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3 years ago

Fourteen is 28% of what number? A. 7 B. 50 C. 14 D. 42

Mathematics

2 answers:

Norma-Jean [14]3 years ago

8 0

The answer is b 50 hope this helps

statuscvo [17]3 years ago

4 0

Fourteen is 28% of B) 50

Why, you ask? Let's see -

Let the unknown number be n, for 'number'.

Let's put the unknown n next to 28%

.28n% = 14

(Think, what times .28 equals 14? Look below to find out! ↓ )

Let's now do the opposite of the actual equation - divide the two known numbers by each other -

.28 ÷ 14 = 50

When in doubt, check your answer:

To check this, we would do the opposite - Multiplication

Multiply .28 by 50 to get 14 (At least you should get 14)

.28 × 50 DOES EQUAL 14! We were right! :D

↑ ↑ ↑ Hope this helps! :D

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A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

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Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

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zmey [24]

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4)20
5)8.9
6) 4.2
7)5.2
8)15
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