Answer:
the answer is 58
Step-by-step explanation:
Answer:
\frac{9\sqrt{3}}{2}\approx 7.79 square units.
Step-by-step explanation:
320 miles in 6 hours
Answer:
53 1/3 mph.
I divided 320 by 6 and got this answer.
This solution represents a unit rate problem in Mathematics. The problem is finding how many miles are covered in one hour when 320 miles are covered in 6 hours. The answer, calculated by dividing the distance (320 miles) by the time (6 hours), is around 53.33 miles/hour.
The question is asking you to find the unit rate for the relationship given, which is 320 miles in 6 hours. A unit rate tells you how many of the first quantity correspond to one unit of the second quantity. In this case, to find the unit rate, we need to find out how many miles are covered in one hour.
To do this, we divide the total distance by the total time. So, 320 miles/6 hours = 53.33 miles/hour. This means that for every hour, approximately 53.33 miles are being covered. This is your unit rate for this relationship.
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2. two bars of soap at RS.25 each
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Answer:
The maximum height that the ball reaches is sf = 31 feet from the ground
Step-by-step explanation:
Solution:-
- A ball is thrown vertically upwards from an initial elevation of si = 6 feet from the ground.
- The velocity with which the ball was thrown up, vi = 40 ft/s
- We can determine the maximum height of the ball when it is thrown vertically up by using the 3rd kinematic equation of motion.
vf^2 - vi^2 = 2*g*( sf - si )
Where,
vf : The final velocity of the ball, for maximum height it is = 0
sf : The final height of the ball from the ground
g : Gravitational constant = -32 ft/s^2
0 - 40^2 = -2*32*( sf - 6 )
sf - 6 = 25
sf = 31 feet
- The maximum height that the ball reaches is sf = 31 feet from the ground.
Answer:
The horizontal distance from the starting point when rocket lands is:
168.83 meters.
Step-by-step explanation:
We are given a function that models the height of the rocket when it is 'x' meters away from the starting point.
The function is:
Now we are asked to find the horizontal distance from the starting point when the rocket will land.
i.e. we are asked to find 'x' ; when y=0
i.e.
Let y=0
on solving the quadratic equation we obtain:
As the distance can't be negative.
Hence, the viable solution is:
x=168.83