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3 years ago

Please Help Calculate!!!

Mathematics

1 answer:

Darina [25.2K]3 years ago

4 0

Answer:

1) 35

2) 1.51937984

Step-by-step explanation:

$\frac{4/\frac{4}{7}•0.3}{2.88 ÷ 4.8}$

$\frac{21}{0.6}$

21 ÷ 0.6

$\frac{0.5 + \frac{1}{18}}{(1\frac{1}{6}-\frac{7}{18}}÷2.8$

$\frac{\frac{5}{9} }{0.36564625}$

$\frac{5}{9}$ ÷0.36564625

1.51937984

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What’s (3b-5c)(2x+5y) squared

lukranit [14]

Answer: In your problem, you said that they are squared. However, it is not written as squared. We just have to multiply them.

We can use the FOIL method.

(3b - 5c)(3x + 5y)

6bx + 15by - 10cx - 25cy

There are no like terms, so that expression is our final answer.

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3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

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3 years ago

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slavikrds [6]

Reflection over the y axis Then a translation of 2 units down

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3 years ago

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Shalnov [3]

64 hours per hour average
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LekaFEV [45]

Answer:

£1920

Step-by-step explanation:

The amount she spent in 2016 is 20% less than 2017. 20% less than is the same as 80% of the amount she spent in 2017. So, to find how much she spent in 2016, multiply 2400 by 0.80 (the decimal form of 80%):

2400 * 0.8 = 1920

Emily spent £1920 on holiday in 2016.

I hope this helps :)

8 0

3 years ago