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Angelina_Jolie [31]
3 years ago
10

A circle has an area of 3.14 cm^2 . What is its radius?

Mathematics
2 answers:
Anni [7]3 years ago
6 0

Answer:

Cant be determined

Step-by-step explanation:

You would need the diameter to figure this out

Mice21 [21]3 years ago
3 0

The radius of the circle is 1 cm, if the circle has an area of 3.14 cm^{2}.

Step-by-step explanation:

         The given is,

                  Area of the circle is 3.14 cm^{2},

Step:1

        The formula for Area of circle,

                    A = πr^{2}....................(1)

       where,

                   π - 3.14

                    r -  Radius of circle

Step:2

         From the given,

               Area, A = 3.14 cm^{2}

        Eqn (1) becomes,

                  3.14 = 3.14 × r^{2}                (∵ π = 3.14 )

                     \frac{3.14}{3.14} = r^{2}

                     r^{2} = 1

                    r = \sqrt{1}

                      = 1

                    r = 1 cm

           Radius of circle, r = 1 cm

Result:

      Thus the radius of the circle is 1 cm, if the area of circle is 3.14 cm^{2}.

       

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Answer:

Kayla's reasoning is not correct.

Step-by-step explanation:

The locker problem is as follows:

Imagine you are at a school that has student lockers. There are 200 lockers, all shut and unlocked, and 200 students. Suppose the first student goes along the row and opens every locker. The second student then goes along and shuts every other locker beginning with number 2. The third student changes the state of every third locker beginning with number 3. (If the locker is open the student shuts it, and if the locker is closed the student opens it). The fourth student changes the state of every fourth locker beginning with number 4. Imagine that this continues until the 200 students have followed the pattern with the 200 lockers. At the end, which lockers will be open and which will be closed? Why?

Solution:

So from the information we know that the first student goes along the row and opens every locker.

Then the second student shuts every other locker, i.e. locker numbers 2, 4, 6, 8, 10, ..., 196, 198 and 200.

Then the third students changes the state of every third locker, i.e. he/she closes an open locker and opens a closed locker.

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Then the fourth students changes the state of every fourth locker.

So the open lockers are: 1, 4, 5, 6, 8,....

So, on we will observe that the open lockers have a perfect square number such as, 1, 4, 9, 16,....

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This is because the square numbers always have an odd number of factors, which leads them to be open at the end.

Take any locker number, 40, for example. Its state (open or closed) is changed for every student whose number in line is a factor of the locker number.

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     1                    Open

     2                   Close

     4                   Open

     5                   Close

     8                   Open

     10                  Close

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    40                  Close

Like all other lockers numbered with non-square numbers, it ends up closed after all the students have gone through the line because it has an even number of factors.

Consider the locker number 16:

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Thus, we can conclude that all the doors with square numbers on them will remain open because all square numbers have an odd number of factors and the doors with non-square numbers on them will remain close because they have even number of factors.

There will be a total of 14 lockers open.

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So, if there are 10 lockers open in the first 100 lockers there must be only 4 other lockers opened in the next 100.

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