All Answers 358087 Answers

Thanks Chris,     

I wish I had time to do all your great questions Mellie   

It's 100.9

Thank you, CPhill and I value all your opinions  :)

However, Chris and I have very limited power on this forum.

The owner and developer of this site, Mr Massow,  is the one who makes all decisions such as these.

He is nearing the completion of a new forum version.  Hopefully, when it is implemented, many of our recent problems will be reduced very markedly.   

Here is a neat little graph.

you can play with the power too :)

I like that 

Heather Feather.

LOL     

$$\\(-5a^5b^4c^0)^3\\\\
=(-5a^5b^4)^3\\\\
=(-5)^3(a^5)^3(b^4)^3\\\\
=-125a^{15}b^{12}\\\\$$

Your question doesn't make sense - something is missing. 

4x²+2x-115.162049=0

Thanks Heureka,

That web site that you have referenced looks very interesting.

I know that there are a lot of features on Pascal's triangle that I am not aware of or that I do not retain in my memory.  So thank you.  I think I will store that site away in our reference material :)

thanks 

That's it !

how would I do this? A competitive high school swimmer takes 52 seconds to swim 100 yards. What is his rate in km/hr? (1 m = 1.0936 yards)

$$\small{\text{
$
\begin{array}{l}
\dfrac{100~\rm{yard}}{52~\rm{s} } \cdot
\dfrac{ 1~\rm{m} } { 1.0936~\rm{yard} } \cdot
\dfrac{ 1~\rm{km} } { 1000~\rm{m} } \cdot
\dfrac{ 60~\rm{s} } { 1~\rm{min} } \cdot
\dfrac{ 60~\rm{min} } { 1~\rm{h} } \\\\
=\dfrac{100\cdot 60\cdot 60}{52\cdot 1.0936\cdot 1000}\cdot \dfrac{ \rm{km}}{ \rm{h} } \\\\
=6.33053851781\cdot \dfrac{ \rm{km}}{ \rm{h} } \\\\
\approx 6.33 \cdot \dfrac{ \rm{km}}{ \rm{h} }
\end{array}
$}}$$

So x≈0.701838730514401 (this isn't the exact value, the number being irrationnal)

It's 5*500,000/100=$25,000

Example: It took me 3 hours to run 15 km: so my average velocity (v) is equal to: $$v=\frac {15}{3}=5$$

So my average velocity is equal to 5 km/h

Thanks EinsteinJr

The web2 calc will calculate log base 4 but most will not.

This is a more usual approach

4^2x=7

log4^2x=log7

2x*log4=log7

2x=log7/log4

x=log7/(2*log4)

the log can be any base so long as they are both the same base .

could you post an example equation please?

Congratulations ! You earned a cookie:

$$\boxed{ g=\dfrac{m \cdot G}{r^2} }$$

$$\\\small{\text{$g_{earth}=G*\dfrac{m_{earth}}{r_{earth}^2}\qquad G= 6.673 * 10^{-11} * \frac { N*\ m^2 } { kg^2 } \qquad m_{earth}=5.972\cdot 10^{24}\ kg\qquad r_{earth}=6.371\cdot 10^6\ m$}}\\\\
\small{\text{$g_{earth}=6.673 \cdot 10^{-11}*\dfrac{ 5.972\cdot 10^{24} }{ ( 6.371\cdot 10^6 )^2}\cdot \frac{N\cdot \ m^2 \cdot\ kg}{ kg^2\cdot\ m^2 } = \dfrac{ 6.673\cdot 5.972\cdot 10^{-11+24-12}} { 6.371^2 } \cdot \frac { N } { kg } $}}\\\\
\small{\text{$g_{earth}= \dfrac{ 6.673\cdot 5.972\cdot 10^{1}} { 6.371^2 } \cdot \frac { \dfrac{kg\cdot \ m }{s^2} } { kg } $}}\\\\
\small{\text{$g_{earth}= 9.81806072145\cdot \dfrac{ m }{s^2}$}}$$

You're welcome

thank you EJ     

That is quite complicated.  I am glad I have a trusty calculator  LOL

4^2x+1=8

4^2x=7

2x=log₄(7)

x=$$\frac{\log_4(7)}{2}$$