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xz_007 [3.2K]
2 years ago
7

A Lua reflete parte da luz solar que nela incide. Admite que: a luz refletida pela Lua demora 1,28 segundos a chegar à Terra; en

tre a Lua e a Terra, a luz percorre 300 000 000 de metros em cada segundo; o trajeto da luz é retilíneo. Determina a distância da Lua à Terra. Apresenta o resultado em metros, escrito em notação científica. Mostra como chegaste à tua resposta.
Mathematics
1 answer:
Anuta_ua [19.1K]2 years ago
4 0

Answer:

3.84 * 10^8 meters

Step-by-step explanation:

As the speed of light is 300,000,000 meters/second and the time that the light takes to go from the moon to the Earth is 1.28 seconds, we can find the distance using the formula:

Distance = speed * time

So we have that:

Distance = 300,000,000 * 1.28 = 384,000,000  meters

In cientific notation, we need the first number to be in the interval from 1 to 10, so we have 3.84, and we have to multiply by 10^8 to reach our result, so we have:

Distance = 3.84 * 10^8 meters

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If a is an integer and n is a positive integer, we define a mod n to be the remainder when a is divided by n. The integer n is c
dezoksy [38]

Answer:

Divisor

Step-by-step explanation:

Given

a\ mod\ n

Required

What does n represents

In arithmetic, the number that divides another is referred to as divisor;

Take for instance;

5\ mod\ 3 = 2

This implies that when 5 is divided by 3, the remainder is 2

The bold text (divided by 3) means that 3 is the divisor of 5;

When compared to a\ mod\ n

We have that

n = 3

<em>Reason being that, n and 3 are in the same position base on order of appearance;</em>

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4 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

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3 years ago
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3 years ago
Please Help!<br><br><br><br> B belongs to AC D belongs to lineAF
Anvisha [2.4K]

Answer:

x = 27

Step-by-step explanation:

∠ CAD and ∠ EDF are corresponding angles and are congruent, so

∠ CAD = 2x

The sum of the 3 angles in Δ ABD = 180°

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3 years ago
Look at pic 10 pts will mark brainilest
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Answer:

<em>For part one...</em>

<em><u>you will multiply 4.5 by 3 to get 13.5 . </u></em>

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<u><em> you will multiply 3m by 4 to get 12.</em></u>

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