She used half a yard worth of ribbon per each bow, you can tell this as 16 divided by 8 is 2. Therefore for every 2 ribbons she was using a yard of ribbon. Split that and you get half a yard
Answer:
PQ = 
Step-by-step explanation:
given MQR = 60, PQR = 75
MQP = 75 -60= 15
WE KNOW THAT MQR +QRM +QMR = 180
MPQ = MQP AS ANGLE OPPOSITE TO EQUAL SIDES ARE EQUAL
THEREFORE QMR = 90 - 60 = 30
therefore PQM = 75 - 60 = 15
PM = PQ because M is the mid point
therefore PR = PM + 
PR = 

PQ = 
From the table we can conclude that after 2 minutes the height of the water is increased by 4.
Example: In minute 2 it was 8 inches, in minute 4 it was 12 inches. (12-8=4)
This means that if the height of the water is increased by 4 inches for 2 minutes, then for 1 minute is increased 4/2=2 inches.
The following s<span>tatement best describes how the slope relates to the height of the water in the pool:
</span><span>The height of the water increases 2 inches per minute.</span>
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.