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maxonik [38]
3 years ago
15

A person makes a mistake when copying information regarding a particular triangle. The copied information is as follows: Two sid

es of a triangle are the same length. The third side is 10 feet less than three times the length of one of the other sides. The perimeter of the triangle is 5 feet. What is the mistake?
Mathematics
1 answer:
irakobra [83]3 years ago
6 0

Answer:

  No such triangle can exist. Any of the described relations may be in error.

Step-by-step explanation:

Taken at face value, the sum of the lengths of the sides is ...

  x + x + (3x-10) = 5

  5x = 15

  x = 3

so the third side is 3·3 -10 = -1 in length. No such triangle can exist.

If one side is actually 3x-10, the perimeter must be greater than 6 2/3 and less than 40.

If the perimeter is actually 5, then the third side must be 3x-7.5 or an expression with an even smaller constant.

So, the mistake could be in copying any of the numbers, or in copying the multiplier in the description of the 3rd side. There is insufficient information to tell exactly what the mistake is.

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How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
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Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
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That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

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Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
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Hello : 
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answer :

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explanation:

sa calculator lang yan haha jk

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