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
Tan^-1 (tan x) = 42/25
x=tan^-1 42/25
x= 59.237 degrees.
Answer:
The option "StartFraction 1 Over 3 Superscript 8" is correct
That is 
is correct answer
Therefore ![[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}](https://tex.z-dn.net/?f=%5B%282%5E%7B-2%7D%29%283%5E4%29%5D%5E%7B-3%7D%5Ctimes%20%5B%282%5E%7B-3%7D%29%283%5E2%29%5D%5E2%3D%5Cfrac%7B1%7D%7B3%5E8%7D)
Step-by-step explanation:
Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared
The given expression can be written as
![[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2](https://tex.z-dn.net/?f=%5B%282%5E%7B-2%7D%29%283%5E4%29%5D%5E%7B-3%7D%5Ctimes%20%5B%282%5E%7B-3%7D%29%283%5E2%29%5D%5E2)
To find the simplified form of the given expression :
( using the property 
)
( using the property 
( combining the like powers )
( using the property 
)
( using the property 
)
Therefore
Therefore option "StartFraction 1 Over 3 Superscript 8" is correct
That is is correct answer
Answer:
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