Answer: Choice A) subtracting; subtractingThe goal is to find the legs of the right triangle, so you can find the hypotenuse. The hypotenuse is the distance between the two points. See the attached image.
We have been given that the upstream speed is 
and the down stream speed is 
.
Let the fisherman traveled a distance of x miles.
The total time of the trip is 6 hours.
Relations between speed (s), time (t) and distance (d) is 
Hence, 
Time upstream + time downstream = 6
lcm = 10
=> 
Cross multiplying, 
x = 20 miles .. Answer.
<h2>✒️Area Between Curves</h2>
![\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}](https://tex.z-dn.net/?f=%5Csmall%5Cbegin%7Barray%7D%7B%20%7Cc%7Cc%7D%20%5Chline%20%5Cbold%7BArea%5C%20Between%5C%20Curves%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BSolving%20for%20the%20intersection%20of%20%7D%5Crm%20y%20%3D%20x%5E2%20%2B%202%5Ctextsf%7B%20and%20%7D%5C%5C%20%5Crm%20y%20%3D%204%2C%20%5C%5C%20%5C%5C%20%5Cqquad%20%5Cbegin%7Baligned%7D%20%5Crm%20y_1%20%26%3D%5Crm%20y_2%20%5C%5C%20%5Crm%20x%5E2%20%2B%202%20%26%3D%5Crm%204%20%5C%5C%20%5Crm%20x%5E2%20%26%3D%20%5Crm%202%20%5C%5C%20%5Crm%20x%20%26%3D%5Crm%20%5Cpm%20%5Csqrt%7B2%7D%20%5Cend%7Baligned%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BWe%20only%20need%20the%20first%20quadrant%20area%20bounded%7D%20%5C%5C%20%5Ctextsf%7Bby%20the%20given%20curves%20so%20the%20integral%20for%20the%20area%7D%20%5C%5C%20%5Ctextsf%7Bwould%20then%20be%7D%20%5C%5C%20%5C%5C%20%5Cboldsymbol%7B%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B%5C%20a%7D%5E%7B%5C%20b%7D%20%7B%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%5Ctext%7Bupper%7D%20%5C%5C%20%5Ctext%7Bfunction%7D%5Cend%7Barray%7D%20%5Cright%29%20-%20%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20%5Ctext%7Blower%7D%20%5C%5C%20%5Ctext%7Bfunction%7D%20%5Cend%7Barray%7D%20%5Cright%29%5C%20dx%7D%7D%20%5C%5C%20%5C%5C%20%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%5CBig%5B4%20-%20%28x%5E2%20%2B%202%29%5CBig%5D%5C%20dx%20%5C%5C%20%5C%5C%20%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%282%20-%20x%5E2%29%5C%20dx%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%20%5Cleft%5B2x%20-%20%5Cdfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%202%5Csqrt%7B2%7D%20-%20%5Cdfrac%7B%5Cbig%28%5Csqrt%7B2%7D%5Cbig%29%5E3%7D%7B3%7D%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%202%5Csqrt%7B2%7D%20-%20%5Cdfrac%7B2%5Csqrt%7B2%7D%7D%7B3%7D%20%5C%5C%20%5C%5C%5Cred%7B%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%20%5Crm%20A%20%3D%20%5Cdfrac%7B4%5Csqrt%7B2%7D%7D%7B3%7D%5Ctextsf%7B%20sq.%20units%7D%20%5C%5C%20%5Ctextsf%7Bor%7D%20%5C%5C%20%5Crm%20A%20%5Capprox%201.8856%5Ctextsf%7B%20sq.%20units%7D%20%5Cend%7Barray%7D%7D%7D%20%5C%5C%5C%5C%5Chline%5Cend%7Barray%7D)
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Answer:
0.75 - 0.8
Step-by-step explanation:
4.5/6 = 0.75
The closest number to round 0.75 to is 0.8
(42 - 14sqrt2)/5 because rationalizing means getting the root out of the denominator, which is done by multiplying the denominator by a negative or positive root depending on the situation
