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3 years ago

Drag the correct steps into order to evaluate 42 + t/6 for t = 12.

Mathematics

2 answers:

evablogger [386]3 years ago

4 0

I actually don't know

Darya [45]3 years ago

3 0

Ok so substitute 12 where it says t in the equation. So first step is to divide 12 by 6 and that = 2 and then add that to 42 which = 44. Hopes that helped

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Given that f(x) = 5x - 10 and g(x) = x + 3, find f(g(-1)).

musickatia [10]

$\bf \begin{cases} f(x) = 5x-10\\ g(x) = x + 3 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ g(-1)=(-1)+3\implies \boxed{g(-1) = 2} \\\\\\ f(~~g(-1)~~)\implies f\left( \boxed{2} \right) = 5(2)-10\implies \stackrel{f(~g(-1)~)}{f(2)}=0$

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Why do you think city streets often are designed to be parallel or perpendicular?

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Answer:

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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:

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

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2 years ago

If four out of ten people vote than a town with two hundred thousand people how many will vote

oksano4ka [1.4K]

Answer:

80,000 people will vote. I'm pretty sure

Step-by-step explanation:

You first divided 200,000 by 10 then multiply it by 4 and you get 80,000. You're welcome, if its right.

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Solve for x. . Please wait for photo to load.

hammer [34]

Your anwer is option B. x=1/2

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2 years ago