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seraphim [82]
3 years ago
13

Josh and Anthony have a lemonade stand. They charge $1 for 2 cups of lemonade. They sell 14 cups each afternoon. Is it reasonabl

e to say Josh and Anthony earnd more than $50 in 3 afternoons?
Mathematics
2 answers:
fgiga [73]3 years ago
7 0
Yes because 2x14 is 28 and 28x3 is 84 since they earned 28 dollars each afternoon and the problem asked did they earn more than $50 in 3 afternoons so if they earned 28 in a afternoon you should multiply 28 and 3 to get the product of 84! (I hope this helped!)
Andru [333]3 years ago
5 0
No, because 14 cups of lemonade divided by 2 is 7, or, $7. $7 multiplied by 3 ( 3 afternoons) is 21, or, $21.
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Can someone check whether its correct or no? this is supposed to be the steps in integration by parts​
Gwar [14]

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

\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:

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

Divide both sides by 2:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}

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}

5 0
2 years ago
7×9-8+11-6 [5+2 (-3)]
kupik [55]
Ok, MissWalker! Please try your best to understand this, it may get confusing.
I have solved this on a different website, and here is my solving in a picture. I hope I helped!!! :D

8 0
3 years ago
Samantha has chocolate bars and lollipops in the the ratio 4:5. If she has 12 chocolate bars,how many lollipops does she have?
labwork [276]
4 • 3 = 12
5 • 3 = 15
she has 15 lollipops
8 0
3 years ago
HELP FAST DUE IN TEN MINUTES!!
goldenfox [79]

Answer:

Someone is spending 80 cents daily from their bank account, starting out with $5.6

5 0
2 years ago
Help please!!<br> asap no wrong answers<br> will give the crown symbol
Svetllana [295]

Answer:

4Hz

Step-by-step explanation:

Standard form of a sine or cosine function,

y = acos(b(x+c))

where a is the amplitude, b is the value to find the period. and c is the phase shift.

Period = \frac{2\pi}{b}

From the equation given in the question,

y = 3cos(8\pi \: t +  \frac{\pi}{2} ) \\ y = 3cos(8\pi(t +  \frac{1}{16} )) \:  \: (factorising \: 8\pi \: out)

We can see:

Amplitude = 3,

Period = \frac{2\pi}{8\pi} = 1 / 4

Phase Shift = 1 / 16

Now we want to find the frequency.

Frequency = 1 / Period

= 1 / (1/4)

= 4Hz

4 0
2 years ago
Read 2 more answers
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