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scoundrel [369]
2 years ago
11

Find the area of each triangle with the given heights and bases. a. h = 6 inches; b = 10 inches b. h = 9 centimeters; b = 4 cent

imeters c. h = 13 yards; b = 20 yards
Mathematics
1 answer:
dsp732 years ago
6 0

The area of triangles is a. 30 inches²,   b.18 inches², and c. 130 yards² for the given triangles.

<h3>What is a triangle?</h3>

A triangle is a closed, 2-dimensional shape with 3 sides, 3 angles, and 3 vertices.

In a triangle, the sum of all three angles is 180°

Triangle is a very common figure to deal with our daily life for example in our daily life to measure the height of any tower then there is a huge application of the right angle triangle.

Area of a triangle = (1/2) × height × base

a) Given

h = 6 inches; b = 10 inches

Area = 6 × 10/2 = 30 inches².

b) Given

h = 9 centimeters; b = 4 centimeters

Area = 9 × 4/2 = 18 inches².

c) Given

h = 13 yards; b = 20 yards

Area = 13 × 20/2 = 130 yards².

Hence the area of triangles is a. 30 inches²,   b.18 inches², and c. 130 yards² .

For more information about triangle

brainly.com/question/2773823

#SPJ1

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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
What are the coordinates of the midpoint of AB IF A (6,4) and B (-8,8)
marissa [1.9K]

Answer:

-I haven't done this, but I have copied and pasted from another question and answer from the user syed514:

Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.

If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.

Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.

Solving for the coordinates gives x2 = 10, y2 = 13

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