For the square to fit inside the circle without touching it, the diagonal of the square needs to be less than the diameter of the circle.
Using the Pythagorean theorem we can calculate the diagonal of the square:
X = SQRT(5^2 + 5^2)
X = SQRT(25 + 25)
X = SQRT(50)
X = 7.07
The diagonal of the square is 7.07 cm, which is less than the diameter of the circle, 9cm, so it will fit .
2x2=4
3x3=9
5x5=10
Because with powers it is the number times itself not the number times the power. So if i had 5 to the third power it would be 5x5x5=? 5x5=10x5=50 therefore 5 to the third power is 50
I hope this helps!
Answer:
The solutions are x=1 and x=-2/3
Step-by-step explanation:
we have
![3x^{2}-x-2=0](https://tex.z-dn.net/?f=3x%5E%7B2%7D-x-2%3D0)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
![3x^{2}-x=2](https://tex.z-dn.net/?f=3x%5E%7B2%7D-x%3D2)
Factor the leading coefficient
![3(x^{2}-(1/3)x)=2](https://tex.z-dn.net/?f=3%28x%5E%7B2%7D-%281%2F3%29x%29%3D2)
Complete the square. Remember to balance the equation by adding the same constants to each side
![3(x^{2}-(1/3)x+(1/36))=2+(1/12)](https://tex.z-dn.net/?f=3%28x%5E%7B2%7D-%281%2F3%29x%2B%281%2F36%29%29%3D2%2B%281%2F12%29)
![3(x^{2}-(1/3)x+(1/36))=25/12](https://tex.z-dn.net/?f=3%28x%5E%7B2%7D-%281%2F3%29x%2B%281%2F36%29%29%3D25%2F12)
Rewrite as perfect squares
![3(x-(1/6))^{2}=25/12](https://tex.z-dn.net/?f=3%28x-%281%2F6%29%29%5E%7B2%7D%3D25%2F12)
![(x-(1/6))^{2}=25/36](https://tex.z-dn.net/?f=%28x-%281%2F6%29%29%5E%7B2%7D%3D25%2F36)
square root both sides
![(x-(1/6))=(+/-)(5/6)](https://tex.z-dn.net/?f=%28x-%281%2F6%29%29%3D%28%2B%2F-%29%285%2F6%29)
![x=(1/6)(+/-)(5/6)](https://tex.z-dn.net/?f=x%3D%281%2F6%29%28%2B%2F-%29%285%2F6%29)
![x=(1/6)(+)(5/6)=1](https://tex.z-dn.net/?f=x%3D%281%2F6%29%28%2B%29%285%2F6%29%3D1)
![x=(1/6)(-)(5/6)=-2/3](https://tex.z-dn.net/?f=x%3D%281%2F6%29%28-%29%285%2F6%29%3D-2%2F3)
Answer:
b
Step-by-step explanation:
15/8