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aivan3 [116]
3 years ago
12

What is 10 times as much as 0.03

Mathematics
1 answer:
katen-ka-za [31]3 years ago
3 0
0.3 is your answer to your question

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Which is one of the transformations applied to the graph of f(x)=x^2 to change it into the graph of g(x)=4x^2+24x*30?
AfilCa [17]

Answer:

Changing g(x) to this form: a(x-h) + k, we have:

g(x) = 4 (x+3)^2 - 6

Comparing this to the original equation, f(x) = x^2, we have the following transformations:

The graph is widened.

The graph is shifted left 3 units.

5 0
3 years ago
Lee graphs the function y=2x2+1 y = 2 x 2 + 1 . Which statement explains whether the function is linear or not linear?
Sveta_85 [38]

Answer:

B

Step-by-step explanation:it in slope intercept form

4 0
3 years ago
What does four over five take away two over three eqall
BartSMP [9]
The answe to this question is 0.133333333
5 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
The diagram shows a sector of a circle radius 10 cm. the angle is 135 degrees
Black_prince [1.1K]

Answer:

The perimeter is 43.6 cm

Step-by-step explanation:

In this question, we are tasked with calculating the perimeter of the sector.

Firstly, we define what a sector is. A sector is part of a circle which is is blinded by two radii and an arc. Hence we say a sector contains two radii.

Thus, to calculate the perimeter of the sector, we need the length of the arc added to 2 * length of the radius

Let’s calculate the length of the arc.

Mathematically, this is theta/360 * 2 * pi * r

where theta is the angle subtended at the middle of the circle which is 135 according to the question, and our radius is 10cm

Thus, we have

135/360 * 2 * 22/7 * 10 = 23.57 cm

Adding two radii to this, we have;

23.57 + 2(10)

= 23.57 + 20 = 43.57 = 43.6 cm to 1 decimal place

5 0
3 years ago
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