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3 years ago

What is 3000 divided by 200 to the 6 power

Mathematics

2 answers:

Grace [21]3 years ago

8 0

3000 divided by 200 is 15 multiply 15 by 6 and you get 90

lys-0071 [83]3 years ago

4 0

11390625 that is what i got.<span> </span>

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Which is the equation of a line that is perpendicular to 6x – 4y = 9?

natulia [17]

6x – 4y = 9

4y = 6x - 9

y = 6/4 x - 9/4

y = 3/2 x - 9/4 this line has slope = 3/2

perpendicular line, slope is opposite and reciprocal so slope of this line = -2/3

answer:

A. y= - 2/3 x - 4 (only this line has slope = -2/3)

8 0

2 years ago

Which of the following have two congruent parallel bases? Check all that apply. A. Cone B. Circle C. None of these D. Cylinder E

koban [17]

Answer: D. cylinder F. prism

Step-by-step explanation:

From the given figures in the options , only cylinder and prism have two congruent parallel bases.

A cylinder is a solid figure which has the two congruent parallel bases i.e. circles.

A prism is a solid shape that has two parallel congruent sides which are called bases and they are joined by the lateral faces that are parallelograms.

Rest of other options do not have congruent parallel bases.

4 0

2 years ago

Read 2 more answers

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

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