can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx

http://i74.photobucket.com/albums/i2...ng104/scan.jpg

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- Jun 4th 2006, 09:14 AMkansaineed help gr12
can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx

http://i74.photobucket.com/albums/i2...ng104/scan.jpg - Jun 4th 2006, 09:30 AMmalaygoel
hello, I could give answer to your first question

when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle

2^(1/2)side of square = diagonal of square

area of square = 1/2 * diagonal * diagonal

S1 = 2 (diagonal of S1 =2)

when a circle is inscribed in a square, length of square = diameter of circle

diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)

S2 = 1/2 * 2 = 1

Sn = 2*[(1/2)^(n-1)] - Jun 4th 2006, 02:44 PMThePerfectHackerQuote:

Originally Posted by**kansai**

You know that,

$\displaystyle t_1=-2$

$\displaystyle t_2=1$

And that,

$\displaystyle t_3=2t_2-t_1$

thus,

$\displaystyle t_3=2(1)-(-2)=2+2=4$

And that,

$\displaystyle t_4=2t_3-t_2$

thus,

$\displaystyle t_4=2(4)-1=8-1=7$

And that,

$\displaystyle t_5=2t_4-t_3$

thus,

$\displaystyle t_5=2(7)-1=13$ - Jun 4th 2006, 08:01 PMSoroban
Hello, kansai!

Here's #5 . . .

Quote:

5. For the matrices: $\displaystyle A = \begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix}$ and $\displaystyle B = \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

(a) State the value of: $\displaystyle \text{(i) }a_{12}\qquad\text{(ii) }a_{22}\qquad\text{(iii) }a_{31}$

(b) Calculate: $\displaystyle \text{(i) }A - 3B\qquad\text{(ii) }2A + B\qquad\text{(iii) }A'$

$\displaystyle \text{(a) }\;a_{12}$ means the element in the $\displaystyle 1^{st}$ row and the $\displaystyle 2^{nd}$column. .Hence: $\displaystyle a_{12} = 5$

. . . .Similarly: $\displaystyle a_{22} = 0,\;a_{31} = 2$

The elements of matrix $\displaystyle B$ would have been written: $\displaystyle b_{12},\;b_{22},\;b_{31}$

$\displaystyle \text{(b)(i) }A - 3B \:=\:\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - 3\cdot\begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

. . . $\displaystyle =$ $\displaystyle \;\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - \begin{bmatrix}6&0&-3\\3&0&3\\9&-6&0\end{bmatrix} \;=\;\begin{bmatrix}-5&5&3\\0&0&1\\-7&6&-1\end{bmatrix}$

$\displaystyle \text{(b)(ii) }2A + B \;= \;2\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

. . . $\displaystyle =\;\begin{bmatrix}2&10&0\\6&0&8\\4&0&-2\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix} \;= \;\begin{bmatrix}4&10&-1\\7&0&9\\7&-2&-2\end{bmatrix}$

$\displaystyle \text{(b)(iii) }A'$ . . . I assume this means the*inverse*of $\displaystyle A.$

The steps are long and messy . . . so I'll omit them.

. . . $\displaystyle A'\;=\;\frac{1}{55}\begin{bmatrix}0&5&20\\11&-1&-4\\0&10&-15\end{bmatrix} $

- Jun 4th 2006, 11:45 PMSoroban
Hello, kansai!

Here's #3 . . .

Quote:

For the map given on the right

(a) Represent the map with a network.

(b) Find the degree of each vertex.

(c) State whether the network is traceable and explain the reason.

Code:`(a)`

A - - - - - B

| / | \

| / | \

| C | \

| / | \

| / | \

D - - - - - E - - -F

. . .B: degree 4

. . .C: degree 2

. . .D: degree 3

. . .E: degree 3

. . .F: degree 2

(c) A network is traceable if it has at most two vertices of odd degree.

This network has two odd vertices (D and E) and hence is traceable. - Jun 5th 2006, 05:50 AMSoroban
Hello, kansai!

I don't "trust" #4 . . . It has some disturbing errors and the meaning is unclear.

. . Can you provide the*original*wording?

Quote:

4. The*NYS school*has four committees.

Each of these committees meet once a month.

Membership on these committees is as follows:

. . Committee A: Anna, John, Steven, Tom

. . Committee B: Debra, Jane John

. . Committee C: Annie, Debra, Larry, Tom

. . Committee D: Bobby, Debra, Grace, Mary, Steven

. . Committee E: Anna, Larry, Tom.

(a) Draw a network to illustrate the*connectivity*among the committees. .**?**

(b) Design a schedule with*minimum timeslots*for the committee meeting without conflicts. .**??**

I would assume that Annie is the same person as Anna, that someone mistyped the names.

. . But do I*dare*make that assumption?

The most glaring errors is that there are**five**committees.

(a) Connectivity is a term used in network theory, but not like this.

. . I assume they mean "common memberships" ?

(b) Minimum timeslots is a**very**sloppy term.

Does it mean "the shortest meetings"?

. . How about: "I call the meeting to order. .I will entertain a motion to adjourn . . ."

Does it mean "the shortest time-span" for the four (five?) meetings?

. .Schedule them consecutively with a one-minute break in between

. . (to allow the ten people to switch seats).

It is obvious that no two committees can meet at the same time.

. . So what is the question?

If I am correct about part (a), this should suffice:Code:`B`

/ | \

/ | \

/ | \

A - - - + - - - D

/ \ | /

/ \ | /

/ \ | /

E - - - - - C

- Jun 5th 2006, 03:55 PMwenlongURGENT can any one do question 6,7,8?
URGENT can any one do question 6,7,8?? and plz show ur steps and explation..............thnx

- Jun 5th 2006, 04:58 PMQuickQuote:

Originally Posted by**malaygoel**

__General high school math-need help gr12__ - Jun 5th 2006, 06:01 PMQuickQuestion 1
From my understanding of this question it works out like this.

the diagnol of the square (d) = the diameter of the circle (d), so the area =

$\displaystyle s$ = the length of one of the square's side

$\displaystyle d=\sqrt{ s^2 +s^2} $

$\displaystyle d^2=s^2 +s^2 $

$\displaystyle d^2=2s^2 $

$\displaystyle \frac{d^2}{2}=s^2 $ since s^2 is the area of the square this is the answer for $\displaystyle S_1 $.

Now the next circle has the diameter equal to the height of the square, so...

$\displaystyle \sqrt{\frac{d^2}{2}}=s=d_2 $

and now you apply the first formula to find the area of $\displaystyle S_2 $

$\displaystyle \sqrt{\frac{d^2}{2}}=\sqrt{s^2 +s^2} $

$\displaystyle \sqrt{\left(\frac{d^2}{2}\right)^2}=s^2 +s^2 $

$\displaystyle \frac{d^2}{2}=2s^2 $

$\displaystyle \frac{\frac{d^2}{2}}{2}=s^2 $

$\displaystyle \frac{d^2}{4}=s^2 $

$\displaystyle \frac{d^2}{4}=s^2 $

This shows us that the area of $\displaystyle S_n = \frac{d^2}{2^n} $

Considering me and malaygoel got different answers, could a third party check my work? - Jun 5th 2006, 06:08 PMmalaygoelQuote:

Originally Posted by**Quick**

- Jun 5th 2006, 06:22 PMQuickQuote:

how did you get d=s+s

- Jun 5th 2006, 06:25 PMmalaygoelQuote:

Originally Posted by**Quick**

- Jun 5th 2006, 06:41 PMQuickQuote:

diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)

mine= $\displaystyle \frac{d^2}{2^n} $ yours= $\displaystyle 2\left(\frac{1}{2^{(n-1)}}\right) $

However, there is a certain value of d that would make our equations the same.

$\displaystyle \frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right) $

$\displaystyle \frac{d^2}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{2^2}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{4}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}} $

but why would you say the diameter is 2? - Jun 5th 2006, 07:04 PMwenlong
friend got these answer for #1

plz tell me is it correct?

area of 1st square = 2

area of 2nd square= 1

area of 3rd square= 2^-1

area of Sn square= 2^2-n

is my friends answer correct? - Jun 5th 2006, 07:22 PMQuick
assuming that the diameter of the first circle is 2, then yes.

$\displaystyle \frac{d^2}{2^n} $

$\displaystyle \frac{2^2}{2^n} $

$\displaystyle \frac{4}{2^n} $

$\displaystyle \frac{4}{2^n} $

$\displaystyle \frac{1}{2^{(n-2)}} $

$\displaystyle 2^{(-1)(n-2)} $

$\displaystyle 2^{(-n+2)} $

$\displaystyle 2^{(2-n)} $