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Alla [95]
3 years ago
14

Quickkkk hurryyyyy!!!!!!

Mathematics
1 answer:
Darina [25.2K]3 years ago
8 0
B would be the second one
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If the area of a square is 36 square unit what si the length of a side of the square​
kodGreya [7K]

Answer:

The length of the square is 6 units.

Step-by-step explanation:

Given that all sides of the square are the same and the area of square formula is A = length×breadth. So let's make x as the sides :

area = length \times breadth

let \: area = 36 \\ let \: sides \:  = x

36 = x \times x

{x}^{2}  = 36

x = ± \sqrt{36}

x = 6 \: units

x =  - 6 \: (rejected)

6 0
2 years ago
Read 2 more answers
W=5, X=3, Y=4, Z=8<br><br> 9X=
vova2212 [387]

Answer:

27

Step-by-step explanation:

9(3)=27

For my answer I got 27 because brackets means multiply in maths so I multiply 9 by 3 and I got 27

6 0
2 years ago
What is the point of symmetry for the circle <br> (x + 2)2 + (y – 4)2 = 25?
amm1812

Answer:

The centre is (-2,4)

Answer a.

Marks as Brainlist

6 0
3 years ago
Read 2 more answers
1/9(8 1/3(21x-51y)1x-108y+2x^2)
Ann [662]

Answer:

569/ 9  x^2−153xy−12y

Step-by-step explanation:

7 0
3 years ago
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
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