Answer:
wrong it is 667.95
Step-by-step explanation:
<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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<h2>Answer</h2>
f(x) = 5(1.25)x + 4
<h2>Explanation</h2>
To solve this, we are going to use the standard exponential equation:

where
is the initial amount
is the growth rate in decimal form
is the time (in months for our case)
Since the hours of classic music remain constant, we just need to add them at the end. We know form our problem that Sue initially has 5 hours of pop, so
; we also know that every month onward, the hours of pop music in her collection is 25% more than what she had the previous month, so
. Now let's replace the values in our function:



Now we can add the hours of classical music to complete our function:

You will need to work 120 hours in order to earn 300$ to get the ps4.
A,C,and D are all correct