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den301095 [7]
3 years ago
13

Is the square root of 2+1 greater or less than or equal to the square root of 2+2

Mathematics
1 answer:
goldenfox [79]3 years ago
5 0

Answer:

\sqrt{2 + 1} < \sqrt{2 + 2}

Step-by-step explanation:

\sqrt{2 + 1} = \sqrt{3} = 1.73205...

\sqrt{2 + 2} = \sqrt{4} = 2

Since 1.73205... < 2, then

\sqrt{2 + 1} < \sqrt{2 + 2}

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Answer:

75.5 is not equivalent because 75 1/2% to a decimal is 0.775

Step-by-step explanation:

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2 years ago
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The value of (a+b)² - (a-b)² is​
ANEK [815]

Answer:

4ab

Step-by-step explanation:

(a+b)^2-(a-b)^2

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3 years ago
tell me which 4 letters are correct! A? B? C? D? E? F? if you get them right i will give you brainliest!!
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Answer:

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Step-by-step explanation:

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2 years ago
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Snezhnost [94]
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3 0
3 years ago
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In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th
iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ )  = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ;  4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal           ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

=  61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from  0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ =  0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

6 0
3 years ago
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