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svlad2 [7]
3 years ago
5

A rettangle is 10 inches smaller than it is wide. The perimeter of the

Mathematics
1 answer:
meriva3 years ago
6 0

Answer:

A. Width = 25   Length = 15

Step-by-step explanation:

Length = L = Width - 10

Width = W

Perimeter of a rectangle

=> 2 (L + B) = 80

=> 2 (W - 10 + W) = 80

=> 2 (2W - 10) = 80

=> 4W -20 = 80

=> 4W -20 + 20 = 80 + 20

=> 4W = 100

=> 4W/4 = 100/4

=> W = 25

Width = 25

Length = 25 - 10 = 15

So, the answer is A.

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1. Determine whether the graphs of the given equation are Parallel, perpendicular, or neither.
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1. A. parallel, because when you get y by itself it is y=-2x+7,  the two equations have the same slope which means they are parallel.

2. B. sometimes, in order for two lines to be parallel they have to have the same slope so, it is possible that two lines say with a slope of 3, those would be parallel BUT it doesn't always have to have a positive slope for it to be parallel like in question 1. 

3. C. never, If they have the same slope they are parallel despite whatever the y-intercept is.
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3 years ago
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-10+16 divided by (-2)+7<br><br> *Please show work!*
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Here’s my work to your question!

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2 years ago
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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
A flower bed has the shape of a rectangle 9 yards long and 3 yards wide. What is its area in square feet?
Zanzabum

Answer:

243 ft2

Step-by-step explanation:

9*3= 27

3*3= 9

27*9=243 ft2

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4 years ago
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