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dangina [55]
2 years ago
10

Help explain!!!! please

Mathematics
1 answer:
sammy [17]2 years ago
6 0
Bring the shape on the other side of the y axis. A mirror image.
It will be on the top right. That is quadrant 1
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How many pages can Jorge print in 3/5 of an hour if he can print 24 pages every 15 minutes?
MariettaO [177]

Answer:

57.6 pages

Step-by-step explanation:

First, you need to identify how many minutes are in 3/5 of an hour:

1 hour = 60 minutes

3/5 of an hour = 3/5 of 60 minutes

= 3/5(60)

= 36

So, you need to find out how many pages can be printed in 36 minutes. To figure this out, a ratio can be used:

15 : 24

36 : p

p represents pages in 36 minutes.

Now, you need to solve the ratio by figuring out what you need to multiply 15 by to get 36 and then multiplying that by 24:

15 * 36/15 = 36

36/15 = 2.4

24*2.4

= 57.6 pages

5 0
3 years ago
Find the surface area of th prism.
shutvik [7]
<span> 2(lh + lw + wh) = 96
lh + lw + wh = 48

l(h + w) + wh = 48
3(3 + 4) + 12

Length = 3
Width = 3
Height = 4
</span>
8 0
3 years ago
Read 2 more answers
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
Finding missing angles
Zolol [24]

Answer:

95

Step-by-step explanation:

x \degree = 95 \degree \\(vertical \: angles)  \\  \\ x = 95

8 0
3 years ago
Evaluate the Expression<br> When X=2, 3x-1=
Alex777 [14]

Answer:

5

Step-by-step explanation:

If X equals 2, then 3 times X would be 6. This is because if a number and a variable are put together, you have to multiply them. 6-1 is 5.

4 0
3 years ago
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