A cube with side length 1 unit is called a unit cube
Answer:
You can find its average speed.
Step-by-step explanation:
To find average speed you can divide distance by time which has been given.
Answer:
The factorization of
is ![(9x^{5} +10)(81x^{10} -90x^{5} +100)](https://tex.z-dn.net/?f=%289x%5E%7B5%7D%20%2B10%29%2881x%5E%7B10%7D%20-90x%5E%7B5%7D%20%2B100%29)
Step-by-step explanation:
This is a case of factorization by <em>sum and difference of cubes</em>, this type of factorization applies only in binomials of the form
or
. It is easy to recognize because the coefficients of the terms are <u><em>perfect cube numbers</em></u> (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).
Let's solve the factorization of
by using the <em>sum and difference of cubes </em>factorization.
1.) We calculate the cubic root of each term in the equation
, and the exponent of the letter x is divided by 3.
![\sqrt[3]{729x^{15}} =9x^{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B729x%5E%7B15%7D%7D%20%3D9x%5E%7B5%7D)
then ![\sqrt[3]{10^{3}} =10](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B10%5E%7B3%7D%7D%20%3D10)
So, we got that
which has the form of
which means is a <em>sum of cubes.</em>
<em>Sum of cubes</em>
![(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})](https://tex.z-dn.net/?f=%28a%5E%7B3%7D%20%2Bb%5E%7B3%7D%20%29%3D%28a%2Bb%29%28a%5E%7B2%7D%20-ab%2Bb%5E%7B2%7D%29)
with
y ![b=10](https://tex.z-dn.net/?f=b%3D10)
2.) Solving the sum of cubes.
![(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )](https://tex.z-dn.net/?f=%289x%5E%7B5%7D%29%5E%7B3%7D%20%2B%20%2810%29%5E%7B3%7D%3D%289x%5E%7B5%7D%20%2B10%29%28%289x%5E%7B5%7D%29%5E%7B2%7D-%289x%5E%7B5%7D%29%2810%29%2B10%5E%7B2%7D%20%29)
![(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)](https://tex.z-dn.net/?f=%289x%5E%7B5%7D%29%5E%7B3%7D%20%2B%20%2810%29%5E%7B3%7D%3D%289x%5E%7B5%7D%20%2B10%29%2881x%5E%7B10%7D-90x%5E%7B5%7D%2B100%29)
.
The pool can hold 65.84 ft³ of water
<u>Explanation:</u>
Given:
Shape of pool = octagonal
Base area of the pool = 22 ft²
Depth of the pool = 3 feet
Volume, V = ?
We know:
Area of octagon = 2 ( 1 + √2) a²
22 ft² = 2 ( 1 + √2 ) a²
![\frac{11}{1+\sqrt{2} } = a^2](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B1%2B%5Csqrt%7B2%7D%20%7D%20%3D%20a%5E2)
a² = ![\frac{11}{2.42}](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B2.42%7D)
a² = 4.55
a = 2.132 ft
Side length of the octagon is 2.132 ft
We know:
Volume of octagon = ![2(1+\sqrt{2} ) X (a)^2 X h](https://tex.z-dn.net/?f=2%281%2B%5Csqrt%7B2%7D%20%29%20X%20%28a%29%5E2%20X%20h)
![V = 2(1+\sqrt{2})X (2.132)^2 X 3\\ \\V = 2 ( 2.414) X 4.5454 X 3\\\\V = 65.84 ft^3](https://tex.z-dn.net/?f=V%20%3D%202%281%2B%5Csqrt%7B2%7D%29X%20%282.132%29%5E2%20X%203%5C%5C%20%5C%5CV%20%3D%202%20%28%202.414%29%20X%204.5454%20X%203%5C%5C%5C%5CV%20%3D%2065.84%20ft%5E3)
Therefore, the pool can hold 65.84 ft³ of water
The answer is 1.9 gallons