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s344n2d4d5 [400]
3 years ago
15

2. At noon, a computer virus infects 1 byte of a computer's memory. Every minute

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

At 12:18, eighteen minutes after noon, the number of bytes infected be the same in both computers.

Step-by-step explanation:

1. Lets name PC1 the first computer and PC2 the second computer, so, at noon 1 byte of PC1 is infected with the virus, that is:

        PC1

Hour       Bytes infected

12:00                1

Then, every minute thereafter, the number of bytes is multiplied by two, that is:

         PC1

Hour         Bytes infected

12:01                  2

12:02                 4

12:03                 8

12:04                16

12:05                32

12:06                64

12:07                128

12:08               256

12:09               512

12:10                1024

12:11                 2048

12:12                4096

2. In the minute 12 after noon, 1 byte of PC2 is infected, meanwhile PC1 has 4096 bytes infected as shown above, that is:

             PC1                                                            PC2

Hour               Bytes infected                       Bytes infected  

12:12                      4096                                           1

And each minute thereafter, the number of bytes infected in PC2 multiplied by eight, that is:

            PC1                                                            PC2

Hour               Bytes infected                       Bytes infected  

12:13                     8192                                            8

12:14                    16384                                          64

12:15                    32768                                         512

12:16                    65536                                        4096

12:17                    131072                                       32768

12:18                    262144                                     262144

Therefore, at 12:18, eighteen minutes after noon the number of bytes infected be the same in both computers.

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ANSWER

Frequency=2

Midline =3

Amplitude =5

Period =π

EXPLANATION

We have:

f(x) = 5 \cos(x)  + 3

The given function is of the form;

f(x) = a\cos(bx)  + c

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a=|5| =5 is the amplitude

and b=2 is the frequency.

and 2π/b=2π/2=π

The range of the given function is

8≤y≤-2

The midline is

\frac{8  + - 2}{2}  =  \frac{6}{2}  = 3

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If (ax^2 + 3x + 2b) - (5x^2+bx-3c)=3x^2-9, what is the value of A + B + C?
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Answer:

We have the equation:

(ax^2 + 3x + 2b) - (5x^2+bx-3c)=  3x^2 - 9

First, move all to the left side.

(ax^2 + 3x + 2b) - (5x^2+bx-3c) - 3x^2 + 9 = 0

Now let's group togheter terms with the same power of x.

(a - 5 - 3)*x^2 + (3 - b)*x + (2b + 3c + 9) = 0.

This must be zero for all the values of x, then the things inside each parenthesis must be zero.

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a - 5 - 3 = 0

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