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3 years ago

15

This figure consists of a rectangle and semicircle. What is the perimeter of this figure? Use 3.14 for pi.

2 answers:

Triss [41]3 years ago

8 0

Perimeter of figure = 3 sides of rectangle + perimeter of semi-circle.

3 sides of rectangle = 15 + 12 + 15 = 42

Perimeter of semi-circle = πr = 3.14 * 6 = 18.84

In short, Your Answer would be: 42 + 18.84 = 60.84 m

Hope this helps!

CaHeK987 [17]3 years ago

3 0

Circle:
12m = diameter

Circumference of a circle:
C = d * pi
C = 12m * 3.14
C = 37.68m

Then cut it in half because the whole circle isn't a part of the outside (only half is) You'd get 18.84m.

Rectangle:
Is 12m by 15m.

Circumference of a rectangle:
C = 2L + 2W
C = 2(12m) + 2(15m)
C = 24m + 30m
C = 54m.
C = 54m - 12m = 42m.

Now add the answer of the circle and answer of the rectangle to get 60.84.

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The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

$a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0$

Will have a characteristic equation of the form:

$a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0$

Where solutions $r_1,r_2...,r_n$ are the roots from which the general solution can be found.

For real roots the solution is given by:

$y(t)=c_1e^{r_1t} +c_2e^{r_2t}$

For real repeated roots the solution is given by:

$y(t)=c_1e^{rt} +c_2te^{rt}$

For complex roots the solution is given by:

$y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)$

Where:

$r_1_,_2=\lambda \pm \mu i$

Let's find the solution for $x''+x=0$ using the previous information:

The characteristic equation is:

$r^{2} +1=0$

So, the roots are given by:

$r_1_,_2=0\pm \sqrt{-1} =\pm i$

Therefore, the solution is:

$x(t)=c_1cos(t)+c_2sin(t)$

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of $x(t)$ in order to find the constants $c_1$ and $c_2$ :

$x'(t)=-c_1sin(t)+c_2cos(t)$

Evaluating the initial conditions:

$x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1$

And

$x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2$

Now we have found the value of the constants, the solution of the second-order IVP is:

$x=-cos(t)+2sin(t)$

3 0

3 years ago