Serena’s average speed during her training run was 8 kilometers per hour over the 2 hours that she ran. What was the total dista
2 answers:
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Answer:
The fraction is
Option (B) is correct.
and
Step-by-step explanation:
Let n denotes numerator and d denotes denominator of the fraction.
Given : The ratio of the numerator to the denominator of a fraction is 2 to 3.
That is
Cross multiply , we get,
Or, ........(1)
Also, given : . If both the numerator and the denominator are increased by 2, the fraction becomes 3/4
That is
Cross multiply , We get,
.........(2)
.......(3)
Thus, from (1) and (2) , option (B) follows.
Solving equation (1) and (3) to get the original fraction using elimination method,
3n - 2d = 0 ............(1)
and 3d - 4n = 2 .........(3)
Multiply equation (1) by 3 , we get ,
9n - 6d = 0 ..........(4)
Multiply equation (3) by 2 , we get ,
-8n + 6d = 4 ..........(5)
Adding (4) and (5) , we get,
9n - 6d -8n + 6d = 4 + 0
⇒ n = 4
Put n = 4 in (1), we get
3n - 2d = 0 ⇒ 3(4) - 2d = 0 ⇒ 12 -2d = 0 ⇒ 12 = 2d ⇒ d = 6
so, the numerator is 4 and the denominator is 6.
Thus, the fraction is
Here we can first find the area of yard and then subtract the area of vegetable garden to get the required area.
The formula to find area is given by:
Now we find area of yard:
length is 25 ft and width is 15 ft.
Area of yard = 375 ft²
Area of vegetable garden:
length = 8ft and width =8ft
So the vegetable garden is a square.
Area = 8*8
Area of vegetable garden =64 ft²
Area of backyard to be laid with grass = Area of yard-Area of vegetable garden
Area required = 375 -64 = 311 ft²
Answer : The area of backyard in which grass is to be laid is 311 square feet.
The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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