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3 years ago

Simplify fraction 80/25

Mathematics

2 answers:

Yakvenalex [24]3 years ago

8 0

80 / 25 = 3 5/25

3 5/25

scoundrel [369]3 years ago

3 0

the simplified answer to this question, 80/25, is 1/5

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IN THE 2000 UNITED STATES PRESIDENTIAL ELECTION, 59.3 MILLION WOMEN VOTED. THAT NUMBER WAS 56% OF THE NUMBER OF THE WOMEN WHO WE

katrin [286]

D - 15.4 million.

59.3m is 56% of 105.89m whereas 51.6m is 57% of 90.52m

105.89 - 90.52 = 15.37

8 0

3 years ago

7th grade math, shouldn't be this hard.

Temka [501]

Answer:

x = 12

Step-by-step explanation:

Step 1: Add 9 to both sides.

$7x - 9 + 9 = 75 + 9$
$7x = 84$

Step 2: Divide both sides by 7.

$\frac{7x}{7} = \frac{84}{7}$
$x = 12$

Step 3: Check if solution is correct.

$7(12) - 9 = 75$
$84 - 9 = 75$
$75 = 75$

Therefore, x = 12.

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2 years ago

Read 2 more answers

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

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