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3 years ago

I really need help with this!!!

Mathematics

1 answer:

netineya [11]3 years ago

5 0

2 because is isisisisisisisisis

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Box A contains 4 Blue counters and 3 white counters. Box B contains 5 white counters and 3 blue counters. Lauren takes a counter

Georgia [21]

Answer: 20/9

Step-by-step explanation:

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4 years ago

Each child is riding on a pony with a saddle

sergeinik [125]

U got 22 riders....each has 1 saddle....so u got 22 saddles...each saddle has 2 saddle bags....so u have 22(2) = 44 saddle bags...each saddle bag has 3 water bottles....so 44(3) = 132 water bottles

7 0

3 years ago

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What is the area, in square units, of the parallelogram shown below? A parallelogram ABCD is shown with height 7 units and base

enot [183]

Answer:

35 square units

Explanation:

Area of parallelogram is calculated as follows:

Area = base * height square units

In the given question, we have:

base = 5 units

height = 7 units

Substitute with the givens in the above equation to get the area:

Area of parallelogram = 5 * 7 = 35 square units

Hope this helps :)

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4 years ago

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Find a common denominator for the following fraction pairs

Vaselesa [24]

A) The least common denominator is 12. Both 4 and 3 go into 12 evenly.
b) The least common denominator is 40. Both 5 and 8 go into 40 evenly.

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3 years ago

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Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

3 years ago