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hodyreva [135]
2 years ago
6

A dog and a cat are 200 meters apart when they see each other. The dog can run at a speed of 30 m/sec, while the cat can run at

a speed of 24 m/sec. How soon will the cat catch the dog
Mathematics
1 answer:
Rzqust [24]2 years ago
7 0

Answer:

A dog and a cat are 200 meters apart when they see each other. The dog can run at a speed of 30 m/sec, while the cat can run at a speed of 24 m/sec. How soon will they meet if they simultaneously start running towards each other?

Let x =distance travelled by dog and 200-x be the distance travelled by cat. Inorder to meet, the sum of their travelled distances shall be equal to 200 meters as they are far apart by 200 meters.

Dog: Distance =Speed x time: x=30t

Cat: Distance = 200-x=24t

The sum of their distance travelled shall be 200 meters: 30t+ 24t= 200;54t=200;t=200/54=3.7secs

So they will meet after 3.7 seconds

To check: Distance travelled by dog:x=30(3.7)=111.11 meters

Distance travelled by cat: 24t=24(3.7)=88.89 meters

Therefore since they are travelling towards each other and their distance apart is 200 meters, after 3.7 seconds they will meet because dog travelled 111.11 meters and cat travelled 88.89 meters and they covered already 200 meters as the sum of 111.11+ 88.89=200 meters

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the area of a rectangular window is 6216 cm^2. if the length of the window is 84 cm, what is it's width?​
Rus_ich [418]

Answer:

<h3>The answer is 74 cm</h3>

Step-by-step explanation:

The area of a rectangle = length × width

Since we are finding the width

We have

width  = \frac{area}{length}  \\

From the question

area = 6216 cm²

length = 84 cm

The width is

width =  \frac{6216}{84}  \\

We have the final answer as

<h3>74 cm</h3>

Hope this helps you

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