Answer:

Step-by-step explanation:
Given (64 y Superscript 100 Baseline) Superscript one-half.
Let us write it into an equation.

Apply radical rule:
and 
![\begin{aligned}\left(64 y^{100}\right)^{\frac{1}{2}} &=\sqrt[2]{64 y^{100}} \\&=\sqrt[2]{8^{2} y^{50} y^{50}} \\&=\sqrt[2]{8^{2}\left(y^{50}\right)^{2}} \\&=8 y^{50}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cleft%2864%20y%5E%7B100%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%26%3D%5Csqrt%5B2%5D%7B64%20y%5E%7B100%7D%7D%20%5C%5C%26%3D%5Csqrt%5B2%5D%7B8%5E%7B2%7D%20y%5E%7B50%7D%20y%5E%7B50%7D%7D%20%5C%5C%26%3D%5Csqrt%5B2%5D%7B8%5E%7B2%7D%5Cleft%28y%5E%7B50%7D%5Cright%29%5E%7B2%7D%7D%20%5C%5C%26%3D8%20y%5E%7B50%7D%5Cend%7Baligned%7D)
Hence,
is equivalent to (64 y Superscript 100 Baseline) Superscript one-half.
<h3>Answer:

</h3>
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Explanation:
The instructions don't mention it, but I'm assuming that the height of 4 feet is the max height the of the ball.
If so, then this is the vertex of the parabola.
The ball is on the ground when x = 0 and x = 10. The x coordinate of the vertex is the midpoint of those two roots. So it's at x = 5.
Overall, the vertex is (h,k) = (5,4)
The equation
y = a(x-h)^2 + k
becomes
y = a(x-5)^2 + 4
Next, we plug in the root (x,y) = (10,0) since the ball hits the ground when x = 10. Let's solve for 'a'
y = a(x-5)^2 + 4
0 = a(10-5)^2 + 4
0 = 25a + 4
25a = -4
a = -4/25
We could have used (x,y) = (0,0) and we'd end up with the same 'a' value.
Therefore, the height function is

Answer:
It is given that red dot is about halfway between 12 and 13.
Draw a number line.Mark points on it on as 1,2,3,.....12,13,...20.
Label 12 as point M and label 13 as N.
Mid point of MN=
=25/2
=12.5
Mark 12.5 as T.
MT =NT
The Red dot is at point N at a distance of 12.5 units from the origin.
Yes. Parenthesis first, then if any exponents, then multiplication or division, then addition and subtraction. PEMDAS.
<u>Given</u>:
Line segment DM with end points D(-4,5) and M(6,0) is reflected about the x - axis to give line segment D'M'.
We need to determine the coordinates of the point D'.
<u>Coordinates of the point D':</u>
The general to reflect the coordinates over the x - axis is given by

To determine the coordinates of the point D', let us substitute the coordinates of the point D in the above rule.
Thus, we get;

The reflection of the coordinate D(-4,5) about the x - axis to give the point D' is (-4,-5)
Thus, the coordinates of the point D' is (-4,-5)