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Akimi4 [234]
3 years ago
5

How do I solve this problem if x mean the missing number? 16\3 = 12\x

Mathematics
1 answer:
katovenus [111]3 years ago
3 0
Set this up as a fraction, then cross multiply and solve. In our case, cross multiplying will get us 16x=36, which gets us x=9/4.

:)
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The price of gold rose from $790 per ounce to $860 per ounce what percent increase does this amount represent round to the neare
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Now, use your research to decide which business is the best choice for this community. Explain your choice in two to three sente
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Elena and Jada are 25 miles apart when they start moving toward each other. Elena runs at a constant speed of 6 mph and Jada wal
Setler [38]

<u>Answer:</u>

The Time taken by Elena and Jade  to meet up is 2.5 hours

<u>Explanation:</u>

Given,

Elena moves with a speed of 6mph  while Jada walks at a speed of 4 mph

Also, Distance can be calculated by the formula

Distance = Speed * Time

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3 0
3 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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