Answer:
13
Step-by-step explanation:
Mean would mean average, which in this case is 
so your equation would look like 
Next multiply both sides by to clear the fraction 
to get 
Subtract 69 from both sides 
you get 
Answer:

Step-by-step explanation:
Hope this work helps
Answer:
3.3
Step-by-step explanation:
If the limit of f(x) as x approaches 8 is 3, can you conclude anything about f(8)? The answer is No. We cannot. See the explanation below.
<h3>What is the justification for the above position?</h3>
Again, 'No,' is the response to this question. The justification for this is that the value of a function does not depend on the function's limit at a given moment.
This is particularly clear when we consider a question with a gap. A rational function with a hole is an excellent example that will help you answer this question.
The limit of a function at a position where there is a hole in the function will exist, but the value of the function will not.
<h3>What is limit in Math?</h3>
A limit is the result that a function (or sequence) approaches when the input (or index) near some value in mathematics.
Limits are used to set continuity, derivatives, and integrals in calculus and mathematical analysis.
Learn more about limits:
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Answer:
<h2>a) 0.5367feet</h2><h2>b) 0.5223feet</h2><h2>c) 0.7292feet</h2>
Step-by-step explanation:
Given the rate at which an eucalyptus tree will grow modelled by the equation 0.5+6/(t+4)³ feet per year, where t is the time (in years).
The amount of growth can be gotten by integrating the given rate equation as shown;

a) The number of feet that the tree will grow in the second year can be gotten by taking the limit of the integral from t =1 to t = 2
![\int\limits^2_1 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_1\\= [0.5(2)-3(2+4)^{-2}] - [0.5(1)-3(1+4)^{-2}]\\= [1-3(6)^{-2}] - [0.5-3(5)^{-2}]\\ = [1-\frac{1}{12}] - [0.5-\frac{3}{25} ]\\= \frac{11}{12}-\frac{1}{2}+\frac{3}{25}\\ = 0.9167 - 0.5 + 0.12\\= 0.5367feet](https://tex.z-dn.net/?f=%5Cint%5Climits%5E2_1%20%7B0.5%20%2B%20%5Cfrac%7B6%7D%7B%28t%2B4%29%5E%7B3%7D%20%7D%20%20%7D%20%5C%2C%20dt%20%3D%20%5B0.5t-3%28t%2B4%29%5E%7B-2%7D%5D%5E2_1%5C%5C%3D%20%5B0.5%282%29-3%282%2B4%29%5E%7B-2%7D%5D%20-%20%5B0.5%281%29-3%281%2B4%29%5E%7B-2%7D%5D%5C%5C%3D%20%5B1-3%286%29%5E%7B-2%7D%5D%20-%20%5B0.5-3%285%29%5E%7B-2%7D%5D%5C%5C%20%3D%20%5B1-%5Cfrac%7B1%7D%7B12%7D%5D%20-%20%5B0.5-%5Cfrac%7B3%7D%7B25%7D%20%5D%5C%5C%3D%20%5Cfrac%7B11%7D%7B12%7D-%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7B3%7D%7B25%7D%5C%5C%20%20%20%3D%200.9167%20-%200.5%20%2B%200.12%5C%5C%3D%200.5367feet)
b) The number of feet that the tree will grow in the third year can be gotten by taking the limit of the integral from t =2 to t = 3
![\int\limits^3_2 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^3_2\\= [0.5(3)-3(3+4)^{-2}] - [0.5(2)-3(2+4)^{-2}]\\= [1.5-3(7)^{-2}] - [1-3(6)^{-2}]\\ = [1.5-\frac{3}{49}] - [1-\frac{1}{12} ]\\ = 1.439 - 0.9167\\= 0.5223feet](https://tex.z-dn.net/?f=%5Cint%5Climits%5E3_2%20%7B0.5%20%2B%20%5Cfrac%7B6%7D%7B%28t%2B4%29%5E%7B3%7D%20%7D%20%20%7D%20%5C%2C%20dt%20%3D%20%5B0.5t-3%28t%2B4%29%5E%7B-2%7D%5D%5E3_2%5C%5C%3D%20%5B0.5%283%29-3%283%2B4%29%5E%7B-2%7D%5D%20-%20%5B0.5%282%29-3%282%2B4%29%5E%7B-2%7D%5D%5C%5C%3D%20%5B1.5-3%287%29%5E%7B-2%7D%5D%20-%20%5B1-3%286%29%5E%7B-2%7D%5D%5C%5C%20%3D%20%5B1.5-%5Cfrac%7B3%7D%7B49%7D%5D%20-%20%5B1-%5Cfrac%7B1%7D%7B12%7D%20%5D%5C%5C%20%20%3D%201.439%20-%200.9167%5C%5C%3D%200.5223feet)
c) The total number of feet grown during the second year can be gotten by substituting the value of limit from t = 0 to t = 2 into the equation as shown
![\int\limits^2_0 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_0\\= [0.5(2)-3(2+4)^{-2}] - [0.5(0)-3(0+4)^{-2}]\\= [1-3(6)^{-2}] - [0-3(4)^{-2}]\\ = [1-\frac{1}{12}] - [-\frac{3}{16} ]\\= \frac{11}{12}+\frac{3}{16}\\ = 0.9167 - 0.1875\\= 0.7292feet](https://tex.z-dn.net/?f=%5Cint%5Climits%5E2_0%20%7B0.5%20%2B%20%5Cfrac%7B6%7D%7B%28t%2B4%29%5E%7B3%7D%20%7D%20%20%7D%20%5C%2C%20dt%20%3D%20%5B0.5t-3%28t%2B4%29%5E%7B-2%7D%5D%5E2_0%5C%5C%3D%20%5B0.5%282%29-3%282%2B4%29%5E%7B-2%7D%5D%20-%20%5B0.5%280%29-3%280%2B4%29%5E%7B-2%7D%5D%5C%5C%3D%20%5B1-3%286%29%5E%7B-2%7D%5D%20-%20%5B0-3%284%29%5E%7B-2%7D%5D%5C%5C%20%3D%20%5B1-%5Cfrac%7B1%7D%7B12%7D%5D%20-%20%5B-%5Cfrac%7B3%7D%7B16%7D%20%5D%5C%5C%3D%20%5Cfrac%7B11%7D%7B12%7D%2B%5Cfrac%7B3%7D%7B16%7D%5C%5C%20%20%20%3D%200.9167%20-%200.1875%5C%5C%3D%200.7292feet)