Answer:
D : 510 units
Step-by-step explanation:
NOTE:
Something to consider when solving problems like this is to break the large shape down into smaller, more managable shapes. So for this problem, you can break down this irregular shape into two rectangles. This will make solving problems similar to this easier in the future :)
WORK:
I broke down this shape into two rectangles with the following dimensions:
- 12 meters by 5 meters
- 3 meters by 14 meters
You also know that the depth has to be 5 feet (the problem itself did not account for differences in feet and meters, as when I converted the 5 feet to meters and solved that way, none of the answers were correct)
Using this information, you can now solve for the volume of each of the rectangles
12*5*5 = 300 units
3*14*5 = 210 units
Then, you simply add the two volumes together to find the total volume needed to fill the pool which equals
510 units
Answer:
27 balls
Step-by-step explanation:
If Dianna and Becky stopped 9 out of 10 shots made against them.
It means they allowed in 1 out of 10 shots made against them.
This is best solved using ratios
Proportion of Shots Stopped : Proportion of Shots allowed=9:1
If the other team scores 3 points, it means the number of shots allowed =3
Let x be the number of shots stopped
Number of Shots Stopped : Number of Shots allowed=x:3
Therefore:
9:1=x:3

Cross multiplying
x=9 X 3 =27
Dianna and Becky stopped 27 balls from going into the net,
If $.72 per 1lb, then $x per 2.7lb, by using proportional property:
.72/1 = x/2.7 cross multiply
x = (.72)*(2.7) = $1.94
First, find the slope of the initial line;
m=(y₁-y₂)/(x₁-x₂)
m=(4-0)/(11-8)
m=4/3
If two lines are parallel, the slope is the same although the intercept can differ.
If this parallel lines passes through the point (4, 5);
y=(4/3)x+c <-- plug in coordinates to solve for c
5=(4/3)(4)+c
(15/3)=16/3+c
(15/3)-(16/3)=c
-1/3=c
Therefore, the final equation should be y=4/3x-1/3
Hope I Helped :)
He would need to find the area of the room (area of the rectangle) by multiplying the length of the room by the width of the room.