Hope I was able to help :-)
First, solve for the volume (V) of each cylinder by the equation,
V = πd²h / 4
where d and h are diameter and height, respectively. Substituting the given values,
V = π(6.19 cm)² x (6 cm) / 4 = 180.56 cm³
To determine the number of cylinders needed, divide the volume of the solid block by the volume of the cylinder,
n = 360 cm³ / 180.56 cm³ = 1.99
Thus, approximately 2 cylinders are needed to mold one block of gold.
Answer: Step-by-step explanation:
What is 5.56666666667 as a fraction?
To write 5.56666666667 as a fraction you have to write 5.56666666667 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
5.56666666667 = 5.56666666667/1 = 55.6666666667/10 = 556.666666667/100 = 5566.66666667/1000 = 55666.6666667/10000 = 556666.666667/100000 = 5566666.66667/1000000 = 55666666.6667/10000000 = 556666666.667/100000000 = 5566666666.67/1000000000 = 55666666666.7/10000000000 = 556666666667/100000000000
And finally we have:
5.56666666667 as a fraction equals 556666666667/100000000000
Answer:
24
Step-by-step explanation:
Four total places someone can come - 1st, 2nd, 3rd or 4th. Anyone can come 1st - 4 people
Now only the remaining 3 can come second - 3 people
Next, only the remaining 2 people can come third - 2 people
Finally, one person comes last - 1 person
4 x 3 x 2 x 1 = 24
Answer:
1.60 
Step-by-step explanation:
Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.
i.e Area = 
So that,
for the inner part of the pipe,
r = 
= 
= 2.5 feet
Area of the inner part of the pipe =
= 3.14 x 
= 19.625
Are of the inner part of the pipe is 19.63 .
Total diameter of the pipe = 5 + 0.2
= 5.2 feet
r = 
= 2.6 feet
Area of the pipe =
= 3.14 x 
= 21.2264
Area of the pipe is 21.23 .
Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe
= 21.2264 - 19.625
= 1.6014
The cross sectional area of the cement pipe is 1.60 .
