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PIT_PIT [208]
3 years ago
9

A painter 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 brea

k, 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?
Mathematics
2 answers:
kotegsom [21]3 years ago
8 0
145.31 ft2 will be painted on his fifth break. 
lutik1710 [3]3 years ago
4 0

Answer:

145.31 ft2

Step-by-step explanation:

The fraction painted by the painter is an absolute value when the fraction is applied to the total area of the wall.

Total area =  150 ft2

He decides to paint half of the wall and then take a break. This means that he painted

= 1/2 × 150 ft2

=75 ft2

Area left unpainted = 150 ft2 - 175 ft2 = 75 ft2

After his break, he paints half of the remaining unpainted portion, This means that he painted

= 1/2 × 75 ft2

=37.5 ft2

Area left unpainted = 75 ft2 - 37.5 ft2 = 37.5 ft2

At this rate, the portion unpainted after the 5th break can be so determined;

a = 150 where a is the original area unpainted

r = common ratio = 1/2 ( the rate of painting as a function of the area)

Area unpainted after first break = ar = 150 × 1/2 = 75 ft2

Area unpainted after the second break = ar^2 = 150(1/2)^2 = 37.5 ft2

Thus area unpainted after the 5th break

=ar^5 = 150(1/2)^5 = 4.6875

Portion of the original wall will be painted when he takes his fifth break

= 150 - 4.6875

= 145.3125  ft2

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Read 2 more answers
How do I solve this?
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\overline{XS}\cong\overline{YT}\Rightarrow 3m+7=4.2m+5\ \ \ |-7\\\\3m=4.2m-2\ \ \ \ |-4.2m\\\\-1.2m=-2\ \ \ \ |:(-1.2)\\\\m=\dfrac{2}{1.2}\\\\m=\dfrac{20}{12}\\\\m=\dfrac{5}{3}\to m=1\dfrac{2}{3}


\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2

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