Apainter is painting a wall with an area of 150 ft2. he decides to paint half of the wall and then take a break. after his break, he paints half of the remaining unpainted portion and then takes another break. if he continues to paint half of the remaining unpainted portion between breaks, approximately what portion of the original wall will be painted when he takes his fifth break?
112.50 ft2
145.31 ft2
147.66 ft2
290.63 ft2

The answer is 141.35 ft²

Before the first break, it was painted:
150 ft² ÷ 2 = 75 ft²
Now it's left:
150 ft² - 75 ft² = 75 ft²

Before the second break, it was painted:
75 ft² ÷ 2 = 37.5 ft²
Now it's left:
75 ft² - 37.5 ft² = 37.5 ft²

Before the third break, it was painted:
37.5 ft² ÷ 2 = 18.75 ft²
Now it's left:
37.5 ft² - 18.75 ft² = 18.75 ft²

Before the fourth break, it was painted:
18.75 ft² ÷ 2 = 9.375 ft²
Now it's left:
18.75 ft² - 9.375 ft² = 9.375 ft²

Before the fourth break, it was painted:
9.375 ft² ÷ 2 = 4.6875 ft²
Now it's left:
9.375 ft² - 4.6875 ft² = 4.6875 ft²

Now, we will sum what he painted for now:
75 ft² + 37.5 ft² + 18.75 ft² + 9.375 ft² 4.6875 ft² = 141.3125 ft² ≈ 141.35 ft²

When the painter takes his fifth break, there will be 141.35 ft² of the wall painted.

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)