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3 years ago

What is 80% of blank games is 32 games

1 answer:

5 0

So basically you are saying that 80% of ____ games is 32 games.

So 32 is the 80% of the games played.

so we can divide 32 by 8, and that will give us 4 ( you might be confused why I did this, just be patient please)

so 4 is 10% of the games played.

We have to find out 20%, so we just multiply 4 by 2, which is 8

too find out 100%, we just add 80% and 20%, so 32 was 80%, and 8 was 20%, and 32 + 8 = 40

So the answer would be 40.

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Which point on the coordinate grid best represents the ordered pair (-5/2, -6/3)?

zhenek [66]

Answer:

point d

Step-by-step explanation:

-5/2 = -2.5 and -6/3 = -2

4 0

2 years ago

A 11-inch board is cut into 6 equal length pieces. How long is each piece?

Dennis_Churaev [7]

Answer:

1.8inches or 1.83in repeating.

Step-by-step explanation:

divide 11 by 6. you get the decimal 1.83333333. you can either round to 1.8 or do 1.83 with a repeating symbol over the three (small dash above the number).

4 0

3 years ago

Leslie is in charge of packing snacks for her class she has 30 cookies and 20 apples she wants to put the same number of apples

Lady_Fox [76]

She can put 1 cookie and 1 apple in each bag. 20 packs

8 0

3 years ago

Read 2 more answers

NEED HELP ASAP!!

Solnce55 [7]

Answer:

Only Cory is correct

Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

$F =G\times \dfrac{M \cdot m}{r^{2}}$

Where;

G = The universal gravitational constant

M = The mass of one object

m = The mass of the other object

r = The distance between the centers of the two objects

For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km

Therefore, for an astronaut in the international Space Station, r = 6,800 km

The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

$\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2} = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx 1.14$

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity

Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

6 0

3 years ago

Simplify 2√2-√3<br> ---------------<br> √2+√3

serious [3.7K]

Step-by-step explanation:

$= \frac{2 \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$