Answer:
The equivalent expression for the given expression ![\sqrt[3]{256x^{10}y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D)
is
![4x^{3} y^{2}(\sqrt[3]{4xy} )](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%20%29)
Step-by-step explanation:
Given:
Solution:
We will see first what is Cube rooting.
![\sqrt[3]{x^{3}} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20x)
Law of Indices
Now, applying above property we get
![\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%5Csqrt%5B3%5D%7B%284%5E%7B3%7D%5Ctimes%204%5Ctimes%20%28x%5E%7B3%7D%29%5E%7B3%7D%5Ctimes%20x%5Ctimes%20%28y%5E%7B2%7D%29%5E%7B3%7D%5Ctimes%20y%20%20%20%29%7D%20%5C%5C%5C%5C%5Ctextrm%7BCube%20Rooting%20we%20get%7D%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204%5Ctimes%20x%5E%7B3%7D%5Ctimes%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29%20%5C%5C%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204x%5E%7B3%7Dy%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29)
∴ The equivalent expression for the given expression is
8.8 rounding the remainder
42 divided by 63 is 0.66666666666
Are you sure it's not 63 divided by 42?
Answer:
79, 80, 81
Step-by-step explanation:
<u>Creating an equation.</u>
let a = first number , a + 1 = second number and a + 2 = third number
the sum of these three consecutive numbers is 240
this means that , a + ( a + 1 ) + ( a + 2 ) = 240
<u>Solving for x using the equation created.</u>
a + ( a + 1 ) + ( a + 2 ) = 240
==> remove parenthesis
a + a + 1 + a + 2 = 240
==> combine like terms
3a + 3 = 240
==> subtract 3 from both sides
3a = 237
==> divide both sides by 3
a = 79
<u>Finding the 3 numbers</u>
First number = a
a = 79 so first = 79
Second number = a + 1
a = 79 so second number = 79 + 1 = 80
Third number = a + 2
a = 79 so third number = 79 + 2 = 81
The three numbers are 79, 80 and 81
