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3 years ago

If x<5 and y<6, then x+y__ 11 is it <,>,=?

1 answer:

7 0

To solve for this, let's try the highest number x and y can be.

X < 5, so it can be 4.
Y < 6, so it can be 5.

4 + 5 = 9.
9 < 11, so x + y < 11.

Your missing inequality symbol is:

<

I hope this helps!

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Determine the volume and the total surface area of the square pyramid.if its perpendicular height is 12cm and square length is 5

Vika [28.1K]

Answer:

Volume = 100 cm³

Surface area of the square pyramid = 145 cm²

Step-by-step explanation:

Given:

Perpendicular height = 12cm

Square length = 5cm

Find:

Volume

Surface area of the square pyramid

Computation:

Area of base = side²

Area of base = 5²

Area of base = 25 cm²

Volume = (1/3)(A)(h)

Volume = (1/3)(25)(12)

Volume = 100 cm³

Surface area of the square pyramid = A + 1/2(P)(h)

Perimeter square pyramid = 4(s)

Perimeter square pyramid = 4(5)

Perimeter square pyramid = 20 cm

Surface area of the square pyramid = 25 + 1/2(20)(12)

Surface area of the square pyramid = 145 cm²

3 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

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

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

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

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$