Answer:

Step-by-step explanation:
Volume of cone = 
Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius
=
=
Surface area of cone including 1 base = 
Since r = h
So, area = 
=
= 
Ratio of volume of cone to its surface area including base :



Rationalizing


Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is 
Answer:
<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>
Step-by-step explanation:
Pratap Puri rowed 10 miles down a river in 2 hours, but the return trip took him 2.5 hours.
We know that, 
So, the <u>speed of Pratap with the current</u> will be:
miles/hour
and the <u>speed of Pratap against the current</u> will be:
miles/hour.
Suppose, the rate of Pratap in still water is
and the rate of current is
.
So, the equations will be........

Adding equation (1) and (2) , we will get......

Now, plugging this
into equation (1), we will get.....

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.
Answer:
x=75
Step-by-step explanation:
The triangle is isosceles, meaning that the angles V and U are congruent
Answer:
723 students participated in the event
Explanations:
The total number of students in Hanley's school = 1000
0.723 of the students participated in an event
Number of students that participated in the event = 0.723 x 1000
Number of students that participated in the event = 723