Jeannine needs to decide what size to make a rectangular
garden in her yard. The dimensions must be natural numbers.
Jeannine wants the perimeter of her Chapter Reference
garden to be 50 dm. She wants the
width to be an even number of decimeters. How many
different combinations are possible? (Length is always longer than or equal to width.)

Question

11-07-2021
Mathematics

contestada

Jeannine needs to decide what size to make a rectangular
garden in her yard. The dimensions must be natural numbers.
Jeannine wants the perimeter of her Chapter Reference
garden to be 50 dm. She wants the
width to be an even number of decimeters. How many
different combinations are possible? (Length is always longer than or equal to width.)

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lublana lublana · Answer 1 · 2021-07-16T08:48:25+02:00

Answer:

Total number of possible combinations are 6

Length width

23 dm 2m

21 dm 4 dm

19 dm 6 dm

17 dm 8 dm

15 dm 10 dm

13 dm 12 dm

Step-by-step explanation:

We are given that

Perimeter of rectangular garden=50 dm

Width is even number.

Length is always longer than or equal to width.

Let length of rectangular garden=x

Width of rectangular garden=y

We have to find the possible number of combinations .

Perimeter of rectangular garden=[tex]2(x+y)[/tex]

[tex]2(x+y)=50[/tex]

[tex]x+y=50/2[/tex]

[tex]x+y=25[/tex]

If y=2 dm

x=25-2=23 dm

If y=4 dm

x=25-4=21 dm

If y=6 dm

x=25-6=19 dm

If y=8 dm

x=25-8=17 dm

If y=10 dm

x=25-10=15 dm

If y=12 dm

x=25-12=13 dm

If y=14 dm

x=25-14=11 dm

x<y

It is not possible

Then, possible combinations are 6