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4 years ago

5.

Mathematics

1 answer:

TiliK225 [7]4 years ago

3 0

Answer:

Step-by-step explanation:

You would scale up the recipe to make a greater portion to accomodate for 75 crepes.

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Use the formula for the difference of cubes to show that 125 - 27 = 98

IrinaVladis [17]

The formula is

a^3 - b^2 = (a - b)(a^2 + ab + b^2) so plugging in the given values:-

a = cube root of 125 = 5 and b = cube root of 27 = 3 , we get:-

125 - 27 = (5 - 3)(5^2 + 5*3 + 3^2)

= 2 * 49

= 98 answer

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3 years ago

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4 years ago

when the point (-3, 2) is rrflected across the -axis, what are the corrdinates of the resulting point?

Mekhanik [1.2K]

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3 years ago

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Answer:

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3 years ago

Uranus has a mass of 8.68 1025 kg and a radius of 2.56 107 m. Assume it is a uniform solid sphere. The distance of Uranus from t

Lelechka [254]

Answer

Given,

Mass of the Uranus, M = 8.68 x 10²⁵ Kg

Radius of Uranus, R = 2.56 x 10⁷ m

Distance of Uranus, D = 2.87 x 10¹² days

a) Rotational Kinetic energy of the Uranus

moment of inertia of the Uranus

$I = \dfrac{2}{5}MR^2$

$I = \dfrac{2}{5}\times 8.68\times 10^{25}\times (2.56\times 10^7)^2$

I = 22.75 x 10³⁹ kg.m²

Angular speed

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{17.3\times 3600}$ \

$\omega = 1 \times 10^{-4}$

Rotational Kinetic energy

$KE = \dfrac{1}{2}I\omega^2$

$KE = \dfrac{1}{2}\times 22.75\times 10^{39}\times (10^{-4})^2$

$KE = 11.38\times 10^{31}\ J$

b) Rotational Kinetic energy of Uranus in its orbit around sun

moment of inertia of the Uranus

$I = \dfrac{2}{5}MR^2+ Ma^2$

$I = 22.75\times 10^{39}+ 8.68\times 10^{25}\times (2.87\times 10^{12})^2$

I = 7.15 x 10⁵⁰ kg.m²

Angular speed

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{3.08\times 10^4\times 3600\times 24}$ \

$\omega =2.36\times 10^{-9}$

Rotational Kinetic energy

$KE = \dfrac{1}{2}I\omega^2$

$KE = \dfrac{1}{2}\times 7.15\times 10^{50}\times (2.36\times 10^{-9})^2$

$KE = 1.99\times 10^{33}\ J$

6 0

3 years ago