<h3>Given</h3>
- room height is x feet
- room length is 3x feet
- room width is 3x feet
- a door 3 ft wide by 7 ft tall
<h3>Find</h3>
- The net area of the wall, excluding the door
<h3>Solution</h3>
The area of the wall, including the door, is the room perimeter multiplied by the height of the room. The room perimeter is the sum of the lengths of the four walls.
... gross wall area = (3x +3x +3x +3x)·x = 12x²
The area of the door is the product of its height and width.
... door area = (7 t)×(3 ft) = 21 ft²
Then the net wall area, exclusive of the door is ...
... net wall area = gross wall area - door area
... net wall area = 12x² -21 . . . . square feet
Isabella drove 600m.
Mia drove 300 more than Isabella=(600+300)m
=900m.
David drove=8.5km=8500m
Therefore, Distance David drove in meters combined of both distances of Mia and Isabella =[8500-(600+900)]m
=(8500-1500)
=7000 meters or 7 km.
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Answer:
B
Step-by-step explanation:
The SA of a cone is A=pi*r^2+pi*r*s
So instead of pi*5^2, it should be pi*3*5 (5 is the length of the slant).
Answer:
Yes, he earned a Varsity letter because he played in 61% of the games.
Step-by-step explanation:
To know the percent of games he played out of the total, we can do it by dividing the games he played by the total of games and multiply this by 100% to get the percent of games he played:

We solve:
0.611111*100% = 61.11%
So we know that he earned a letter because he played in 61.11% of the games, more than he needed.