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kaheart [24]
2 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]2 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]2 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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Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.
Rashid [163]

Answer:

The following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Step-by-step explanation:

Lets solve all the expressions to match the results.

  • 5\left(2t+1\right)+\left(-7t+28\right)

<em>Solving the expression</em>

5\left(2t+1\right)+\left(-7t+28\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

5\left(2t+1\right)-7t+28

10t+5-7t+28

3t+33

Therefore, 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33

  • 3\left(3t-4\right)-\left(2t+10\right)

<em>Solving the expression</em>

3\left(3t-4\right)-\left(2t+10\right)

9t-12-\left(2t+10\right)

9t-12-2t-10

7t-22

Therefore, 3\left(3t-4\right)-\left(2t+10\right) = 7t-22

  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

<em>Solving the expression</em>

\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)

4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)

4t-\frac{8}{5}-3+\frac{4t}{3}

As

-3-\frac{8}{5}:\quad -\frac{23}{5}    and  \frac{4t}{3}+4t:\quad \frac{16t}{3}

So,

4t-\frac{8}{5}-3+\frac{4t}{3} will become \frac{16t}{3}-\frac{23}{5}

Therefore, \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}

  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

<em>Solving the expression</em>

\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

-\frac{9}{2}t+3+\frac{7}{4}t+33

\mathrm{Group\:like\:terms}

\frac{9}{2}t+\frac{7}{4}t+3+33

\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t

-\frac{11}{4}t+3+33

-\frac{11}{4}t+36

Therefore, \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Thus, the following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

#learnwithBrainly

5 0
3 years ago
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