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X = 7

If x^2-49=0

\(\dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\sqrt{x-1}}\cdot (x-3x^2) } { x-1 }\)

\(\small{ \begin{array}{rcll} &=& \left[ \dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { x-1 } \right] \cdot \dfrac{ 2\cdot \sqrt{x-1} } { 2\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ \sqrt{x-1}\cdot 2\cdot \sqrt{x-1}\cdot(1-6x)- \frac{2\cdot \sqrt{x-1}}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot \left(\sqrt{x-1}\right)^2 \cdot(1-6x)-(x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( x-6x^2-1+6x )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( 7x-6x^2-1 )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 14x-12x^2-2 - x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 13x-9x^2-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot (x-1)^{\frac12} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot(x-1)^{\frac32} } \\\\ \end{array} }\)

How do I find inverse log?

The inverse log of \(\log_b{(x)}\) is \(b^x\), because \(\log_b{(b^x)} =x\cdot \log_b{b}= x\) , and we have  \(\log_b{(b)}=1\)

Use the \(y^x\) Button.

If you have \(\log_{10}{(x)}\) use the \(10^x\) Button.

If you habe \( \ln{(x)}\) use the \(e^x\) Button.

Hi again Namodesto,

A part of the problem is that you presented the question incorrectly.  1/(sqrt(x-1)*2) the bottom needed to be in brackets. 

You did not hve brackets which meant your 2 was up the top when it should  have been down the bottom.

I have been called away.  I will be back in about a half hour.  sorry.

I got t there answer except I got -2 anstead of +2    

I will look more in a while :))  

Must run .

Ok I am back and I can see that Heureka is already preparing another answer for you.

That +2 on  the first answer on this web page that you have since inserted  :-

http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1

is a typo, the answer further down is -2   and it is the same as I got when I redid your expression with the 2 down the bottom instead of up the top.  :)

I'll leave it now untill we both see what Heureka is preparing for you :))

How can i find the inverse of log3

Hi Melody, 

http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1

I have problem understanding the last step. The following problem above was taken from an example of this website. How did he get to the final answer being -9x^2 + 13x - 2 from what I've written above? 

Hi Namodesto,

You cannot solve it because it is not an equation.  

You can only simplify it

\({ \sqrt{x-1}*(1-6x)-(1/ \sqrt{x-1}*2)*(x-3x^2) \over x-1}\\ =\left[\sqrt{x-1}*(1-6x)-(\frac{2}{ \sqrt{x-1}})*(x-3x^2) \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\frac{x-6x^2-1+6x-2 x+6x^2}{\sqrt{x-1}*(x-1)} \\ =\frac{5x-1}{(x-1)^{3/2}} \\ \)

Well learn from these fellas.  :))

3^2 = 3*3 = 9 

 so, 

9 + 4+ 2 = 15 :) 

Press shift, or type acos(

you can also type cos^-1(

:D

A neutron has a mass of 1.67*10^-27kg and an electron has a mass of 9.11*10^-31kg. Determine how many times smaller the mass of the electron is than the mass of the neutron. (Scientific Notation FYI)

neutron has a mass \(1.674927471\cdot 10^{-27}\ kg\)

electron has a mass \(9.109 383 56\cdot 10^{−31}\ kg\)

\(\small{ \begin{array}{rcl} \dfrac{ \text{neutron}_\text{mass} }{ \text{electron}_\text{mass} } &=& \dfrac{1.674927471\cdot 10^{-27}\ kg}{9.109 383 56\cdot 10^{-31}\ kg} \\\\ &=& \dfrac{1.674927471\cdot 10^{-27} \cdot 10^{31} }{9.109 383 56} \\\\ &=& \dfrac{1.674927471\cdot 10^{4} }{9.109 383 56} \\\\ &=& 0.1838761084\cdot 10^{4} \\\\ &=& 1.838761084\cdot 10^{3} \\\\ &=& 1838.761084 \\\\ \end{array} }\)

The mass of the eletron is \(1.838761084\cdot 10^{3}\) times smaller than the mass of the neutron.

Sorry, young ladd/lass. I took the mass to be a neutron star, NOT the elementary particle, the Neutron!!!!.Of course the calculation will quite different:

9.11 X 10^-31/ 1.67 X 10^-27=5.45508 X 10^-4Kg. Or a neutron's mass is:

=1,833.15 times that of the electron.

C(6,5) * C(3,2)

this is read, 6 choose 5 times 3 choose 2

\(\binom{6}{5} \cdot \binom{3}{2} = \binom{6}{1} \cdot \binom{3}{1} = 6\cdot 3 = 18\)

7.77777777778 to fraction

=7.7777777778=7 7/9

Apples cost $1.29 per pound. How much would a bag of apples weighing 4.7 cost?(Round your answer to the nearest cent.)

Just multiply $1.29 X 4.7=$6.063 the cost of 4.7 pound bag.

You are doing just fine. Congrats and keep it up.

I submitted my answer and it was correct. Thank you for your quick replies. I only have a certain number of attempts for each question in the lesson. So Im being over paranoid. Thanks!

OK. I think I know what you have done and it looks right.

what is the inverse tangent of 77.6/81.3

Tan=0.954489............etc. Then the arctangent or the inverse tangent is:

=43.67 degrees.

Ok, this is progress then, thank you.

Ac in my formula is Amplitude * Speed of light = 4.32e5

Amplitude = 432km (converted to 432e+5m)
Speed of light = 2.998e8m/s

Correct?

Yes, it's right the way you have calculated. The only thing I don't see is the speed of light, in your calculations. The answer that you got is very close to the mass of an average planet like Earth.

Hi

Thanks for the replies.

Ill provide for information.


I'm using the formula Mp = Mstar (Ac/r)

I know that Mstar = 2.8e30kg

I know the Radius of the planet = 1.35e11m  (This is how far the planet is from the star)
The amplitude (A) I have is 432km (converted to 432e+5m)
The speed of light (c) is 2.998e8m/s


So based off my formula It should be

Mass Of Planet = 2.8e30 X (4.32e5/1.35e11)

The answer I get = 8.96e24

Am I doing this correctly?
 

The mass of your neutron star is: 1.67 X 10^27Kg NOT 10^-27.

The ratio of the mass of the electron to the mass of the neutron is:                                                                                             9.3 X 10^-31 / 1.67 X 10^27=5.5688622 X10^-58 Kg.

Yes, I do not really understand your problem either, you can put it straight into a calculator but maybe you want to understand the scientific notation better??

\(2.8*10^{30}(1.29*10^{14}/(1.35*10^{11})\\ =2.8*10^{30}*\frac{1.29*10^{14}}{1.35*10^{11}}\\ =(2.8*1.9 / 1.35)\;\;\;*\;\;\;10^{30}*\frac{10^{14}}{10^{11}}\\ =(2.8*1.9 / 1.35)\;\;\;*\;\;\;10^{30+14-11}\\ =3.94074\;\;\;*\;\;\;10^{33}\\\)

You may try it, if you wish. If you do it right, and get a different answer, then I will pay you $1,000 Aus.