The standard form for the equation of a circle is :
<span><span><span> (x−h)^</span>2</span>+<span><span>(y−k)^</span>2</span>=<span>r2</span></span><span> ----------- EQ(1)
</span><span> where </span><span>handk</span><span> are the </span><span>x and y</span><span> coordinates of the center of the circle and </span>r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :
((−10+(−8))/2,(1+5)/2)=(−9,3)
So the point (−9,3) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(−10−(−9))^2+(1−3)^2=1+4=5
⇒r=√5
Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :
(x+9)^2+(y−3)^2=5
When substituting, you want to take the y value from one equation and plug it into the y variable in the other equation to find the x value. When you find the c value, you plug the number into one of the equations to get your y value.
3x-5=1
+5 +5
3x=6
3 3
X= 2
Answer:
6
Step-by-step explanation:
216 raised to the one third power is the same as asking for the cubed root of 216. 6x6x6=216
hope this helps :)
(2x -5y) = -10 that’s your answer -10
