if we take 260 to be the 100%, what is 52 off of it as a percentage?
The perimeter of a rectangular field is 322 m.
1 answer:
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FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE.
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2
I think you meant to say
(as opposed to <em>x</em> approaching 2)
Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:
Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2: