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2 years ago

The letters of ALABAMA and LAMB were placed on separate slips of paper and put into two different bags.

Mathematics

1 answer:

labwork [276]2 years ago

5 0

Answer:

45%

Step-by-step explanation:

There are 5 A's out of a total of 11 letters which is 5/11. 5/11 as a decimal is 0.45 which, as a percentage, is 45%

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A=(whole number less than 6)

OLEGan [10]

Answer:

then the answer is

A=4 or 5

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3 years ago

What is the solution to the equation 1/3 (9x-6)2x=3

jarptica [38.1K]

$\frac{1}{3} (9x-6)2x=3 \\ \\ \frac{(9x-6) \times 2x}{3} = 3 \\ \\ \frac{3(3x - 2) \times 2x}{3} = 3 \\ \\ \frac{6x (3x-2)}{3} = 3 \\ \\ 2x(3x-2) = 3 \\ \\ 6x^2 - 4x = 3 \\ \\ 6x^2 - 4x - 3 = 0 \\ \\ x = \frac{4+2 \sqrt{22} }{12} , \frac{2- \sqrt{22} }{12} \\ \\ x = \frac{2+ \sqrt{22} }{6} , \frac{2- \sqrt{22} }{6}$

5 0

3 years ago

Can you guys help me?

ioda

Answer:

61.93°

Step-by-step explanation:

sin x=15/17

=0.8824

sin-1 0.8824

=61.93°

5 0

2 years ago

HELP ME PLEASE!!!!!!!!

garri49 [273]

Perimeter is adding all the outside dimensions. The left side of the shoae is a rectangle, so the bottom line is the same as the top line, 10 feet.

Perimeter = 8 + 10 + 8.9 + 10 + 4 = 40.9 ft.

4 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

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

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago