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2 years ago

Volume of triangular prism

Mathematics

1 answer:

Licemer1 [7]2 years ago

8 0

Answer:

48 cm^2

Step-by-step explanation:

The area of one of the triangular ends is A = (1/2)(4 cm)(3 cm) = 6 cm^2.

Multiply this "base area" by 8 cm, the length of the prism. We get:

volume of triangular prism = (6 cm^2)(8 cm) = 48 cm^2

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If a cone has a volume of 60pi cubic meters then what is the radius, height, and diameter

algol [13]

Answer:

60pi = ⅓ × pi × r² × h

180 = r²h

Assuming r and h are whole numbers,

r = 3 m , h = 180/3² = 20 m

Diameter = 6 m

Or,

r = 6 m, h = 180/6² = 5 m

Diameter = 12 m

8 0

3 years ago

Read 2 more answers

labwork [276]

Domain is the numbers you can use
range is the result of inputing the domain

an interesting fact is that the inverse of a function switches the domain and range

basically
the domain of f(x) becomes the range of f^-1(x)
the range of f(x) becomes the domain of f^-1(x)
so just find the domain and range of f(x)

$f(x)=3x- \frac{1}{2}$
there are no restrictions
all real numbers can be used
all real numbers can result

so the answer is domain and range for both is all real numbers

D is answer

7 0

3 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

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