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2 years ago

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36 fifth graders are on a team with the same number on students on each team, if there are more then then 1 but less then 20 tea

ms, how many students can be on each team?

1 answer:

frosja888 [35]2 years ago

7 0

Depending on the number of teams, the possible numbers of students per team are in the set:

{2, 3, 4, 6, 9, 12, 18}

<h3>How many students can be on each team?</h3>

We know that there are 36 students divided equally into N teams.

Such that the number of teams is larger than 1 and smaller than 20.

Then we could have 2 teams, such that the number of students in each team is given by the quotient between the total number of students and the number of teams.

36/2 = 18

There are 18 students in each team.

Notice that the numbers of teams can only be factors of 36, where:

36 = 6*6 = 2*2*3*3

So the factors are:

2, 3, 2*3, 3*3, etc...

If there are 3 teams we have:

36/3 = 12 students per team.

If there are 6 teams we have:

36/6 = 6 students per team.

if there are 9 teams:

36/9 = 4 students per team.

If there are 12 teams:

36/12 = 3 students per team.

If there are 18 teams:

36/18 = 2 students per team.

These are all the possibilities.

If you want to learn more about factors:

brainly.com/question/219464

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Read 2 more answers

Please help me with this

denis23 [38]

Answer:

20) $\displaystyle [4, 1]$

19) $\displaystyle [-5, 1]$

18) $\displaystyle [3, 2]$

17) $\displaystyle [-2, 1]$

16) $\displaystyle [7, 6]$

15) $\displaystyle [-3, 2]$

14) $\displaystyle [-3, -2]$

13) $\displaystyle NO\:SOLUTION$

12) $\displaystyle [-4, -1]$

11) $\displaystyle [7, -2]$

Step-by-step explanation:

20) {−2x - y = −9

{5x - 2y = 18

⅖[5x - 2y = 18]

{−2x - y = −9

{2x - ⅘y = 7⅕ >> New Equation

__________

$\displaystyle \frac{-1\frac{4}{5}y}{-1\frac{4}{5}} = \frac{-1\frac{4}{5}}{-1\frac{4}{5}}$

$\displaystyle y = 1$ [Plug this back into both equations above to get the x-coordinate of 4]; $\displaystyle 4 = x$

_______________________________________________

19) {−5x - 8y = 17

{2x - 7y = −17

−⅞[−5x - 8y = 17]

{4⅜x + 7y = −14⅞ >> New Equation

{2x - 7y = −17

_____________

$\displaystyle \frac{6\frac{3}{8}x}{6\frac{3}{8}} = \frac{-31\frac{7}{8}}{6\frac{3}{8}}$

$\displaystyle x = -5$ [Plug this back into both equations above to get the y-coordinate of 1]; $\displaystyle 1 = y$

_______________________________________________

18) {−2x + 6y = 6

{−7x + 8y = −5

−¾[−7x + 8y = −5]

{−2x + 6y = 6

{5¼x - 6y = 3¾ >> New Equation

____________

$\displaystyle \frac{3\frac{1}{4}x}{3\frac{1}{4}} = \frac{9\frac{3}{4}}{3\frac{1}{4}}$

$\displaystyle x = 3$ [Plug this back into both equations above to get the y-coordinate of 2]; $\displaystyle 2 = y$

_______________________________________________

17) {−3x - 4y = 2

{3x + 3y = −3

__________

$\displaystyle \frac{-y}{-1} = \frac{-1}{-1}$

$\displaystyle y = 1$ [Plug this back into both equations above to get the x-coordinate of −2]; $\displaystyle -2 = x$

_______________________________________________

16) {2x + y = 20

{6x - 5y = 12

−⅓[6x - 5y = 12]

{2x + y = 20

{−2x + 1⅔y = −4 >> New Equation

____________

$\displaystyle \frac{2\frac{2}{3}y}{2\frac{2}{3}} = \frac{16}{2\frac{2}{3}}$

$\displaystyle y = 6$ [Plug this back into both equations above to get the x-coordinate of 7]; $\displaystyle 7 = x$

_______________________________________________

15) {6x + 6y = −6

{5x + y = −13

−⅚[6x + 6y = −6]

{−5x - 5y = 5 >> New Equation

{5x + y = −13

_________

$\displaystyle \frac{-4y}{-4} = \frac{-8}{-4}$

$\displaystyle y = 2$ [Plug this back into both equations above to get the x-coordinate of −3]; $\displaystyle -3 = x$

_______________________________________________

14) {−3x + 3y = 3

{−5x + y = 13

−⅓[−3x + 3y = 3]

{x - y = −1 >> New Equation

{−5x + y = 13

_________

$\displaystyle \frac{-4x}{-4} = \frac{12}{-4}$

$\displaystyle x = -3$ [Plug this back into both equations above to get the y-coordinate of −2]; $\displaystyle -2 = y$

_______________________________________________

13) {−3x + 3y = 4

{−x + y = 3

−⅓[−3x + 3y = 4]

{x - y = −1⅓ >> New Equation

{−x + y = 3

________

$\displaystyle 1\frac{2}{3} ≠ 0; NO\:SOLUTION$

_______________________________________________

12) {−3x - 8y = 20

{−5x + y = 19

⅛[−3x - 8y = 20]

{−⅜x - y = 2½ >> New Equation

{−5x + y = 19

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$\displaystyle \frac{-5\frac{3}{8}x}{-5\frac{3}{8}} = \frac{21\frac{1}{2}}{-5\frac{3}{8}}$

$\displaystyle x = -4$ [Plug this back into both equations above to get the y-coordinate of −1]; $\displaystyle -1 = y$

_______________________________________________

11) {x + 3y = 1

{−3x - 3y = −15

___________

$\displaystyle \frac{-2x}{-2} = \frac{-14}{-2}$

$\displaystyle x = 7$ [Plug this back into both equations above to get the y-coordinate of −2]; $\displaystyle -2 = y$

I am delighted to assist you anytime my friend!

7 0

3 years ago