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andriy [413]
3 years ago
14

40% is what number is 82?

Mathematics
2 answers:
Licemer1 [7]3 years ago
8 0
32.8 because.40•82=32.8
kicyunya [14]3 years ago
6 0
Do 10% of 82 which is 8.2 
 then x it by 4 which would be 32.8 
<u>
</u>the answer is 32.8

#swag1famalam
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Solve for x and explain?
zhannawk [14.2K]

Answer:

x = 7

Step-by-step explanation:

When 2 secants are drawn from an external point to a circle.

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5(5 + x) = 6(6 + 4) ← distribute and simplify both sides

25 + 5x = 6 × 10

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x = 7

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3 years ago
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<span>Not included first digit (6), there are 7 digits so 

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Read 2 more answers
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

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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

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\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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