Answer:6 days* 24 hours per day= 144 hours
Final answer: 144 hours
Step-by-step explanation:
Answer: Yes
Step-by-step explanation: If V=12
Then we plug in for the inequality and get 12>7
2.8.1
![f(x) = \dfrac4{\sqrt{3-x}}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-x%7D%7D)
By definition of the derivative,
![f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D)
We have
![f(x+h) = \dfrac4{\sqrt{3-(x+h)}}](https://tex.z-dn.net/?f=f%28x%2Bh%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-%28x%2Bh%29%7D%7D)
and
![f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}](https://tex.z-dn.net/?f=f%28x%2Bh%29-f%28x%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-%28x%2Bh%29%7D%7D%20-%20%5Cdfrac4%7B%5Csqrt%7B3-x%7D%7D)
Combine these fractions into one with a common denominator:
![f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}](https://tex.z-dn.net/?f=f%28x%2Bh%29-f%28x%29%20%3D%20%5Cdfrac%7B4%5Csqrt%7B3-x%7D%20-%204%5Csqrt%7B3-%28x%2Bh%29%7D%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%7D)
Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:
![f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}](https://tex.z-dn.net/?f=f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B%5Cleft%284%5Csqrt%7B3-x%7D%20-%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B%5Cleft%284%5Csqrt%7B3-x%7D%5Cright%29%5E2%20-%20%5Cleft%284%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%5E2%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B16%283-x%29%20-%2016%283-%28x%2Bh%29%29%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B16h%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D)
Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :
![\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%28x%2Bh%29-f%28x%29%7Dh%20%3D%20%5Cdfrac%7B16%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7Dh%20%3D%20%5Cdfrac%7B16%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-x%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-x%7D%5Cright%29%7D%20%5C%5C%5C%5C%20%5Cimplies%20f%27%28x%29%20%3D%20%5Cdfrac%7B16%7D%7B4%5Cleft%28%5Csqrt%7B3-x%7D%5Cright%29%5E3%7D%20%3D%20%5Cboxed%7B%5Cdfrac4%7B%283-x%29%5E%7B3%2F2%7D%7D%7D)
3.1.1.
![f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E5%20-%20%5Cdfrac1%7B4x%5E2%7D%20%2B%20%5Csqrt%5B3%5D%7Bx%7D%20-%20%5Cpi%5E2%20%2B%2010e%5E3)
Differentiate one term at a time:
• power rule
![\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4](https://tex.z-dn.net/?f=%5Cleft%284x%5E5%5Cright%29%27%20%3D%204%5Cleft%28x%5E5%5Cright%29%27%20%3D%204%5Ccdot5x%5E4%20%3D%2020x%5E4)
![\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac1%7B4x%5E2%7D%5Cright%29%27%20%3D%20%5Cdfrac14%5Cleft%28x%5E%7B-2%7D%5Cright%29%27%20%3D%20%5Cdfrac14%5Ccdot-2x%5E%7B-3%7D%20%3D%20-%5Cdfrac1%7B2x%5E3%7D)
![\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%27%20%3D%20%5Cleft%28x%5E%7B1%2F3%7D%5Cright%29%27%20%3D%20%5Cdfrac13%20x%5E%7B-2%2F3%7D%20%3D%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D)
The last two terms are constant, so their derivatives are both zero.
So you end up with
![f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cboxed%7B20x%5E4%20%2B%20%5Cdfrac1%7B2x%5E3%7D%20%2B%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D%7D)
620.75-(-55.03)= 620.75+55.03 = A difference of 675.78ft
Answer:
35
Step-by-step explanation:
370+50=420
420 divided by 12 equals 35