A family swimming pool holds 60m2 of water. The function below gives the cubic
1 answer:
Answer:
58.74 cubic meters
Step-by-step explanation:
The function is actually
Here x is the number of days and the corresponding value of "f(x)" would be the water REMAINING after x days.
Since, we want to figure out how much water (from 60 m^3) is remaining after 7 days, we simply have to plug in x = 7 into the function. Whatever number we get, is the AMOUNT OF WATER REMAINING AFTER 7 DAYS.
So,
Thus, the amount of water remaining in the pool after 7 days is 58.74 cubic meters
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For this case we have the following equations:
x + y = 3
y = 2x - 15
Whose solution is given by:
x = 6
y = -3
Or equivalently the ordered pair:
(x, y) = (6, -3)
Answer:
(x, y) = (6, -3)
See attached image
x + y = 3
y = 2x - 15
Whose solution is given by:
x = 6
y = -3
Or equivalently the ordered pair:
(x, y) = (6, -3)
Answer:
(x, y) = (6, -3)
See attached image
Answer: The domain is in the picture
Step-by-step explanation:
Let
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²