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+2+4+8+16+32+64+128+256+489=1000
- 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.
- 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
, by 2) we know that this would be possible.
The answer would be B. -4
Answer:
<h2>
<em><u>60</u></em></h2>
Step-by-step explanation:
tiny stuffed zebras(tsz)= $3
giant stuffed zebras(gsz)=$14
total sold=$208
208÷3=69.3recurring so that's incorrect as you cannot sell .3 of a zebra so keep going until you find the next whole number. (trial and error)
207÷3=69 but it cant be this as it is not the whole money made
208-14=194
194÷3=64.6recurring (cannot be this)
194-14=180
180÷3= 60 = answer as 60 tsz is $180 and add that to 2 gsz and that totals to 208 which is the total
Answer:
c = 
Step-by-step explanation:
Given
R= 
Clear the radical by squaring both sides
R² = b² - 4ac ( subtract b² from both sides )
R² - b² = - 4ac ( multiply all terms by - 1 )
b² - R² = 4ac ( divide both sides by 4a )
= c
For the first 6 1/3
second is 14/25
third 2 29/42