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2 years ago

A rectangular garden is 20 ft longer than it is wide. Its area is 800 ft 2 . What are its dimensions?

1 answer:

6 0

Our equation for the area of a rectangle is l x w = A (length times width equals area)
We know that l=w+20 so we can plug this into our equation.
It is now, (w+20) x w =800
We can then solve for w
w^2 +20w =800
We can subtract 800 on each side to get
w^2 +20w - 800 =0
We then factor to get
(w-20)(w+40)=0
This gives us w =20 and w = -40
Width cannot be negative so our width will be 20 feet
Since our length is 20 more than this, then our length will be 40feet
So the dimensions of the rectangle are 20 ft wide by 40 ft long

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1 year ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

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3 years ago

Determine whether this statement is true or false. if two lines have the same slope, their equations describe the same line.

maksim [4K]

Answer:

False

Step-by-step explanation:

When you say "slope", Im guessing your referring to the gradient (the steepness of the line).

If two lines have the same gradient their equations won't usually describe the same line because their 'y intercepts' will be different (the point in which the line crosses the y axis)

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3 years ago

Which steps can be used in order to determine the solution to Negative 1.3 + 4.6 x = 0.3 + 4 x?

Mars2501 [29]

Answer:

x=8/3 OR 2.7

Step-by-step explanation:

-1.3+4.6x=0.3+4x

4.6x-4x=0.3+1.3

0.6x=1.6

x=1.6/0.6=8/3

x=8/3 OR 2.7

Hope this helps!

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