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aleksley [76]
3 years ago
12

Mr. Evens Is paid $9.20 per hour for the first 40 hours he works in a week he is paid 1.5 times that rate for each hour after th

at. last week mr. evens worked 42.25 hours he says he earned $388.70 last week do you agree support your answer
Mathematics
2 answers:
spin [16.1K]3 years ago
7 0
I don't agree with his statement. The first part says that he was paid $9.20 for the first 40 hours so we multiply them (9.20 x 40) to get 368. Then it says that he was paid 1.5 times that rate for every hour after that. The rate is "$9.20 per hour" so you find 1.5 times $9.20. You multiply $9.20 by 1.5 to get $13.80 per hour. Now you have to find how much money he got after the initial 40 hours. It says he worked 42.25 hours, and we already figured out how much money he got the first 40 hours ($368). So for the remaining 2.25 hours, Mr.Evens is paid $13.80 an hour. So we multiply 2.25 by $13.80 to get $31.05. To find out the total amount of money accumulated, we add. $368 + $31.05 is $399.05 which means that Mr.Evens is not correct. Hope this helped :]
Misha Larkins [42]3 years ago
3 0
Ok so the answer is No becuse if you times 40 × 9.20 =368 + 42.25 its over then divided 1.5 it equals $276.35
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Answer:

\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

Step-by-step explanation:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

Given integral:

\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x

\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:

\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x

Using <u>integration by parts</u>:

\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

Therefore:

\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}

\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:

\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}

Therefore:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x

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Divide both sides by 2:

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Rewrite in the same format as the given integral:

\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

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