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3 years ago

5

The amount of money a student spends at the end of year carnival can be represented by the given equation. Graph the equation on

the coordinate plane.

1 answer:

Rom4ik [11]3 years ago

7 0

Answer: See the graph attached.

Step-by-step explanation:

The Slope-intercept form of the equation of the line is:

$y=mx+b$

Where "m" is the slope and "b" is the intersection of the line with the y-axis.

Given the equation $y=\frac{1}{2}x+3$ , you can identify that:

$m=\frac{1}{2}\\\\b=3$

The line intercepts the y-axis when $x=0$ , therefore, you can know that the line passes through the point:

(0,3) → Point of intersection with the y-axis.

Now, substitute $y=0$ and solve for "x" to find the point of intersection with the x-axis:

$0=\frac{1}{2}x+3\\\\-3=\frac{1}{2}x\\\\(-3)(2)=x\\\\x=-6$

Now, you know that the line also passes through the point:

(-6,0) → Point of intersection with the x-axis.

Draw the a line that passes through these points.

Therefore, you get the graph attached.

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Parameterize the lateral face $T_1$ of the cylinder by

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where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
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$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

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3 years ago

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alekssr [168]

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