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masha68 [24]
3 years ago
9

To transmit information on the internet, large files are broken into packets of smaller sizes. Each packet has 1,500 bytes of in

formation. An equation relating packets to bytes of information is given by b=1,500p where p represents the number of packets and b represents the number of bytes of information.
How many packets would be needed to transmit 30,000 bytes of information?
How much information could be transmitted in 30,000 packets?
Each byte contains 8 bits of information. Write an equation to represent the relationship between the number of packets and the number of bits.
Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

Step-by-step explanation:

Jjhu

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Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.
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Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, \overline{CP} ⊥ \overline{AM}, AP = 9, and PM = 16

\overline{AC}² + \overline{CM}² = \overline{AM}²

\overline{AM} = \overline{AP} + PM = 9 + 16 = 25

\overline{AM} = 25

\overline{AC}² = \overline{AP}² + \overline{CP}² = 9² +  \overline{CP}²

∴ \overline{AC}² = 9² +  \overline{CP}²

Similarly we get;

\overline{CM}² = 16² + \overline{CP}²

Therefore, we get;

\overline{AC}² + \overline{CM}² = 9² +  \overline{CP}² + 16² + \overline{CP}² = \overline{AM}² = 25²

2·\overline{CP}² = 25² - (9² + 16²) = 288

\overline{CP}² = 288/2 = 144

\overline{CP} = √144 = 12

From \overline{AC}² = 9² +  \overline{CP}², we get

\overline{AC} = √(9² +  12²) = 15

\overline{AC} = 15

From, \overline{CM}² = 16² + \overline{CP}², we get;

\overline{CM} = √(16² + 12²) = 20

\overline{CM} = 20.

3 0
2 years ago
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