Answer:
Step-by-step explanation:
<u>Volume of a cylinder:</u>
<u>Container A</u>
- r = 10 ft, h = 11 ft
- V = π*10²*11 = 1100π ft
<u>Container B</u>
- r = 7 ft, h = 19 ft
- V = π*7²*19 = 931π ft²
<u>After pump out the container A has the level of:</u>
<u>Level % in the container A:</u>
- 169π/1100π×100% = 15.4% rounded to the nearest tenth
Answer:
Yes, 1 kg of paint is enough to completely paint the inside and outside surfaces of the vessel.
Step-by-step explanation:
For the given vessel without a lid,
diameter = 0.4 x 1.5
= 0.6 m
radius = =
= 0.3 m
Total external surface area of the vessel = 2rh +
= [2 x x 0.3 x 1.5] + [ x ]
= 2.8286 + 0.2829
= 3.1112
Total external surface area of the vessel is 3.1112 square meter.
Total internal and external surface area = 2 x 3.1112
= 6.2224
But for 1 square meter, 150 g of paint is consumed. Thus;
for 6.2224 = 6.2224 x 150 g
= 933.36 g
Thus, 6.2224 of area would require 933.36 g of paint.
So that,
= 0.93336 kg
Therefore, 1 kg of paint is enough to completely paint the inside and outside surfaces of the vessel.
Answer:
9) -25/21
11) 14
Step-by-step explanation:
In each case, subtract the y-coordinates. Subtract the x-coordinates in the same order. Divide the difference in y by the difference in x.
9)
y-coordinates: -10 - 15 = -25
x-coordinates = 6 - (-15) = 21
slope = (difference in y)/(difference in x) = -25/21
11)
y-coordinates: -20 - 8 = -28
x-coordinates = 3 - 5 = -2
slope = (difference in y)/(difference in x) = -28/(-2) = 14
X = - b/2a
Solving for a,
a = -b/2x
If b is the numerator, the denominator is -2x
The second option is correct.
C=2πr and we are told that C=28π so
2πr=28π
r=14m (assuming that 3.14 was a weak approximation of π)