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2 years ago

I need help with this

Mathematics

2 answers:

natima [27]2 years ago

8 0

Answer:

Step-by-step explanation:

Simora [160]2 years ago

4 0

Maybe try d? i really don’t know

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Cual es la mitad de 1/4 kg de miel

Evgesh-ka [11]

Answer:

1/8 Kg de miel.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

PLZ PLZ PLZZZZZZ HELP ME IMMEDIATLEY

earnstyle [38]

Answer:

×=64

I HOPE IT CAN HELP:(

8 0

2 years ago

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

What is the value of the expression shown when x = 5?

sergey [27]

3(2(5)+4)-5
3(10+4)-5
30+12-5
42-5
37

4 0

3 years ago

What is the longest line segment that can be drawn in a right rectangular prism that is 14cm long, 13cm wide, and 11cm tall?

vlabodo [156]

The longest line segment that can be drawn in a right rectangular prism that is 14cm long, 13cm wide and 11cm tall is 19.1cm.

<h3>What is a right rectangular prism?</h3>

A right rectangular prism is a three dimensional solid shape formed by 6 rectangles.

it is also called the cuboid.

Analysis:

The diagonal of the face of the prism with dimensions 14cm long and 13cm wide is the longest line segment that can be drawn.

Since rectangles have 90° on each vertex, we can use Pythagoras theorem to calculate for the length of the diagonal.

$(diagonal)^{2}$ = $(length)^{2}$ + $(width)^{2}$

$(diagonal)^{2}$ = $(14)^{2}$ + $(13)^{2}$

= 196 + 169 = 365

$(diagonal)^{2}$ = 365

diagonal = $\sqrt{365}$ = 19.1cm

In conclusion, the length of the longest diameter is 19.1cm

Learn more about Right rectangular prism: brainly.com/question/3317747

#SPJ1

4 0

2 years ago