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inna [77]
3 years ago
12

14 Bolve for the value of z: 5 Z 25

Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

The answer is 15

Step-by-step explanation:

Thats what I got

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maria purchased a new jacket for 78.12. the orginal price of the jacket was discounted 20% and Maria had to pay 5% sales tax. wh
prohojiy [21]
20 percent of the discount minus 5 percent sales tax equals to 15 percent.
know calculate the 15 percentage out of 78,12 so x on 78,12 equals to 20 on 100 so 78,12 equals to 100 percent and x equals to 15 percent so 78,12 times 15 divided by 100 is 15 percent or the number you found discount and and sales tax combined know you substact 15 percent out of your original price.


7 0
3 years ago
I wrote it down better
Elenna [48]
number one is 280 because 4/2 is 2 + 2 is 4 then you multiply 7 and that is 28 ^ 2 which is 280
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What’s the correct answer for this?
Aleksandr [31]

Answer:

Step-by-step explanation:

52units

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What is six ninths minus one third
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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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