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2 years ago

What was the average rate in miles per hour (problem shown in picture)

Mathematics

1 answer:

Vladimir79 [104]2 years ago

6 0

Answer:

39 miles/1 hr

Step-by-step explanation:

13m=20min

20x3=60min

60min=1hour

13x3=39

39m=1hour

unit rate:

39 miles/ 1 hr

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Which container will hold the most?

elena55 [62]

Answer:

cylinder lookin one .. right ?

6 0

3 years ago

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Karl earned scores of 58, 88, 80, and 70 on his first 4 Chemistry tests. What is the minimum score he needs to earn on his fifth

OLga [1]

Answer:

Step-by-step explanation:

(58+88+80+70+ x)/5=75

75*5=296x

375=296x

x=375-296

x=79

8 0

2 years ago

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Which are the solutions of x2 = -5x + 8?

Oksana_A [137]

Answer:

1.27, -6.27 to the nearest hundredth,

or if you require it in exact form,

-2.5 + √14.25, -2.5 - √14.25.

Step-by-step explanation:

x^2 = -5x + 8

x^2 + 5x = 8

Competing the square:

(x + 2.5)^2 - 6.25 = 8

(x + 2.5) = 14.25

x + 2.5 = +/-√14.25

x = -2.5 + √14.25, -2.5 - √14.25

x = -2.5 + 3.77, -2.5 - 3.77

= 1.27, -6.27.

6 0

3 years ago

A bus is going to a museum 135 miles away. The bus will travel at 30 miles per hour. How long will it take the bus to get

TEA [102]

Answer:

4.5 hours

Step-by-step explanation:

distance equals rate times time, a relationship that can be solved for time:

time = distance / rate

Here the time is (135 miles) / (30 mph) = 4.5 hours

The bus requires 4.5 hours to get to the museum.

7 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