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Elena L [17]
3 years ago
7

A breeder of horses wants to fence two adjacent rectangle grazing areas along a river with 600 meters of fence.

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
3 0
"Adjacent grazing areas" to me means that they share a fence.  In other words, it's just one big area with a fence in the middle.  I'm going to consider that fence in the middle a width.  The amount of fence we would need is the perimeter (2L + 2W) plus the middle fence (so 2L + 3W instead).  The area of such an area would be L times W
2L + 3w = 600A = LW2L = 600 - 3WL = 300 - (3/2)W
A=(300-1.5W)W=W(300-1.5W)
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Find the area of a triangle whose base is 10 mm, and its height is 15 mm.
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Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same poi
skelet666 [1.2K]

Answer:

Alan: 200 m/min

Brian: 150 m/min

Step-by-step explanation:    

Given : Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same point. Alan races around the track in 45 seconds before Brian does and overtakes him every 9 minutes.

To find : What are their rates, in meters per minute?

Solution :

Let n represent the number of laps that Alan completes in 9 minutes.

Then n-1 is the number of laps Brian completes.

45 seconds = 3/4 minutes.

The difference in their lap times in minutes per lap is

\frac{9}{(n -1)}-\frac{9}{n} =\frac{3}{4}

Solving the equation we get,

\frac{9n-9n+9}{n(n -1)} =\frac{3}{4}

9\times 4=3\times n(n -1)

36=3\times n(n -1)

n^2-n-12=0

n^2-4n+3n-12=0

n(n-4)+3(n-4)=0

(n-4)(n+3)=0

n=4,-3

Neglecting n=-3

So, n=4

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S=\frac{D}{T}

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S_a=200 m/min

Brian completes 3 laps in that 9-minute time, so his rate is

S_b=\frac{3\times 450}{9}

S_b=150 m/min

Therefore, Alan: 200 m/min

Brian: 150 m/min

4 0
4 years ago
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