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I need help with this one, is by the method of Undetermined Coefficients.

Maybe you can explain to me... 

4y''+9y=15

\(ay''+by=c\)

2 th-order differential equation: \(4y''+9y=15\)

1. characteristic equation:

\(\begin{array}{|rcll|} \hline 4\lambda^2 + 9 &=& 0 \\ 4\lambda^2 &=& -9 \\ \lambda^2 &=& -\frac{9}{4} \\ \lambda^2 &=& \frac{9}{4} \cdot(-1) \\ \lambda &=& \pm \frac{3}{2} i \\ \mathbf{\lambda} & \mathbf{=} & \mathbf{ \underbrace{0}_{=\alpha} \pm \underbrace{\frac{3}{2}}_{=\beta} i } \\ \hline \end{array}\)

2. Homogeneous Equations \(y_h =\ ?\)

\(\large{ \begin{array}{|rcll|} \hline y_h & =& e^{\alpha\cdot x}\Big(c_1\sin(\beta x)+c_2\cos(\beta x) \Big) \quad & | \quad \lambda \text{ complex!} \\ &=& e^{0\cdot x}\Big(c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) \Big) \\ \mathbf{y_h} &\mathbf{=}& \mathbf{ c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) }\\ \hline \end{array} } \)

3. Particular Solutions \(y_p = \ ?\)

\(\begin{array}{|rcll|} \hline y_p = c \qquad y_p^{'} = 0 \qquad y_p^{''} = 0 \\\\ 4y_p^{''}+9y_p &=& 15 \\ 4\cdot 0 +9 \cdot c &=& 15 \\ 9 \cdot c &=& 15 \\ c &=& \frac{15}{9} \\ \mathbf{c=y_p} &\mathbf{=}& \mathbf{\frac{5}{3}} \\ \hline \end{array} \)

4. \(y(x) = \ ?\)

\(\begin{array}{|rcll|} \hline y(x) &=& y_h + y_p \\ \mathbf{y(x)} &\mathbf{=}& \mathbf{ c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) + \frac{5}{3} } \\ \hline \end{array}\)

The first term of a geometric sequence is 729, and the 7th term is 64.

What is the positive, real value for the 5th term?

\(\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $a =$ term }\\ \text{ $i,j,k = $ indices}\\\\ \large{ a_i^{j-k} \times a_j^{k-i} \times a_k^{i-j} = 1 } \\\\ & i = 1 & a_1 = 729 \\ & j = 7 & a_7 = 64 \\ & k = 5 & a_5 =\ ? \\ & j-k = 7-5 = 2 \\ & k-i = 5-1 = 4 \\ & i-j = 1-7 = -6 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_1^{2} \times a_7^{4} \times a_5^{-6} &=& 1 \quad & | \quad a_1 = 729 \quad a_7 = 64 \\ 729^{2} \times 64^{4} \times a_5^{-6} &=& 1 \\ 729^{2} \times 64^{4} &=& a_5^{6} \quad & | \quad ()^{\frac16} \\ 729^{\frac26} \times 64^{\frac46} &=& a_5^{\frac66} \\ 729^{\frac13} \times 64^{\frac23} &=& a_5 \\ 9 \times 16 &=& a_5 \\ \mathbf{a_5} & \mathbf{=} & \mathbf{144} \\ \hline \end{array}\)

Consider the arithmetic series:

-6 + 4 + 14 + 24 + ...

Write a formula for the sum of the first n terms in this series.

10n + 10

10n + 4

5n²−11n

5n²-11

\(\begin{array}{lrrrrrrrrrr} & -6 && 4 && 14 && 24 && \cdots \\ \text{1. Difference } && 10 && 10 && 10 && \cdots \\ \end{array} \)

Formula arithmetic series:

\(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)d \quad & | \quad a_1 = -6 \qquad d = 10 \\ a_n &=& -6 + (n-1)10 \\ a_n &=& -6 + 10n-10 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{10n-16} \\ \hline \end{array} \)

Formula the sum of a arithmetic series:

\(\begin{array}{|rcll|} \hline s_n &=& \left(\dfrac{a_1+a_n}{2}\right) \cdot n \quad & | \quad a_1 = -6 \qquad a_n = 10n-16 \\ &=& \left(\dfrac{-6+10n-16 }{2}\right) \cdot n \\ &=& \left(\dfrac{10n-22 }{2}\right) \cdot n \\ &=& (5n-11)\cdot n \\ \mathbf{s_n} & \mathbf{=} & \mathbf{5n^2-11n} \\ \hline \end{array} \)

What is the sum of the positive integers that are solutions of $-3n +3 >-11?$

\(\begin{array}{|rcll|} \hline -3n+3 &\gt& -11 \quad & | \quad -3 \\ -3n &\gt& -14 \quad & | \quad :(-3) \\ n &\color{red}\lt& \frac{-14}{-3} \\ n &\lt& \frac{14}{3} \\ n &\lt& 4.\bar{6} \\ \hline \end{array} \)

positive integers:  \(n= 1,2,3,4\)

sum: \(1+2+3+4 =\mathbf{ 10}\)

Marina solved the quadratic equation $9x^2-18x-720=0$ by completing the square.

In the process, she came up with the equivalent equation $$(x+r)^2 = s,

$$where $r$ and $s$ are constants. What is $s$?

\(\begin{array}{|rcll|} \hline 9x^2-18x-720 &=& 0 \quad & | \quad :9 \\ x^2-2x-80 &=& 0 \\ (x-1)^2 -1-80 &=& 0 \\ (x-1)^2 &=& 81 \quad & | \quad (x-r)^2 = s \\ & & \quad & | \quad \mathbf{s=81} \\ \hline \end{array}\)

yea it is 3.20

What is the degree-measure of the acute angle formed by extending sides AB and ED of regular nine-sided polygon ABCDEFGHI until these extensions meet?

----------------------------------------------------------------------

Let the point of intersection of lines AB and ED  be  P .

sum of interior angle measures of nonagon ABCDEFGHI  =  180°(9 - 2)  =  1260°

ABCDEFGHI is a regular polygon, so each interior angle has the same measure.

measure of each interior angle of ABCDEFGHI  =  1260° / 9   =   140°

sum of interior angle measures of heptagon APEFGHI  =  180°(7- 2)  =  900°

 m∠P + 840°   =   900°

 m∠P   =   60°

B sounds good; I think that's the answer.

Alright, the final 2.

5. 

\(x+y+z=6,\ x(y+z)=5, \ \text{and} \ y(x+z)=8\)

From the first equation, we can get:

\(y+z=6-x \ \text{and}\ x+z=6-y\)

We can plug this into equations 2 and 3.

\(x(6-x)=5 \\ 6x-x^2-5=0 \\ x^2-6x+5=0 \\ (x-1)(x-5)=0 \\ x_1=1 \ \text{and} \ x_2=5\)

and

\(y(6-y)=8 \\ 6y-y^2=8 \\ y^2-6y+8=0 \\ (y-2)(y-4)=0 \\ y_1=2 \ \text{and} \ y_2=4\)

We plug our x values and y values to find z. 

Our four solution sets, (x, y, z) are:

(1, 2, 3) ; (1, 4, 1) ; (5, 2, -1) ; (5, 4, -3)

Solution:

6 foot man=20 feet

6/20=1 foot

?= 50 feet

Multiply 6/20*50=3*5=

\(\boxed{15}\)

Oh yeah! Haven't though of it that way!

I just added both expressions, but you just substituted the new expression from the Pythagorean Identity.

Solution:

So, we have \(2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}} \), which at first looks rather scary. But we notice a pattern! Because the nested fractions descend forever, the fraction essentially contains itself! So let \(a\)  equal this fraction:

\( a = 2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}} \)

Thus,

\( a = 2+\frac{4}{1+\frac{4}{a}} \)

So now, we simply need to solve for a.

\(a = 2+\frac{4}{1+\frac{4}{a}} = 2 + \frac{4}{\frac{4+a}{a}}= 2+\frac{4a}{4+a}=\frac{8+2a+4a}{4+a}=\frac{8+6a}{4+a}\)


Getting rid of the fraction by multiplying y \(4+a\), we get

\(4a + a^2 = 8 + 6a \Longrightarrow a^2 - 2a - 8 = 0\)



So now we factor...

\(a^2 - 4a + 2a - 8 = 0\)

\(a(a-4)+2(a-4) = 0\)

\((a+2)(a-4) = 0\)  





So \(a\) can be \(-2\) or \(4\). Since we know, by looking at the fraction, that \(a\)  must be positive, \(a\), and the value of the fraction, must be \(\boxed{4}\).

Tell me if you have any questions,

The first few  terms are  

-6  -2   12   36    70

And using the sum of differences, we have :

-6    -2    12    36     70

    4     14    24    34

        10    10   10

This will be a 2nd power polynomial  in the form an^2 + bn + c

And we have this system

a  +  b  +   c    =  -6

4a  + 2b + c   =  -2

9a +  3b  + c  =  12

Subtract the frist equation from the second   and the  second equation from  the third and we have

3a  + b  = 4

5a + b  =  14        subtract the first equation from the second

2a   = 10

a  = 5

And using  3a  + b  = 4  ⇒    3(5) + b  = 4  ⇒   b  = -11

And using  a + b + c  = -6  ⇒      5  - 11  +  c  = -6   ⇒  -6 + c  = -6  ⇒   c  = 0

So....the polynomial is

5n² - 11n

Thank you, CPhill! 

The slope of  Function 1  is  [ 4 - 1  ]  /  [ 1  - 0 ]  =   3 / 1  = 1

The slope  of Function 2  is  [11 - 7 ]  / [ 2 - 1 ] =   4  / 1    =  4

So.....Function 2 has  a greater rate of change by 1

If  \(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9})=\frac94\) , evaluate

\(\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ \text{____________________________________________} \)  .

By the Pythagorean identity,

\(\sin^2\theta+\cos^2\theta=1\)       so       \(\cos^2\theta=1-\sin^2\theta\)

So

\(\ \qquad\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ =\\ \qquad (\,1-\sin^2(\frac{\pi}{9})\,)+(\,1-\sin^2(\frac{2\pi}{9})\,)+(\,1-\sin^2(\frac{3\pi}{9})\,)+(\,1-\sin^2(\frac{4\pi}{9})\,)\\ =\\ \qquad \phantom{(\,}1-\sin^2(\frac{\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{2\pi}{9})\phantom{\,)}+ \phantom{(\,}1-\sin^2(\frac{3\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{4\pi}{9})\phantom{\,)}\\ =\\ \qquad4-\sin^2(\frac{\pi}{9})-\sin^2(\frac{2\pi}{9})-\sin^2(\frac{3\pi}{9})-\sin^2(\frac{4\pi}{9})\\ =\\ \qquad4-(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9}))\\ =\\ \qquad4-\frac94\\ =\\ \qquad\frac{16}{4}-\frac94\\ =\\ \qquad\frac74 \)

We can check this answer with the calculator by entering:

(cos(pi/9))^2+(cos(2pi/9))^2+(cos(3pi/9))^2+(cos(4pi/9))^2

which does equal  1.75  

Let  a   =  2  +   4

                       ______

                       1 +   4

                              ____

                                 ........

So we have

a =   2  +       4

                  ____

                 1  + 4

                       __

                        a           simplify

a =   2      +    4a

                    _______

                      (a  + 4)

a   =  (2a + 8 + 4a) 

         _____________

             (a + 4)

a( a + 4)  =   6a + 8

a^2 + 4a   = 6a  + 8

a^2 - 2a  - 8   = 0

(a - 4) (a + 2)    = 0

Setting both factors to 0  and solving for   "a"  we  get   that  a  = 4  or a  = -2

We can reject -2 because  a  is positive

So......the real  number is   4

Excellent....!!!

P.S.  - you can always check  your work  by multiplying the factorization back out.....if you get what you started with, you have done it correctly  !!!!

[64/729]^1 / (N - 1)

[64/729]^1/6 = 2/3 Common ratio

5th term =F x R(N - 1)

              =729 x (2/3)^4

              =729 x 16/81

              = 144 - the fifth term of the GS.