Tublu buys a cylindrical water tank of height 1.4m and diameter 1.1m to catch rainwater off his roof. He has a full 2 litres tin of paint in his store and decides to paint the tank (not the base). If he uses 250 ml to cover 1 m2, will he have enough paint to cover the tank with one layer of paint? [take π=3.142] Tublu buys a cylindrical water tank of height 1.4m and diameter 1.1m to catch rainwater off his roof. He has a full 2 litres tin of paint in his store and decides to paint the tank (not the base). If he uses 250 ml to cover 1 m2, will he have enough paint to cover the tank with one layer of paint? [take π=3.142]

Step-by-step explanation:

Height of the cylinder = 1.4 m

diameter of the cylinder = 1.1 m

Tublu has a full 2 litre tin of paint  

Assume that the tank has an open top to allow the capturing of rainwater.

Also, Tublu does not need to paint the base of the tanks.

Tublu needs 250 ml to cover 1 m^2 of the tank body

Only the body of the tank is to be painted, so we find the perimeter of the tank body.

from basic circle mensuration, perimeter of the tank body =  

where d is the diameter of the tank

==> 3.142 x 1.1 = 3.456 m

If we imagine that this perimeter of the tank body is spread out, it will form a rectangle with a height of 1.4 m from the base.

The area of this rectangle that will be formed =  (perimeter of the cylinder body) x ( height of the cylinder)

==> 3.456 x 1.4 = 4.838 m^2

this area is the area of the tank that needs to be painted.

Now we convert the 250 ml to litre,  we'll have

250 ml = 250 x 10^-3 litres = 0.25 litres

Since 250 ml or rather 0.25 litres of paint is needed to cover 1 m^2 area of the tank's body, then, we will need

==> 4.838 x 0.25 = 1.21 litres of paint

Since Tublu has a full 2 litres tin of paint, and he needs 1.21 litres  to cover the body of the tank, then we can say that Tublu has more than enough paint to cover the tank surface in one layer coating.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}