Give me maths problems!

**northdown** · June 29th 2012, 07:04 AM

Yes missing bracket and I appear to be missing a lot. Thank you for your tolerance

**northdown** · June 29th 2012, 07:37 AM

Here are the calculation used in my excel spread sheet I hope this clarifies the two equations

f= (1/(1.74-2*LOG((2*B4)/B5))^2)
f= (1.14+(2*LOG(B5/B4)))^-2

**loplop** · June 29th 2012, 10:04 AM

I need some help with a problem i am currently having trying to transpose a geometric formula. The formula is ar^n and i need to solve it for n.

Someone told me i have to used logarithms to do it but its has been so long since i have done it i cant remember.

In a worked example i have a = 9volts, r = 0.9985 and i used n = 1 which gives me an answer of 8.964v (I know this is correct as this is term 1 in my sequence and 9V is term 0.

But i need to know what the value of n is when the product is 8V. I also know the answer is 29 but i just dont know how to achieve it!!

Any help would be hugely appreciated

**renaee7** · June 29th 2012, 10:48 AM

Questions 14-16: Use the triangle to calculate the specified trigonometric values.
http://is.byu.edu/courses/hs/GEOM-04...math10-929.gif
14.Calculate sin
a. 7/ square root of 13
b. 6/ square root of 13
c. 6/ square root of 85
d. 7/ square root of 85

15.Calculate cos
a. 7/ square root of 13
b. 6/ square root of 13
c. 6/ square root of 85
d. 7/ square root of 85

16.Calculate tan
a. 7/ square root of 13
b. 6/ square root of 13
c. 6/ square root of 85
d. 7/ square root of 85

17. If tan=5/4 calculate the value of cos
a. 5/ square root of 41
b. 4/ square root of 41
c. 5/3
d. 4/3

18. If cos= 2xsquare root of 10/7 calculate the value of sin.
a. 3/7
b. square root of 29/7
c. square root of 69/7
d. square root of 89/7

19. If sin= 1/ square root of 2 find the value of 0.
a. 30º
b. 45º
c. 60º
d. 90º

Questions 20-22: Use a calculator to find the trigonometric value to four decimal places.

20. cos 88º =
a. 0.0354
b. 0.0114
c. 0.9994
d. 0.0349

21. sin 47º =
a. 0.7314
b. 0.6820
c. 0.1236
d. 1.0734

22. tan 8º =
a. 6.7997
b. 0.1405
c. 0.1392
d. 5.6713

**thevinh** · June 29th 2012, 10:51 AM

$\text{Check out these problems.}$

$\text{Solve for \emph{x}.} \mspace{10mu} x^3+1=2\sqrt[3]{2x-1}$

$\text{Solve for \emph{x}.} \mspace{10mu} \log_{1-x}{(2x^2+x+1)}=2$

$\text{Solve for \emph{x}.} \mspace{10mu} \sqrt[5]{27}x^{10}-5x^6+\sqrt[5]{864}=0$

$\text{I am happy to discuss these problems with you. Have fun!}$

**Prove It** · June 29th 2012, 08:13 PM

Originally Posted by loplop

I need some help with a problem i am currently having trying to transpose a geometric formula. The formula is ar^n and i need to solve it for n.

Someone told me i have to used logarithms to do it but its has been so long since i have done it i cant remember.

In a worked example i have a = 9volts, r = 0.9985 and i used n = 1 which gives me an answer of 8.964v (I know this is correct as this is term 1 in my sequence and 9V is term 0.

But i need to know what the value of n is when the product is 8V. I also know the answer is 29 but i just dont know how to achieve it!!

Any help would be hugely appreciated

$\displaystyle \begin{align*} t_n &= a\,r^n \\ \frac{t_n}{a} &= r^n \\ \ln{\left(\frac{t_n}{a}\right)} &= \ln{\left(r^n\right)} \\ \ln{\left(\frac{t_n}{a}\right)} &= n\ln{(r)} \\ n &= \frac{\ln{\left(\frac{t_n}{a}\right)}}{\ln{(r)}} \\ n &= \frac{\ln{\left(t_n\right)} - \ln{(a)}}{\ln{(r)}} \end{align*}$

**Prove It** · June 29th 2012, 08:30 PM

Originally Posted by thevinh

$\text{Check out these problems.}$

$\text{Solve for \emph{x}.} \mspace{10mu} x^3+1=2\sqrt[3]{2x-1}$

$\text{Solve for \emph{x}.} \mspace{10mu} \log_{1-x}{(2x^2+x+1)}=2$

$\text{Solve for \emph{x}.} \mspace{10mu} \sqrt[5]{27}x^{10}-5x^6+\sqrt[5]{864}=0$

$\text{I am happy to discuss these problems with you. Have fun!}$

$\displaystyle \begin{align*} x^3 + 1 &= 2\sqrt[3]{2x - 1} \\ \left(x^3 + 1\right)^3 &= \left(2\sqrt[3]{2x - 1}\right)^3 \\ x^9 + 3x^6 + 3x^3 + 1 &= 8(2x - 1) \\ x^9 + 3x^6 + 3x^3 + 1 &= 16x - 8 \\ x^9 + 3x^6 + 3x^3 - 16x + 9 &= 0 \\ (x - 1)\left(x^2 + x - 1\right)\left(x^6 + 2x^4 + 2x^3 + 4x^2 + 2x + 9 \right) &= 0 \\ x - 1 = 0 \textrm{ or } x^2 + x - 1 &= 0 \textrm{ (the third factor can not be solved over the reals)} \\ x = 1 \textrm{ or }x &= \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} \\ x = \frac{-1 - \sqrt{5}}{2} \textrm{ or } x = 1 \textrm{ or } x &= \frac{1 + \sqrt{5}}{2} \end{align*}$

$\displaystyle \begin{align*} \log_{1 - x}{\left(2x^2 + x + 1\right)} &= 2 \\ 2x^2 + x + 1 &= \left(1 - x\right)^2 \\ 2x^2 + x + 1 &= 1 - 2x + x^2 \\ x^2 + 3x &= 0 \\ x(x + 3) &= 0 \\ x = 0 \textrm{ or } x &= -3 \end{align*}$

**Prove It** · June 29th 2012, 08:43 PM

May I suggest to the mods that this thread be closed please. People are just double posting their problems, and the OP isn't here to answer the questions anyway...

**skeeter** · June 30th 2012, 04:10 AM

Originally Posted by Prove It

May I suggest to the mods that this thread be closed please. People are just double posting their problems, and the OP isn't here to answer the questions anyway...

note that this site no longer has any "moderators"; the new owner has not logged any activity since the end of May.

**cakalaw** · June 30th 2012, 02:03 PM

this may not be the most complicated of maths problems and more of a physics problem but i couldnt access the partner site so thougth i have a try at posting here.
i was involved in a road traffic accident the other day where the other driver attempted what i think was an impossible overtake so i'd appreciate a little help in proving this. i was traveling at 40mph in the outside lane of a dual carriageway, a temporary lane opens up for 116m which the other driver attempted to overtake me in. he was about 4m behind me and traveling at the same speed. what i would like to know is,

1; the speed the other driver would need to be traveling at the end of the maneuver to overtake and pull back in front of me again,
2; the acceleration required by his vehicle to complete the maneuver and
3; (not strictly a maths question) an few examples of vehicles capable of completing this manuver.

many thanks for any help on this as it would be very useful in my insurance claim

**thevinh** · June 30th 2012, 03:44 PM

Originally Posted by Prove It

$\displaystyle \begin{align*} x^3 + 1 &= 2\sqrt[3]{2x - 1} \\ \left(x^3 + 1\right)^3 &= \left(2\sqrt[3]{2x - 1}\right)^3 \\ x^9 + 3x^6 + 3x^3 + 1 &= 8(2x - 1) \\ x^9 + 3x^6 + 3x^3 + 1 &= 16x - 8 \\ x^9 + 3x^6 + 3x^3 - 16x + 9 &= 0 \\ (x - 1)\left(x^2 + x - 1\right)\left(x^6 + 2x^4 + 2x^3 + 4x^2 + 2x + 9 \right) &= 0 \\ x - 1 = 0 \textrm{ or } x^2 + x - 1 &= 0 \textrm{ (the third factor can not be solved over the reals)} \\ x = 1 \textrm{ or }x &= \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} \\ x = \frac{-1 - \sqrt{5}}{2} \textrm{ or } x = 1 \textrm{ or } x &= \frac{1 + \sqrt{5}}{2} \end{align*}$

$\displaystyle \begin{align*} \log_{1 - x}{\left(2x^2 + x + 1\right)} &= 2 \\ 2x^2 + x + 1 &= \left(1 - x\right)^2 \\ 2x^2 + x + 1 &= 1 - 2x + x^2 \\ x^2 + 3x &= 0 \\ x(x + 3) &= 0 \\ x = 0 \textrm{ or } x &= -3 \end{align*}$

$\text{\underline{Problem 1} After you factor out} \mspace{7mu} (x-1), \, \text{how do you know to factor out} \mspace{7mu}$
$(x^2 + x - 1).\; \text{This is how I solved that problem.}$

$\text{Let}\mspace{5mu} u=\sqrt[3]{2x-1}\, \Leftrightarrow\, u^3=2x-1\, \Leftrightarrow\, u^3+1=2x$

$\text{Now we have a system of equation}$

$\displaystyle \begin{cases} x^3+1=2u\\ u^3+1=2x \end{cases} \mspace{-12mu} \Rightarrow\: x^3-u^3=2(u-x)\, \Leftrightarrow\, (x-u)(x^2+ux+u^2+2)=0$

$\Leftrightarrow\, u=x\, \Rightarrow\, x=\sqrt[3]{2x-1}\, \Leftrightarrow\, x^3-2x+1=0 \, \Leftrightarrow\, (x-1)(x^2+x-1)=0$

$\text{\underline{Problem 2} Here}\mspace{5mu} x=0\mspace{5mu} \text{cannot be the solution since}\mspace{5mu} \log_1{1}\neq2.$

**Ben007** · July 1st 2012, 08:53 AM

Hello Mr Neeraj Karn, glad to know you. Please help me out.
In the game of tennis a player has two serves. If the first serve is successful the game continues. If the first serve is not successful the player serves again. If this second service is successful the game continues.
If both serves are unsuccessful the player has served a "double fault" and loses the point.
Gabriella plays tennis. She is successful with 60% of her first serves and 95% of her second serves.
(a) Find the probability that Gabriella serves a double fault.

If Gabriella is successful with her first serve she has a probability of 0.75 of winning the point.
If she is successful with her second serve she has a probability of 0.5 of winning the point.
(b) Find the probability that Gabriella win the point.

**Prove It** · July 1st 2012, 09:44 PM

Originally Posted by Ben007

Hello Mr Neeraj Karn, glad to know you. Please help me out.
In the game of tennis a player has two serves. If the first serve is successful the game continues. If the first serve is not successful the player serves again. If this second service is successful the game continues.
If both serves are unsuccessful the player has served a "double fault" and loses the point.
Gabriella plays tennis. She is successful with 60% of her first serves and 95% of her second serves.
(a) Find the probability that Gabriella serves a double fault.

If Gabriella is successful with her first serve she has a probability of 0.75 of winning the point.
If she is successful with her second serve she has a probability of 0.5 of winning the point.
(b) Find the probability that Gabriella win the point.

Are we assuming that the events are independent?

**Ben007** · July 1st 2012, 10:42 PM

Well, the question says nothing about Independence. But let's assume the events to be independent.

**Prove It** · July 1st 2012, 11:21 PM

If they're independent, then the probability of two events happening is equal to the product probabilities of each event.

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