Answer:
- 31 pencils
- 38 erasers
- 44 sharpeners
Step-by-step explanation:
The number of packets is the greatest common divisor of the given numbers of pencils, erasers, and sharpeners.
It can be helpful to look at the differences between these numbers:
748 -646 = 102
646 -527 = 119
The difference of these differences is 17, suggesting that will be the number of packets possible.
527 = 17 × 31
646 = 17 × 38
748 = 17 × 44
The numbers 31, 38, and 44 are relatively prime (31 is actually prime), so there can be no greater number of packets than 17.
There will be 31 pencils, 38 erasers, and 44 sharpeners in each of the 17 packets.
_____
We may have worked the wrong problem. The way it is worded, the <em>maximum</em> number of items in each packet will be 527 pencils, 646 erasers, and 748 sharpeners in one (1) packet. The <em>minimum</em> number of items in each packet will be the number that corresponds to the maximum number of packets. Since 17 is the maximum number of packets, each packet's contents are as described above.
17 is the only common factor of the given numbers, so will be the number of groups (plural) into which the items can be arranged.
the answer would be 23 because 46 and 2 are both common factors of 23.
.41 dollars a pound
.25 dollars an hour
Answer:
Step-by-step explanation:
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Answer: The coordinate are
( 5 , 2.5 )
Distance = 5
Step-by-step explanation:
The coordinates of the rest stop will be the mid point between the high school is at point (3, 4) and the stadium in Columbus at point (7, 1).
Mid point = (X1 +X2)/2 , (Y1+Y2)/2
(3+7)/2 , (4+1)/2 = ( 5 , 2.5 )
The approximate distance between the high school and the stadium will be
D^2 = (7-3)^2 +( 4-1)^2
Where D = distance
D^2 = 4^2 + 3^2
D = 5
