercept value

m = 1/1 = 1, then y = x + b.
We notice that y intercept is = - 1
And the equation is:

y = x - 1

6 0

3 years ago

Factor −5x2 + 10x. −5x(x + 2) 5(−x2 + 10x) 5x(−x + 2) x(5x + 10)

Hitman42 [59]

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