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Sunny_sXe [5.5K]

3 years ago

9

A flat, rectangular television screen has a height of 25 inches and a diagonal of 32 inches. What is the width of the television

screen? Round your answer to the nearest hundredth of an inch.

1 answer:

Katarina [22]3 years ago

3 0

Answer:

19.97

Step-by-step explanation:

We can use the Pythagorean Theorem to solve for the width.

32 squared minus 25 squared equals 399

Find the square root. it is about 19.97

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svetoff [14.1K]

The answer is C y = 1/3x

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sergey [27]

Answer:

see below for a graph

Step-by-step explanation:

You know the line crosses the x-axis at x=6, so one way to write the equation is by translating the line with slope -1/2 to a point 6 units to the right of the origin.

y = -1/2(x -6)

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3 years ago

Solve for x. Show all your work

olasank [31]

42°

Step-by-step explanation:

$\angle BAE =180\degree -132\degree$

(Angles in linear pair)

$\angle BAE =48\degree$

$\angle AEB =90\degree..(\because \overrightarrow{EC}\perp\overrightarrow{ED})$

$\angle ABE = 180\degree-(48\degree+90\degree)$

(Angle sum postulate of a triangle)

$\implies\angle ABE = 180\degree-138\degree$

$\implies\angle ABE = 42\degree$

$\angle CDE =\angle ABE = 42\degree$

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$\implies x\degree=\angle CDE$

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$\implies x\degree=42\degree$

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2 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

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4 years ago

What is the value of the function y= 3x – 1 when x= -1 ?<br> -8<br> -4<br> 02<br> 6

netineya [11]

Answer:

-4

Step-by-step explanation:

3(-1)-1 PEMDAS

-3-1=-4

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3 years ago