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3 years ago

In the above figure, what's the length of the tangent from the external point D to point B?

Mathematics

2 answers:

choli [55]3 years ago

4 0

It looks like the answer is D) 4.

From C to D is 2, from B to C is 2.

2 + 2 = 4

So, 4 is the answer.

Ira Lisetskai [31]3 years ago

4 0

Answer is 4.

These questions are too easy @lawrencemadisopdp8ld.

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Find the point on the line y = 3x + 2 that is closest to the origin.

anzhelika [568]

1.
the slope of the line y=3x+2 is the coefficient of x, that is 3.

2.
the slope of any line perpendicular to the line y=3x+2, is m, such that:

$3*m=-1$

so $m= \frac{-1}{3}$

3.
so any line with equation y=-1/3x+k is perpendicular to y=3x+2.

moreover, the line y=-1/3x (so k=0) is the line passing through the origin
(0, 0) and perpendicular to y=3x+2

4. This means that the closest point of y=3x+2 to the origin is the intersection point of lines y=3x+2 and y=-1/3x

5.
to find this point:

3x+2=(-1/3)x
x(3+1/3)=-2
x(9/3+1/3)=-2

x*10/3=-2

$x= \frac{-2*3}{10} = \frac{-6}{10}=-0.6$

thus y is : (-1/3)(-6/10)=0.2

Answer: (-0.6, 0.2)

3 0

3 years ago

N equation is shown below:

QveST [7]

12x -6 = 1 then 12x =7

7 0

3 years ago

Wich graph best represents a function with range -4 < y < 3

anygoal [31]

Answer:

Witch craft or Which graph?!!

Step-by-step explanation:

8 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

Identify the slope as positive, negative, zero or undefined

slavikrds [6]

Answer:

Zero

Step-by-step explanation:

6 0

3 years ago