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3 years ago

10

PLEASE HELP ME! Find the cube roots of 125(cos 288° + i sin 288°).

1 answer:

shutvik [7]3 years ago

6 0

◆ COMPLEX NUMBERS ◆

125 ( cos 288 + i sin 288 ) can be written as -

125.e^i( 288)
125.e^i( 288 +360 )
125.e^i( 288+ 720)

[ As , multiples of 360 can be added to an angle without changing any trigonometric functions or sign ]

To find the cube root , take the cube root of above 3 expressions ,

We get -
5 e^( i 96 )
5 e^( i 216 )
5 e^( i 336 )

Now using Euler's formula , We rewrite above as -

5 ( cos 96 + i sin 96 )
5(c os 216 + i sin 216 )
5 ( cos 336 + i sin 336 ) Ans.

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ladessa [460]

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$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

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= $\frac{1}{3} \pi r^{2} \times r$

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

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3 years ago

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