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Vilka [71]
3 years ago
15

How many megabytes can a 4.7 gigabyte DVD store

Mathematics
2 answers:
Yuri [45]3 years ago
6 0
Only about 12 or 13 gb's are very small

SOVA2 [1]3 years ago
6 0
There are 1000 megabyte in 1 gigabyte

4.7 x 1000 = 4700 megabyte

4700 megabyte is your answer

hope this helps
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If you see that the temperature rose in the next 4 hours and the temperture was taken at noon so the answer would be D there was no change in temperature
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3 years ago
Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete
bazaltina [42]

Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}

b) 19 inches and 50 inches

\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}

c) 33 centimeters and 80 centimeters

\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

6 0
3 years ago
What eigen value for this matix <br> (1 -2)<br> (-2 0)
natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

  1. Compute the matrix A' = A-\lambda I, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
  2. Compute the determinant of A'
  3. Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

Finally, we have

\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}

So, the two eigenvalues are

\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}

5 0
3 years ago
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kondor19780726 [428]

Answer:

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Step-by-step explanation:

x = 1/4y + 2

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-x              -x

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A survey of 100 people found that 35 people exercise in the morning, 45 people exercise in the afternoon, and 20 people exercise
Rashid [163]

100 total people ... 35 exercise in the morning ... 45 in the afternoon ... and 20 at night

Jim is correct ...

35/100 is the ratio of people who exercise in the morning to total people.

45/100 is the ratio of afternoon exercisers to total people

20/100 is the ratio of night exercisers to total people

7 0
3 years ago
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