Answer:
Area of the shaded part is 3.14 square unit.
Step-by-step explanation:
Area of the shaded part = Area of large semicircle - (Area of two small semi circles)
Area of large semicircle with center O = 
= 
= 2π
Area of semicircle with center O' = 
= 
Area of semicircle with center O" =
Now substitute these values in the formula,
Area of shaded part = 
= 
= 3.14 square unit
Area of the shaded part is 3.14 square unit.
We can multiply the number of parts Marsha has finished by the miles of each part:
5 × 1/10
= 5/10
=1/2
Therefore, she raced 1/2 miles.
Hope it helps!
Answer:
Jon lost a total of 3 kilograms in these 3 months.
Step-by-step explanation:
Weight that Jon gained in December = 2.2 kilograms
Weight that Jon lost in January = 1.5 kilograms
Weight that Jon lost in February = 3.7 kilograms
Overall Change in his weight = Total Weight he gained - Total weight he Lost
Total weight he gained is 2.2 kilograms as he only gained weight in December. Total weight he lost will be sum of weights he lost in January and February.
So, total weight Jon lost = 1.5 + 3.7 = 5.2 kilograms
Thus,
Total change in Jon's Weight = 2.2 - 5.2 = -3.0 kilograms
This shows that Jon lost a total of 3 kilograms in these 3 months.
Conclusion:
The statement that best describe the total change in his weight is: Jon lost a total of 3 kilograms in these 3 months
Answer: There will enough to paint the outside of a typical spherical water tower.
Step-by-step explanation:
1. Solve for the radius r from the formula for calculate the volume of a sphere. as following:
![V=\frac{4}{3}r^{3}\pi\\\frac{3V}{4\pi}=r^{3}\\r=\sqrt[3]{\frac{3V}{4\pi}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7Dr%5E%7B3%7D%5Cpi%5C%5C%5Cfrac%7B3V%7D%7B4%5Cpi%7D%3Dr%5E%7B3%7D%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%7D%7D)
2. Substitute values:
![r=\sqrt[3]{\frac{3(66,840.28ft^{3})}{4\pi}}=25.17ft](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%2866%2C840.28ft%5E%7B3%7D%29%7D%7B4%5Cpi%7D%7D%3D25.17ft)
3. Substitute the value of the radius into the equation fo calculate the surface area of a sphere, then you obtain that the surface area of a typical spherical water tower is:
3. If a city has 25 gallons of paint available and one gallon of paint covers 400 square feet of surface area, you must multiply 25 by 400 square feet to know if there will be enough to paint the outside of a typical spherical water tower.
As you can see, there will enough to paint the outside of a typical spherical water tower.
