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kaheart [24]
3 years ago
14

Justin has 20 pencils 25 erasers and 40 paper clip he organized them into groups with the same number of item of each group. all

the items in a group will be the same type. how many items can he put in each group?
Mathematics
2 answers:
Kaylis [27]3 years ago
7 0

Justin can put 4 pencils, 5 erasers, and 8 paperclips into 5 different bags. If he hands out the bags, 5 different recipients will have the same things in each bag.                             20/5=4.                                                    25/5=5.                                                    40/5=8.

4vir4ik [10]3 years ago
4 0

Pencils:  

20 /5=4


Erasers:  

25 /5=5


Paper clips:  

40 /5=8


Justin can put 4 pencils, 5 erasers, and 8 paperclips into 5 different bags. If he hands out the bags, 5 different recipients will have the same things in each bag.

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3 years ago
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8 0
4 years ago
Calculate the perimeter of a rectangle which is eleven metres long and four metres wide.​
Leviafan [203]

Step-by-step explanation:

the formulae to find the perimeter of a rectangle is p = 2(l+w)

<em>where</em>

p= perimeter

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6 0
3 years ago
Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?
Vsevolod [243]
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so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ \square }}\quad ,&{{ \square }})\quad &#10;%  (c,d)&#10;&({{ \square }}\quad ,&{{ \square }})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


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Answer:

C

Step-by-step explanation:

8 0
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Read 2 more answers
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