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Blizzard [7]
3 years ago
6

Plz help I need help

Mathematics
1 answer:
Pani-rosa [81]3 years ago
6 0

Answer:

2/3

Step-by-step explanation:

0/0 is 0

1/2 is 1 of 2

2/3 is 2 of 3

3/4 is 3 of 4

hopefully it helps!

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A number sentence is shown below:
d1i1m1o1n [39]

Number sentence is simply a sentence, that uses numbers and arithmetic signs.

  • The context that can be created from the sentence is: <em>How many bits are there, in </em>32\frac 14 bytes
  • The verbal meaning of 32\frac 14 \div \frac 18 = 258 is:  <em>When </em>32\frac 14<em> is divided by </em>\frac 18<em>, the result is </em>258<em />

<em />

Given

32\frac 14 \div \frac 18 = 258

<u />

<u>(a) Context or story</u>

A context is as follows:

<em>There are 8 bits in a byte; How many bits are there, in </em>32\frac 14 bytes

<u />

<u>(b) As a multiplication</u>

32\frac 14 \div \frac 18 = 258

Change the <em>division to multiplication </em>as follows:

32\frac 14 \times \frac 81 = 258

<u />

<u>(c) Verbal explanation how the numbers are related</u>

When 32\frac 14 is divided by \frac 18, the result is 258

Read more about number sentence at:

brainly.com/question/17322770

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.

8 0
4 years ago
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