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2 years ago

14

The grocery store is having a sale on frozen vegetables. 4 bags are being sold for $11.96. At this rate what is the cost of: 9 b

ags?

2 answers:

sertanlavr [38]2 years ago

7 0

Answer:

1 bag = 2.99

9 bags= 26.91

Step-by-step explanation:

Alexandra [31]2 years ago

4 0

Answer:

$26.91

Step-by-step explanation:

11.96 divided 4 = 2.99

2.99 x 9 = $26.91

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Answer:

Both.

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2 years ago

9 out of 10 people at a game are rooting for the home team. What is the probability that exactly 6 of 8 people sitting together

scZoUnD [109]

A random supporter roots the home team with probability 0.9, and the away team with probability 0.1.

Choosing 6 out of 8 supporters who root for the home team has probability

$\displaystyle\binom{8}{6}\cdot 0.9^6\cdot 0.1^2 = 28\cdot0.531441\cdot 0.01=0.14$

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2 years ago

The frequency of a wave traveling through the air of a hot, dry desert is 1,200 hertz. Its wavelength is 0.300 meters. What is t

nekit [7.7K]

$speed=frequency*wavelength$

Given
$frequency=1200\ Hz$
and
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$speed=1200\ Hz*0.300\ m=360\ m/s$

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2 years ago

A playground measures 300 m × 170 m.Find the cost of planting grass on this at the rate of Rs 80 per hectare.

Ierofanga [76]

<h2><u>Answer</u> :-</h2>

To find the cost of planting grass , we have to find area of the field.

Area of field = Length × Breadth

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${=51000 sq. m}$

Now convert into hectare.

1 hectare = 10000m sq.

51000 / 10000

So area field = 5.1 hectare sq.

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<h2><u>___________</u></h2>

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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,

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

6 0

2 years ago