The circumference of a circular garden is approximately 30 feet. What is the approximate area inside the circular garden?
1 answer:
Answer:
or 71.6197
Step-by-step explanation:
This just manipulating formulas.
The formula for the circumference of a circle is , and we know this is equal to 30. That means
. The formula for the area of a circle is
, so we know the area of the circle must be
(
, and we found
to be 15 earlier).
Now, using this information, let's find r in terms of .
Now let's plug in r to the expression we found for area.
The area of the circle is approximately or 71.6197.
Edit- Apparently there is a formula for this:
where
is the circumference of the circle.
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Record of june temperature of last 50 years
Probability of June temperature being above 80° F from last 50 years
It means , out of 30 days ,there are only 20 days , in which temperature is above 80°F.
⇒Probability that it will be above 80° F for 20 or more days this June
where, x is the number of extra days after 20 days in month of june, when temperature is above 80°F.
Value of this Probability will be in between
Now, what Azul wants
To find the probability that it will be above 80° F for 20 or more days this June.
The Simulation given is
⇒There are 30 days in June. If we use a number cube with the numbers 1-6 and assign 1-4 as a temperature above 80° F and 5-6 as a temperature 80° F or below.
Total number of numbers on number cube ={1,2,3,4,5,6}
Favorable Outcome={1,2,3,4}=Temperature above 80° F
Temperature 80° F or below ={5,6}
⇒Probability of getting number {1,2,3,4} on the number cube when rolled once,that is Temperature above 80° F
⇒Probability of getting number {5,6} on the number cube when rolled once, that is Temperature 80° F or below
This is Incorrect Simulation, as there are chances that , there can be more than 20 days out of 30 days when Temperature goes above 80° F.But Here we have got Probability of temperature being above 80° F for 20 days only, not more than 20 days, which is
Not,a number
.
<h3>Explanation:</h3>
1. "Create your own circle on a complex plane."
The equation of a circle in the complex plane can be written a number of ways. For center c (a complex number) and radius r (a positive real number), one formula is ...
|z-c| = r
If we let c = 2+i and r = 5, the equation becomes ...
|z -(2+i)| = 5
For z = x + yi and |z| = √(x² +y²), this equation is equivalent to the Cartesian coordinate equation ...
(x -2)² +(y -1)² = 5²
__
2. "Choose two end points of a diameter to prove the diameter and radius of the circle."
We don't know what "prove the diameter and radius" means. We can show that the chosen end points z₁ and z₂ are 10 units apart, and their midpoint is the center of the circle c.
For the end points of a diameter, we choose ...
- z₁ = 5 +5i
- z₂ = -1 -3i
The distance between these is ...
|z₂ -z₁| = |(-1-5) +(-3-5)i| = |-6 -8i|
= √((-6)² +(-8)²) = √100
|z₂ -z₁| = 10 . . . . . . the diameter of a circle of radius 5
The midpoint of these two point should be the center of the circle.
(z₁ +z₂)/2 = ((5 -1) +(5 -3)i)/2 = (4 +2i)/2 = 2 +i
(z₁ +z₂)/2 = c . . . . . the center of the circle is the midpoint of the diameter
__₁₂₃₄
3. "Show how to determine the center of the circle."
As with any circle, the center is the <em>midpoint of any diameter</em> (demonstrated in question 2). It is also the point of intersection of the perpendicular bisectors of any chords, and it is equidistant from any points on the circle.
Any of these relations can be used to find the circle center, depending on the information you start with.
As an example. we can choose another point we know to be on the circle:
z₄ = 6-2i
Using this point and the z₁ and z₂ above, we can write three equations in the "unknown" circle center (a +bi):
- |z₁ - (a+bi)| = r
- |z₂ - (a+bi)| = r
- |z₄ - (a+bi)| = r
Using the formula for the square of the magnitude of a complex number, this becomes ...
(5-a)² +(5-b)² = r² = 25 -10a +a² +25 -10b +b²
(-1-a)² +(-3-b)² = r² = 1 +2a +a² +9 +6b +b²
(6-a)² +(-2-b)² = r² = 36 -12a +a² +4 +4b +b²
Subtracting the first two equations from the third gives two linear equations in a and b:
11 -2a -21 +14b = 0
35 -14a -5 -2b = 0
Rearranging these to standard form, we get
a -7b = -5
7a +b = 15
Solving these by your favorite method gives ...
a +bi = 2 +i = c . . . . the center of the circle
__
4. "Choose two points, one on the circle and the other not on the circle. Show, mathematically, how to determine whether or not the point is on the circle."
The points we choose are ...
- z₃ = 3 -2i
- z₄ = 6 -2i
We can show whether or not these are on the circle by seeing if they satisfy the equation of the circle.
|z -c| = 5
For z₃: |(3 -2i) -(2 +i)| = √((3-2)² +(-2-i)²) = √(1+9) = √10 ≠ 5 . . . NOT on circle
For z₄: |(6 -2i) -(2 +i)| = √((6 -2)² +(2 -i)²) = √(16 +9) = √25 = 5 . . . IS on circle