Would this be a cone??
2 answers:
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First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:
To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:
We have our answer: approximately 0.14 cm or 1.4 mm per year
To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:
We have our answer: approximately 0.14 cm or 1.4 mm per year
Given :
Length of parade , .
Breadth of parade , .
5 people can fit into a area.
1 mile = 5280 ft .
To Find :
The size of the crowd.
Solution :
Area of parade ,
Also , 5 people can fit in area .
So , number of people fit in A is :
Therefore , 95040 can fit in the parade .
Hence , this the required solution .