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Vinvika [58]
1 year ago
13

Write the equation of the line.

Mathematics
1 answer:
omeli [17]1 year ago
4 0

Answer:

y=x+5

Step-by-step explanation:

The slope of the line is (9-5)/(4-0)=1, and the y-intercept is 5, so the equation is y = x + 5.

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Help I forgot how to do this​
lakkis [162]

Answer:

number 5 is 11 and number 6 is 16

Step-by-step explanation:

you substitute a for 4 bc a equals 4 and than finish the equation 4+7=11

6. you substitute b for 2 and c for 8 and multiply which than is 16

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3 years ago
F(x)=2-3x I need help with this please
Fynjy0 [20]

Answer:

F= 0.5

Step-by-step explanation:

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3 years ago
HURRY PLS HELP ILL GIVE BRAINLYEST Subtract.
natka813 [3]

Answer:

2940

Step-by-step explanation:

You are subtracting a negative number. "Minus a negative is plus."

525 - (-2415) = 525 + 2415 = 2940

6 0
2 years ago
Read 2 more answers
It’s raining outside, and after 1 hour of rain, the water level in Jake’s pond is 2 meters. After 3 hours, the water level is 2.
GREYUIT [131]
Th state of change is 0.4 meters per hour
5 0
2 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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