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3 years ago

6

Can the following expressions be used to determine the volume of the rectangular prism in cubic inches? Select Yes or No for eac

h expression.
8
×
6

Choose...

(
2
×
4
)
+
6

Choose...

4
×
(
6
×
2
)

Choose...

2
×
(
4
+
6
)

Choose...

1 answer:

Airida [17]3 years ago

3 0

8x6 yes 2x4+6 no 4x6x2 yes 2x4+6 no

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3 years ago

Integration questions .

dlinn [17]

<h2>1)</h2>

$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

<h2>2)</h2>

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

$\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C$

<h2>3)</h2>

$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$

<h2 />

8 0

2 years ago

What is the equation of the line that passes through the point (6,14) and is parallel to the line with the following equation? y

Gekata [30.6K]

Answer:

$y=\displaystyle-\frac{4}{3}x+22$

Step-by-step explanation:

Hi there!

<u>What we need to know:</u>

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where <em>m</em> is the slope and <em>b</em> is the y-intercept
Parallel lines always have the same slope (<em>m</em>)

<u>Determine the slope (</u><em><u>m</u></em><u>):</u>

<u /> $y=\displaystyle-\frac{4}{3}x -1$ <u />

The slope of the given line is $\displaystyle-\frac{4}{3}$ , since it is in the place of <em>m</em> in y=mx+b. Because parallel lines always have the same slope, the slope of a parallel line would also be $\displaystyle-\frac{4}{3}$ . Plug this into y=mx+b:

$y=\displaystyle-\frac{4}{3}x+b$

<u>Determine the y-intercept (</u><em><u>b</u></em><u>):</u>

$y=\displaystyle-\frac{4}{3}x+b$

To find the y-intercept, plug in the given point (6,14) and solve for <em>b</em>:

$14=\displaystyle-\frac{4}{3}(6)+b\\\\14=-8+b\\b=22$

Therefore, the y-intercept of the line is 22. Plug this back into $y=\displaystyle-\frac{4}{3}x+b$ :

$y=\displaystyle-\frac{4}{3}x+22$

I hope this helps!

5 0

3 years ago