X=0 y=-2
y=0 x=4
The shortest distance has to be the perp bisector from origin to the midpoint of the line.
midpoint =(2,-1) and distance from the (0,0) origin is sqrt(5) hence, sqrt(5).
Answer:
Yes Billy is correct
Step-by-step explanation:
Think of a pizza in 4 peices on pizza had 1 clice taken out from 4 and one has 3 pieces have been taken away from 4. So therfore -3/4 is less.
Answer:
$1.27
Step-by-step explanation:
You simply just need to subtract 10.00-8.73 to get your answer which is 1.27.
Answer:

Step-by-step explanation:
The expression to transform is:
![(\sqrt[6]{x^5})^7](https://tex.z-dn.net/?f=%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7)
Let's work first on the inside of the parenthesis.
Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:
![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Therefore ![\sqrt[6]{a} = a^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now let's replace
with
which is the algebraic form we are given inside the 6th root:
![\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E5%7D%20%3D%20%28x%5E5%29%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now use the property that tells us how to proceed when we have "exponent of an exponent":

Therefore we get: 
Finally remember that this expression was raised to the power 7, therefore:
[/tex]
An use again the property for the exponent of a exponent:
Answer:
diameter = 16cm, circumference ≈ 50.27
Step-by-step explanation:
The radius is 8 centimeters, so the diameter should be twice of that because the radius is the distance from the middle of the circle to the edge while the diameter is from edge to edge passing through the middle of the circle. Therefore,
8 · 2 = 16cm
The equation for the circumference of a circle is 2πr. In this case, because nothing else is specified about π, we use 3.14. So,
2 · 3.14 · 8 ≈ 50.27
Hope this helped! Have a great day. ^^