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2 years ago

How many shovels of sand need to be added to two shovels of cement?

Mathematics

1 answer:

Alborosie2 years ago

4 0

Answer:

8 sand shovels

Step-by-step explanation:

1 cement needs 4 sand shovels

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Olivia made 14 outfits for her dolls. Each outfit used 3/5 of fabric. If she started with 10 3/4 yards of fabric, how much did s

Zepler [3.9K]

Answer:

0.1 yards of fabric

Step-by-step explanation:

Data provided in the question:

Number of outfits made for the doll = 14

Fabric used in each outfit = $\frac{3}{5}$ yards of fabric

Starting fabric available = 10 × $\frac{3}{4}$ or $\frac{30}{4}$ yards

Now,

The total fabric used in making the outfits of 14 dolls

= 14 × $\frac{3}{5}$ yards of fabric

= $\frac{42}{5}$ yards of fabric

Therefore,

The fabric left after making the outfits

= Fabric available in the start - Fabric used in making the outfits

= $\frac{30}{4}$ yards - $\frac{42}{5}$ yards

= $\frac{30\times5-42\times4}{4\times5}$ yards

= $\frac{150-168}{20}$ yards

= $\frac{2}{20}$ yards

= 0.1 yards of fabric

5 0

3 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

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

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

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

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

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$