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zaharov [31]
4 years ago
14

How did you get 100 squares out of 10 line segments?

Mathematics
1 answer:
elena-14-01-66 [18.8K]4 years ago
7 0

Answer:

If every male on earth got a bonar at the same time, the earth's rotation would slow down. Assume there are about 3.8 billion males, with an average dic-k height of about 80 cm off the ground. The average dic-k weighs about 100 grams.

That's a combined mass of 380,000,000 kg of coccc.

Now we must make an approximation. For simplicity's sake, let us assume the peeens are all evenly lined up in a ring around the equator. The equation for moment of inertia of a ring is I = mass*radius^2. The radius of earth is about 6.371 million meters. Therefore the radius of the approximated dic-k ring is 6,371,000 + 0.80 = 6,371,000.8 meters.

I = 380,000,000*6,371,000.8^2 = 1.5424*10^22

The Earth has a moment of inertia, I = 8.04×10^37 kg*m^2. The Earth rotates at a moderate angular velocity of 7.2921159 ×10^−5 radians/second.

Using Conservation of Angular Momentum we can find the new rotation of the earth.

L=I*omega= 8.04×10^37 kg*m^2 * 7.2921159 × 10^−5 = 5.86286*10^33.

5.86286*10^33 = (8.04×10^37 + 1.

Step-by-step explanation:

bc its like that

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Dmitriy789 [7]

1) 6, 12,18,24

The rule used here is: Arithmetic Progression or Linear sequence

From the data above:

a = 6, d = 6

\begin{gathered} T_{n\text{ }}=\text{ a + (n-1)d} \\ \text{where n = 20} \\ T_{20\text{ }}=\text{ a + (20-1)d} \\ T_{20\text{ }}\text{ = a + 19d} \\ T_{20\text{ }}\text{ = 6 + 19}\times6 \\ T_{20\text{ }}\text{ = 6 + 114} \\ T_{20\text{ }}\text{ = 120} \end{gathered}

2) 3,6,9,12

The rule used here is: Arithmetic Progression or Linear sequence

From the data above:

a = 3, d = 3

\begin{gathered} T_{n\text{ }}=\text{ a + (n-1)d} \\ \text{where n = 20} \\ T_{20\text{ }}=\text{ a + (20-1)d} \\ T_{20\text{ }}\text{ = a + 19d} \\ T_{20\text{ }}\text{ = 3 + 19 }\times3 \\ T_{20\text{ }}\text{ = 3 + 57} \\ T_{20\text{ }}\text{ = 60} \end{gathered}

3) 1,5,9,13

The rule used here is: Arithmetic Progression or Linear sequence

From the data above:

a = 1, d = 4

\begin{gathered} T_{n\text{ }}=\text{ a + (n-1)d} \\ \text{where n = 20} \\ T_{20\text{ }}=\text{ a + (20-1)d} \\ T_{20\text{ }}\text{ = a + 19d} \\ T_{20\text{ }}\text{ = 1 + 19 }\times4 \\ T_{20\text{ }}\text{ = }1\text{ + 76} \\ T_{20\text{ }}\text{ = }77 \end{gathered}

5 0
1 year ago
(1)/(4p)(x-h)^(2)+k=0
erik [133]
Assume that p\ \textgreater \ 0 and <span>multiply the equation \frac{1}{4p} (x-h)^2+k=0 by 4p. Then you obtain the equation (x-h)^2+4pk=0.
</span>
1) If k>0, then 4pk>0 and the equation doesn't have solutions, because <span>(x-h)^2=-4pk and this is unreal.
</span>
<span>2) If k=0, then 4pk=0 and (x-h)^2=0. There is one solution x=h.
</span>
2) If k<0, then 4pk<0 and the equation
<span>(x-h)^2=-4pk>0 has two different solutions x_1=h+ \sqrt{-4pk} and x_2=h- \sqrt{-4hk}.</span>

8 0
3 years ago
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