Answer:
the domain is -11,10 but i didnt know if you wanted that or not
Step-by-step explanation:
![-a^2=(-1)\cdot a\cdot a](https://tex.z-dn.net/?f=-a%5E2%3D%28-1%29%5Ccdot%20a%5Ccdot%20a)
Regardless of the sign of
![a](https://tex.z-dn.net/?f=a)
, we have
![a\cdot a=a^2\ge0](https://tex.z-dn.net/?f=a%5Ccdot%20a%3Da%5E2%5Cge0)
(never negative). But multiplying by -1 makes it negative.
On the other hand,
![(-a)^2=((-1)\cdot a)^2=(-1)^2\cdot a^2=1\cdot a^2=a^2](https://tex.z-dn.net/?f=%28-a%29%5E2%3D%28%28-1%29%5Ccdot%20a%29%5E2%3D%28-1%29%5E2%5Ccdot%20a%5E2%3D1%5Ccdot%20a%5E2%3Da%5E2)
which can never be negative for real
![a](https://tex.z-dn.net/?f=a)
.
We need to determine the radius and diameter of the circle. If the area of the circle is 10 pi in^2, then, according to the formula for the area of a circle,
A = 10 pi in^2 = pi*r^2. Thus, 10 in^2 = r^2, and r = radius of circle = sqrt(10) in.
Thus, the diam. of the circle is 2sqrt(10) in. This diam. has the same length as does the hypotenuse of one of the triangles making up the square.
Thus, [ 2*sqrt(10) ]^2 = x^2 + x^2, where x represents the length of one side of the square. So, 4(10) in^2 = 2x^2. Then:
40 in^2 = 2x^2, or 20 in^2 = x^2, and so the length x of one side of the square is sqrt(20). The area of the square is the square of this result:
Area of the square = x^2 = [ sqrt(20) ]^2 = 20 in^2 (answer). Compare that to the 10 pi sq in area of the circle (31.42 in^2).
Answer:
x=3
Step-by-step explanation:
This is a proportional equation:
Given: x/9=9/27
Cross-multiply: 27x=81
Divide: x=3
So x=3