Answer:
![\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E4%7D)
Step-by-step explanation:
Maybe you want 2^(4/5) in radical form.
The denominator of the fractional power is the index of the root. Either the inside or the outside can be raised to the power of the numerator.
![2^{\frac{4}{5}}=\boxed{\sqrt[5]{2^4}=(\sqrt[5]{2})^4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%3D%5Cboxed%7B%5Csqrt%5B5%5D%7B2%5E4%7D%3D%28%5Csqrt%5B5%5D%7B2%7D%29%5E4%7D)
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In many cases, it is preferred to keep the power inside the radical symbol.
Answer:
=13.73 ft²
Step-by-step explanation:
The diameter of the largest possible circle that can be cut out of a square is equal to the length of the side of the square.
Therefore the diameter of the circle cut from a square of side 8ft =8ft
r=4ft
Area of the remaining board= Area of square- area of circle
=side²-πr²
=8²-π×4²
=64-16π
=13.73 ft²
Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
Total number of different tickets are possible is


Therefore, 4032 different tickets are possible.
1. Place your compass point on X and measure the distance to point Y. Swing an arc of this size above (or below) the segment.
2. Without changing the span on the compass, place the compass point on Y and swing the same arc, intersecting with the first arc.
3. Label the point of intersection as the third vertex of the equilateral triangle.
4. Use you straight edge and connect the points.