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3 years ago

7

How to simplify proper fractions

1 answer:

Kobotan [32]3 years ago

3 0

Find the greatest common factor between the numerator and denominator. The divide the numerator and denominator by the GCF.

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Question 6 Multiple Choice Worth 1 points)

OleMash [197]

Answer:

Quantity A because it changes with gravitational pull.

Explanation:

Mass defines the constitutive matter of an object. It is immutable (does not change). Although it could be converted into energy, the matter is constant.

On the other hand, the weight of an object (gravitational force) depends on the acceleration due to gravity.

That is why an astronaut weighs about three times more on earth than on Mars. The gravitational acceleration on Mars is about one third that on earth.

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2 years ago

What is the common ratio between successive terms in the sequence?<br> 27, 9, 9, 1,<br> 3,<br> of

grin007 [14]

Answer:

What is the common ratio between successive terms in the sequence?

27, 9, 9, 1,

3,

of

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2 years ago

What is the sum of 2.335 and 8.773

GrogVix [38]

Answer:

11.108

Step-by-step explanation:

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2 years ago

Read 2 more answers

Your deck is 3 2/3 feet off

Irina-Kira [14]

Your deck is bigger because it’s 3 and 2/3 and your neighbor is 3 and 8/9 your deck is bigger because 2/3 are bigger then 8/9

6 0

2 years ago

Read 2 more answers

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

2 years ago