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2 years ago

Please help me with this.

Mathematics

1 answer:

JulijaS [17]2 years ago

3 0

The speed intervals such that the mileage of the vehicle described is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h]

<h3>How to determine the range of speed associate to desired gas mileages</h3>

In this question we have a quadratic function of the gas mileage (g), in miles per gallon, in terms of the vehicle speed (v), in miles per hour. Based on the information given in the statement we must solve for v the following quadratic function:

g = 10 + 0.7 · v - 0.01 · v² (1)

An effective approach consists in using a graphing tool, in which a horizontal line (g = 20) is applied on the maximum desired mileage such that we can determine the speed intervals. The speed intervals such that the mileage of the vehicle is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h].

To learn more on quadratic functions: brainly.com/question/5975436

#SPJ1

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How?!? I don't know what to do

Goshia [24]

This is easy, if adding the first number to the second number you get 70.2 then if you want to find the middle number you will have to subtract 32.34 from 70.2 to get 37.86, now add that and 32.34 to check if it gives you 70.2.... When you add them you indeed get 70.2 so your work is correct meaning your middle number is 37.86

Hope this helped

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3 years ago

The temperature in Vancouver is -8ºC, in Montreal it is -4ºC , in Seattle it is -6ºC, and in Buffalo it is -10ºC. Which city is

Marina CMI [18]

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3 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 integration by parts:

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

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

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:

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