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Olin [163]
4 years ago
11

Rosy has 16 pencils and 24 erasers. She is putting together packets that will have an equal number of pencils and erasers in eac

h packet. If Rosy uses all of the pencils and erasers, what the the maximum number of packets she can make?
Mathematics
2 answers:
Effectus [21]4 years ago
8 0

Answer:

If she puts 2 pencils and 3 erasers in each packet,

Rosy can make a total of 8 packets.

tia_tia [17]4 years ago
6 0

Answer:

the max packets she can make is 8

Step-by-step explanation:

16/8=2 and 24/8=3 s and if she make more packets there will be a decimal amount of erasers.

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I need help learning how to solve this problem please.
Radda [10]

we have 4 gals, Stephanie, Andrea, Emily and  Becca.

btw the end point should be (-6, -7) and the start point is (8, 6).

so, if we know the distance between those two points, we can just cut it in 4 equal pieces and each gal will be driving a piece.

now, Emily's turn is after Andrea, BUT, we had first Stephenie driving ¼ of the way, then Andrea driving ¼ of the way too, but after ¼+¼, Emily comes on, but ¼+¼ is really 1/2, so when Emily starts, is really half-way through, or the midpoint of that distance.


\bf ~~~~~~~~~~~~\textit{middle point of 2 points }&#10;\\\\&#10;\stackrel{\textit{Hometown}}{(\stackrel{x_1}{8}~,~\stackrel{y_1}{6})}\qquad&#10;\stackrel{\textit{San Antonio}}{(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-7})}&#10;\qquad&#10;\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)&#10;\\\\\\&#10;\left( \cfrac{-6+8}{2}~~,~~\cfrac{-7+6}{2} \right)\implies \left(\cfrac{2}{2}~,~\cfrac{-1}{2} \right)\implies \stackrel{\textit{Emily starts off}}{\left(1~,-\frac{1}{2} \right)}


Emily's turn ends ¼ of the way later, which will be pretty much half-way in between  (1, -½) and (-6, -7), so is really the midpoint of those two.


\bf ~~~~~~~~~~~~\textit{middle point of 2 points }&#10;\\\\&#10;(\stackrel{x_1}{1}~,~\stackrel{y_1}{-\frac{1}{2}})\qquad&#10;\stackrel{\textit{San Antonio}}{(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-7})}&#10;\\\\\\&#10;\left( \cfrac{-6+1}{2}~~,~~\cfrac{-7-\frac{1}{2}}{2} \right)\implies \left( \cfrac{-5}{2}~,~\cfrac{~\frac{-14-1}{2}~}{2} \right)\implies \left( \cfrac{-5}{2}~,~\cfrac{~\frac{-15}{2}~}{2} \right)&#10;\\\\\\&#10;\left( \cfrac{-5}{2}~,~\cfrac{-15}{4} \right)\implies \left( -2\frac{1}{2}~,~-3\frac{3}{4} \right)

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4 years ago
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