How to calculate the circumference

1 answer:

3 0

The circumference of a circle can be found by multiplying pi ( π = 3.14 ) by the diameter of the circle.
If a circle has a diameter of 4, its circumference is 3.14*4=12.56.
If you know the radius, the diameter is twice as large.

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If the lines a and b in the diagram below are parallelm why do angles 2 and 6 measure 60 degree angles

Bas_tet [7]

Angles 2 and 6 are corresponding angles, which would explain why they are congruent if lines a and b are parallel, but there is nothing in the problem to explain that any of the angles equals 60 degrees since no measurements were given in the diagram.

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In her last basketball game, Carla scored 46 points. In the current game, she has scored 24 points so far. How many more two-poi

KatRina [158]

Answer:

Carla needs to make at least 11 two-pointer shots in the surrent game

Step-by-step explanation:

The first thing we can do is to find the difference between the number of points that Carla scored in her first game and her second game.

This will be 46 - 24 = 22 points difference

Carla needs to make a certain number of two-pointers to get at least the same score she had in her previous game.

We can get this number of two-pointers that needed to be made by dividing the difference in scores by 2

i.e number of two-pointer shots = 22/2 =11 shots

Therefore, Carla needs to make at least 11 two-pointer shots to be able to get the same score in her current game.

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3 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$