In order to find which <u>expression</u> is <u>equivalent</u> to ![\sqrt[4]{x^{10}},](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%2C)
<u>simplify</u> all given expressions:
0.
![\sqrt[4]{x^{10}}=x^{\frac{10}{4}}=x^{\frac{5}{2}}=x^{2.5}.](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%3Dx%5E%7B%5Cfrac%7B10%7D%7B4%7D%7D%3Dx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%3Dx%5E%7B2.5%7D.)
1.
![x^2(\sqrt[4]{x^2})=x^2\cdot x^{\frac{2}{4}}=x^2\cdot x^{\frac{1}{2}}=x^2\cdot x^{0.5}=x^{2.5}.](https://tex.z-dn.net/?f=x%5E2%28%5Csqrt%5B4%5D%7Bx%5E2%7D%29%3Dx%5E2%5Ccdot%20x%5E%7B%5Cfrac%7B2%7D%7B4%7D%7D%3Dx%5E2%5Ccdot%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3Dx%5E2%5Ccdot%20x%5E%7B0.5%7D%3Dx%5E%7B2.5%7D.)
2.
3.
![x^3(\sqrt[4]{x} )=x^3\cdot x^{\frac{1}{4}}=x^3\cdot x^{0.25}=x^{3.25}.](https://tex.z-dn.net/?f=x%5E3%28%5Csqrt%5B4%5D%7Bx%7D%20%29%3Dx%5E3%5Ccdot%20x%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3Dx%5E3%5Ccdot%20x%5E%7B0.25%7D%3Dx%5E%7B3.25%7D.)
4.
Therefore,
Answer: correct choice is A
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Answer:
This proves that f is continous at x=5.
Step-by-step explanation:
Taking f(x) = 3x-1 and 
, we want to find a 
such that 
At first, we will assume that this delta exists and we will try to figure out its value.
Suppose that 
. Then
.
Then, if , then 
. So, in this case, if 
we get that . The maximum value of delta is 
.
By definition, this procedure proves that 
. Note that f(5)=14, so this proves that f is continous at x=5.
The function y = 1100.74x - 1976.47 represent the total number of visitors x years after 2005.
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
Let y represent the total number of visitors x years after 2005. Hence:
y = 1100.74x - 1976.47
The function y = 1100.74x - 1976.47 represent the total number of visitors x years after 2005.
Find out more on equation at: brainly.com/question/13763238
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