Find the circumference of the circle. use 3.14 for 39 cm

Mathematics

2 answers:

sdas [7]3 years ago

8 0

Answer:

circumference (c) , radius (r)

Step-by-step explanation:

c=2 pi r

c=2*3.14 *39

c= 244.92cm

In-s [12.5K]3 years ago

7 0

Answer:

The circumference is about 22.14cm

Step-by-step explanation:

This is because first you take the area and multiply it by pi. Next you take the square root of this number. Now you take this number and multiply it by 2 to get 22.14.

Hope I helped! Have a Great rest of your day! :-)

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Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the gene

nikklg [1K]

Answer:

The general solution of the system is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\$

Step-by-step explanation:

To verify that the system $X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X$ has the solutions given,

First, we determine the Eigen Values β of the matrix of the form $X^{'} =A X$ Using |A-βI|=0, where I is the Identity Matrix, we have:

$|A-\beta I|=0$

$|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0$

Subtracting matrices

$\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0$

Taking the determinant

$(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2$

We determine the eigen vector using $\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$

When $\beta$ = -2,

$\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$ which implies that $3c_{11}+3c_{12}=0$ and thus

$If c_{11}=1, c_{12}=-1$

When $\beta$ = 4,

$\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0$ which implies that $3c_{21}-3c_{22}=0$ and thus