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Kipish [7]
3 years ago
11

How do i do #3? I dont know how to do the whole problem.

Mathematics
1 answer:
lisov135 [29]3 years ago
3 0
You have to put the problem down we cant help you if its not there
sorry but please put equation in or problem
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Step-by-step explanation:

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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
taurus [48]

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.

4 0
3 years ago
What is the vertex of the quadratic function f(x) = (x – 8)(x – 2)?
LekaFEV [45]
Something funny is that the x value of the vertex lies directl in the middle of the x intercepts
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we see the x intercepts or 0's at x=8 and 2
the average is x=5
so find f(5) to find the y value of the vertex
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f(5)=-9

vertex is at (5,-9)

the actual way the teacher wants is to expand then compltete the square to get into the form f(x)=a(x-h)^2+k where the vertex is (h,k)
but whatever


verrtex is at (5,-9)
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4 years ago
In the image, point A is the center of the circle. Which two line segments must be equal in length?
alex41 [277]
The answer is C (EF and HI)

This is because both lines pass through the centre A, meaning both are diameters. Hence, they are equal in length.
6 0
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