Answer: 
<u>Remove "x" and add</u>
-2 + 9 = 7
<u>Add "x" back</u>
7 -----> 7x
Note: When adding/subtracting problems like these remove "x" and then add/subtract normaly. When you get your answer put x back.
 
        
                    
             
        
        
        
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
 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:
![\sqrt[3]{256x^{10}y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D)
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 ![\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
 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)
 
        
             
        
        
        
Answer:
320,000
Step-by-step explanation:
 
        
             
        
        
        
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 ×  -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
brainly.com/question/4318844
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