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4 years ago

The sun and a planet with a period of 5 years would have a distance of

1 answer:

8 0

To solve the problem shown above you must apply the proccedure shown below:

1. You must apply the Kepler's third law, as below:

(T1/T2)²=(R1/R2)³

T1 and T2 are the orbital periods. (T1=1 year; T2=5 years)
R1 and R2 are the orbital radius. (R1=1 Astronomical unit)

2. Now, you must substitute the values shown above into the formula and clear R2:

(1/5)²=(1/R2)³
1/25=1/R2³
R2³/25=1
R2³=25
R2=∛25
R2=2.92 AU

Therefore, the answer is: 2.92 AU

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F:x = x²-4 and g:x = 2/x-2, x≠2, find g²(1)

ArbitrLikvidat [17]

Answer:

Sub to lazer beam

Step-by-step explanation:

6 0

3 years ago

A satellite dish has cross-sections shaped like parabolas. The receiver is located 13 inches from the base along the axis of sym

horsena [70]

Answer:

Depth = 3.3 inches

Step-by-step explanation:

Given that the shape of the satellite looks like a parabola

The equation of parabola is given as follows

$x^2=4\times a\times y$

Where

a= 13

Therefore

$x^2=4\times 13\times y$

$x^2=52\times y$

Lets take (13 , y) is a

Now by putting the values in the above equation we get

$13^2=52\times y$

$y=\dfrac{13^2}{52}=3.25$

y=3.25 in

Therefore the depth of the satellite at the nearest integer will be 3.3 inches.

Depth = 3.3 inches

7 0

3 years ago

Can someone help me with this problem

stepan [7]

The answer is D. 3 and 17.
3 + 17 = 20. They have a sum of 20.
17 - 3 = 14. They also have a difference of 14.
Hope I helped!

7 0

3 years ago

Use completing the square to solve the equation x2+16x=−44 . First determine the number you must add to both sides of the equati

lilavasa [31]

Answer:

We will add 64 to both sides.

x=8±√20

Step-by-step explanation:

Here we are given the equation:

$x^{2}+16x=-44$

Now we take half of the coefficient of x and add it to x.

Coefficient of x is 16, half of 16 is 8.

Also we add square of 8 to both sides.

So new equation becomes:

$x^{2}+16x+64=-44+64$

Completing the square:

$(x+8)^{2}=20$

NOw let us simplify it to get the value of x.

Taking square root on both sides:

$x+8=+/-\sqrt{20}$

Bringing 8 to the right side:

x=8±√20

4 0

3 years ago

3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south

LiRa [457]

Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

5 0

3 years ago