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2 years ago

Write (y^2)^3 without exponents, and fill in the blank

Mathematics

1 answer:

irinina [24]2 years ago

7 0

Answer:

y*y*y*y*y*y

Step-by-step explanation:

$(y^{2})^{3} = y^{6} = y*y*y*y*y*y$

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What value of x makes this proportion true?

Trava [24]

The answer is 5:) you first have to cross multiply then divide both sides from x

8 0

2 years ago

A cone with radius 2 and height 9 has its radius tripled. How many times greater is the volume of the larger cone than the small

SpyIntel [72]

Answer:

When increasing the radius 3 times the volume increases 9 times and when it is reduced to a third the volume decreases 9 times

Step-by-step explanation:

We have that the formula for the volume of a cone is:

Vc = pi * (r ^ 2) * h

We first calculate the original volume, where the radius is 2 and the height is 9, replacing:

Vc = 3.14 * (2 ^ 2) * 9

Vc = 113.04

Now if the radius is tripled it would be: 2 * 3 = 6, the radius would be 6, replacing:

Vc = 3.14 * (6 ^ 2) * 9

Vc = 1017.36

If we compare:

1017.36 / 113.04 = 9

This means that when the radius is tripled, the volume increases 9 times.

When if re reduces to a third the radius would be: 2/3, replacing:

Vc = 3.14 * ((2/3) ^ 2) * 9

Vc = 12.56

113.04 / 12.56 = 9

Which means that by reducing it to a third the volume becomes 9 times smaller.

6 0

2 years ago

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$