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Keith_Richards [23]
3 years ago
10

You are given 1000 one dollar bills and 10 envelopes. Put the bills into the envelopes in such a way that someone can ask you fo

r any amount of money from $1 to $1000 (examples - $532, $619, $88, etc.) and you can give it to them through a combination of the envelopes.
Mathematics
1 answer:
taurus [48]3 years ago
4 0

Answer:

We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489

Step-by-step explanation:

  • Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.

This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.

However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.

Observe that:

  1. 1+2+4+8+16+32+64+128+256+489=1000
  2. Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
  3. If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since x-489\leq511, by 2) we know that this would be possible.

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2 years ago
The lengths of three sides of a quadrilateral are shown below:
julsineya [31]
Answers: 
Part A: 12y² + 10y – 21
Part B: 4y³ + 6y² + 6y – 5 
Part C: See below.

Explanations: 
Part A: 
For this part, you add Sides 1, 2 and 3 together by combining like terms:
Side 1 = 3y² + 2y – 6
Side 2 = 4y² + 3y – 7
Side 3 = 5y² + 5y – 8
3y² + 2y – 6 + 4y² + 3y – 7 + 5y² + 5y – 8
Combine like terms: 
3y² + 4y² + 5y² + 2y + 3y + 5y – 6 – 7 – 8 
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Part B: 
You have the total perimeter and the sum of three of the sides, so you just need that fourth side value, which we can call d.
P = 4y³ + 18y² + 16y – 26
Sides 1, 2 & 3 = 12y² + 10y – 21 
Create an algebraic expression: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
– 12y²                                – 12y²
10y – 21 + d = 4y³ + 6y² + 16y – 26 
– 10y                               – 10y
– 21 + d = 4y³ + 6y² + 6y – 26 
+ 21                                + 21 
d = 4y³ + 6y² + 6y – 5 

Part C: 
If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y², y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms.
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