Answer:
KPF = 9 and H = 129
Step-by-step explanation:
An inscribed quadrilateral in a circle has all diagonal angles add up to 180, so we can use this to find the angles of the quadrilateral.
H= 180-EPK and K = 180-E so H = 129 and K = 60
Now PF and EH are parallel, so PE is a transversal. That means FPE = 180 - E = 60. Now it's pretty easy to solve for KPF = FPE - EPK = 60 - 51 = 9.
Let me know if you don't see how I did any of this and I'll be happy to explain it..
Answer:
![\sqrt{x}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D)
Step-by-step explanation:
If we have that worded expression, we can convert it into actual mathematical terms.
x to the one sixth power: ![x^{\frac{1}{6}}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
That to the power of 3:
![(x^{\frac{1}{6}})^3](https://tex.z-dn.net/?f=%28x%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D%29%5E3)
Using exponent rules, since we have a power to a power, the powers multiply, so:
![x^{\frac{1}{6}\cdot 3}\\\\x^{\frac{3}{6}}\\\\ x^{\frac{1}{2}}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B6%7D%5Ccdot%203%7D%5C%5C%5C%5Cx%5E%7B%5Cfrac%7B3%7D%7B6%7D%7D%5C%5C%5C%5C%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
If we have a number to a fraction power, the denominator becomes the “find the (denominator) root of x.
So:
![x^{\frac{1}{2}} = \sqrt{x}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Csqrt%7Bx%7D)
Hope this helped!
<em>Answer:840 hours</em>
<em>Step-by-step explanation:You have to Add to get your answer.This is how I got 840:</em>
60+60=120
120+720=840
Your answer is:840 hours
hope this was helpful :)
Answer:Every square is a rhombus, and a rhombus can be a square, if all its angles are 90 degrees. Thus, a rhombus can be a rectangle (if the angles of the rhombus are all 90 degrees), and a rectangle can be a rhombus (if the sides of the rectangle are all equal length).
Step-by-step explanation: