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Hope this helps you all feel better. 

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ok Juiemagic but you left out key components of the question and the other question I spent a lot of time on also had bits missing.

Last time I thought it was a badly written, incomplete question but maybe the question was just fine until you gave us the condensed version. 

I certainly accept your appology but please do not give us half questions (or reworded questions) any more. :/

7^11

let x=7^11

\(x=7^{11}\\ logx=log7^{11}\\ logx=11log7\\\)

11*log(7) = 9.2960784401568248

\(logx=9.2960784401568248\\ x=10^{9.2960784401568248}\\ x=10^{0.2960784401568248}\times10^{9}\\ x=1.977326743\times10^{9}\\~\\ 7^{11}=1.977326743\times10^{9}\\\)

Okay, let me clarify this...

The paper reads:

GIVEN:

1)   AOIIQR

2)   PA = AQ

3)   PB = BT

Prove:

1) ABIIQT

2) O is the centre of the circle if PR is the centre line through the circle

3) BORT is a trapezium

Sorry if there was confusion, I did not realize giving it like it actually was on the paper would or could cause a mis-understanding. From my side, please accept my apologies.

LOL That is what I thought the first time I saw it but you did that boo boo a couple of times :)

Did you put your working in the middle of the question as I have asked further down the page?

Please be careful. It is very frustrating to put a lot of time into a question when the question is not presented properly. :/

I thought that statement was a part of Juriemagic's working.

I did not think it was a part of the question!

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

This is the the question, along with the pic of course:

I am working through a paper, and have completed so to speak the majority of problems. I got stuck with another really tricky one. Please if you could, how can I prove "O" is the centre?

GIVEN:

1)   AOIIQR

2)   PA = AQ

3)   PB = BT

No where does it say that PR is a diameter!

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

NOW that I look at it with your eyes I can see that maybe Juriemagic put his working in the middle of his question.

All very confusing!

If an object is moving at a constant rate of 2 m/s, then it has an acceleration of 0 m/s/s.

haha...disects!!..oh my goodness!!!..so sorry, that was an honest boo boo...

Also, This exact question I found on the inernet in an exam question paper. The marks allocated for proving ABIIQT, was 2, and for proving the centre, also just 2. Sooo, I'm thinking going the similar triangle way is a lot of work for just 2 marks, HOWEVER, I do aknowledge that the approach given to me was far superior and most likely, the best approach.

Guys, you have all been a great help with this one, I really admire and love you all!!!

Thank you for this information, I will use it frequently!

Solve for t:
S = P (r t + 1)

Reverse the equality in S = P (r t + 1) in order to isolate t to the left-hand side.
S = P (r t + 1) is equivalent to P (r t + 1) = S:
P (r t + 1) = S

Divide both sides by a constant to simplify the equation.
Divide both sides by P:
r t + 1 = S/P

Isolate terms with t to the left-hand side.
Subtract 1 from both sides:
r t = (S - P)/P

Solve for t.
Divide both sides by r: 

t = (S - P)/(P r)

No assumption Melody.

Read the question.

NOW, prove "O" is the centre of the circle if PR is the centre line of the circle.

A wood Chuck can't chuck wood. Great math question

Hi Juriemagic,

Firstly you disect rats in a laboratory and you bisect intervals when you cut them in half :)

Yes I like your logic, although you would need to discuss similar triangles as Chris did.

I have the same problem with your logic as I do with Chris's, I can see that O bisects PR easily enough but I do not know why either of you has stated that PR is a diameter   

I now see that the original guest answerer has made exactly the same 'assumption' ://

Do you understand Finnish?

Määritä funktio f(x)=4x^2+3 derivaattafunktio?

Definiere die Funktion  f (x) = 4x ^ 2 + 3         Ableitungsfunktion?

Define the function       f (x) = 4x ^ 2 + 3          derivative function?

\(f(x)=4x^2+3\\ f ' =8x \)

  !

What help do you need?

Hi Chris,

I followed your logic right up till you said "And since PR is the diameter of the circle"

Why can you say that PR is the diameter of the circle ?? 

Maybe it is written somewhere and i am blind. I am not being sarcastic, this is always a distinct posibility :)

It is Red's fault! He stole my glasses!

Great! 

Personally, I just want to ace the exams...

A pyramid with volume 40 cubic inches has a rectangular base. If the length of the base is doubled, the width tripled and the height increased by $50\%$, what is the volume of the new pyramid, in cubic inches?

\(\frac{1}{3}lwh=40\\ lwh=120\\~\\ New\;\;volume\;\; =\frac{1}{3}*2l*3w*1.5h\\ =\frac{1}{3}*2*3*1.5*lwh\\ =3lwh\\ =3*120\\ =360\;\;cubic\;\;inches \)

CPhill,

I really wish I had the experience you guys have!!..It's good to see how it's solved, however, please if you do not mind....what in my solution makes it wrong?..I understand my approach was wrong, simply because you say it was, and you gave me the appropriate solution....but why is my approach wrong?..would you kindly spend just a little more time please, and just educate me..please...I need to understand this?..You guys are great, thank you for being out there!!

John computes the sum of the elements of each of the 15 two-element subsets of
\(\{1,2,3,4,5,6\} \).
What is the sum of these 15 sums?

15 two-element subsets:

\(\begin{array}{|llllll|} \hline &1,& 2,& 3,& 4,& 5,& 6 \\ \hline &1,2 & 2,3 & 3,4 & 4,5 & 5,6 \\ &1,3 & 2,4 & 3,5 & 4,6 & \\ &1,4 & 2,5 & 3,6 & \\ &1,5 & 2,6 & \\ &1,6 \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline \text{the numbers are: } && 5\times 1 \\ &+& 5\times 2 \\ &+& 5\times 3 \\ &+& 5\times 4 \\ &+& 5\times 5 \\ &+& 5\times 6 \\ \hline &=& 5\times (1+2+3+4+5+6) \\ &=& 5\times \left(\frac{1+6}{2}\right)\times 6 \\ &=& 5\times \left(\frac{7}{2}\right)\times 6 \\ &=& 105 \\\\ &=& (n-1)\times \left(\frac{1+n}{2}\right)\times n \\ &=& 3\times \binom{n+1}{3} \\ \hline \end{array} \)

hyperbolic functions

\(sinh^{-1} (1)\)

Formula:

\(\begin{array}{|rcll|} \hline sinh(z) &=& \frac{e^z-e^{-z}}{2} \\ z &=& sinh^{-1} (\frac{e^z-e^{-z}}{2} ) \\ \hline \end{array}\)

\(z=\ ?\)

\(\begin{array}{|rcll|} \hline \frac{e^z-e^{-z}}{2} &=& 1 \\ e^z-e^{-z} &=& 2 \\ e^z-\frac{1}{e^{z}} &=& 2 \quad & | \quad \text{substitute } e^z = x \\ x-\frac{1}{x} &=& 2 \quad & | \quad \cdot x \\ x^2-1 &=& 2x \\ x^2-2x-1 &=& 0 \\\\ x &=& \frac{2\pm \sqrt{4-4\cdot(-1)} }{2} \\ &=& \frac{2\pm \sqrt{8} }{2} \\ &=& \frac{2\pm \sqrt{2\cdot 4 } }{2} \\ &=& \frac{2\pm 2\sqrt{2} }{2} \\ \mathbf{x} &\mathbf{=}& \mathbf{1\pm \sqrt{2}} \quad & | \quad x = e^z \\ e^z & = & 1\pm \sqrt{2} \quad & | \quad \ln() \\ z & =& \ln(1\pm \sqrt{2}) \\\\ z_1 &=& \ln(1 + \sqrt{2}) \\ z_2 &=& \ln(1 - \sqrt{2}) \quad & | \quad \text{no solution } \quad 1 - \sqrt{2} < 0! \\\\ \mathbf{sinh^{-1} (1)= z} &\mathbf{=}& \mathbf{\ln(1 + \sqrt{2})} \\ \mathbf{sinh^{-1} (1) } &\mathbf{=}& \mathbf{0.88137358702\ldots} \\ \hline \end{array}\)

How do you find the domain and range of 1/square root of x-9

I think that this is what you intended?

\(f(x)=\frac{1}{\sqrt{x-9}}\)

Let see..

you cannot divide by 0 and the square root of a real number cannot be neg so the range must be greater than 0

x-9 cannot be negative so  x>9

so the range is  f(x)>0

and the domain is x>9

Here is the funtion

Modular Arithmetic

(a) How many positive integers   from 1 to 5000 satisfy the congruence  \(N\equiv5\pmod{12} \)?

\(\begin{array}{|lrcll|} \hline N\equiv5\pmod{12} & \text { or } & N-5 &=& n\cdot 12 \\ & & N &=& n\cdot 12 + 5 \quad & | \quad N_{max} = 5000 \\ & & 5000 &=& n\cdot 12 + 5 \\ & & n &=& \frac{5000-5}{12} \\ & & n &=& [416].25 \\\\ \mathbf{n=416} &\Rightarrow& N &=& 416\cdot 12 + 5 = 4997 \\ \hline \end{array}\)

416  + 1(n=0)  = 417 positive integers from 1 to 5000 satisfy the congruence \(N\equiv5\pmod{12}\)

(b) How many positive integers   from 1 to 5000 satisfy the congruence \( N\equiv11\pmod{13}\) ?

\(\begin{array}{|lrcll|} \hline N\equiv11\pmod{13} & \text { or } & N-11 &=& n\cdot 13 \\ & & N &=& n\cdot 13 + 11 \quad & | \quad N_{max} = 5000 \\ & & 5000 &=& n\cdot 13 + 11 \\ & & n &=& \frac{5000-11}{13} \\ & & n &=& [383].769230769 \\\\ \mathbf{n=383} &\Rightarrow & N &=& 383\cdot 13 + 11 = 4990 \\ \hline \end{array}\)

383 + 1(n=0) = 384 positive integers from 1 to 5000 satisfy the congruence \(N\equiv11\pmod{13}\)

volume of cone  =  (1/3)(area of base)(height)

volume of cone  =  (1/3)  (pi)(radius²) (height)

                                                                               Plug in the information from the problem.

       12 cm³        =  (1/3)  (pi)(radius²) (14 cm)

                                                                               Multiply both sides by  3 .

       36 cm³        =  (pi)(radius²) (14 cm)

                                                                               Divide both sides by  14 cm.

    36/14 cm²      =  (pi)(radius²)

                                                                               Divide both sides by  pi  .

  36/(14pi) cm²   =  (radius²)

                                                                               Take the positive square root of both sides.

\(\sqrt{\frac{36}{14\pi}}\,\text{ cm}\,=\,\text{radius} \\~\\ \text{radius} \,=\,\sqrt{\frac{36}{14\pi}}\,\text{ cm} \,\approx\,\sqrt{\frac{36}{14(\frac{22}7)}}\,\text{ cm}\,=\,\sqrt{\frac{36}{44}}\,\text{ cm}\,\approx\,0.905\,\text{ cm}\)