The Question İs 415=5x^3

First of all

5x^3=5x.5x.5x

5x^3=125x

İf 415=125x, What is X 

$${\frac{{\mathtt{415}}}{{\mathtt{125}}}}$$=$${\frac{{\mathtt{125}}}{{\mathtt{125}}}}$$

X= 415/125=3,32

$$\\2^m = 0.04\\
\mbox{when ever you are solving for a power you use logs (unless it is a very easy one)}\\\\
log\;2^m = log 0.04\\\\
mlog\;2 = log 0.04\\\\
m=\frac{log 0.04}{log\;2}$$

And you can finish it 

$$g(x)= \frac{3x+4}{x-1}\\

We are looking for the domain.\\

Well we can't divide by 0 so x cannot be 1, I can see no other restrictions.\\


Domain: x\;\in R \quad where \quad x\ne1$$

Do you want me to use calculus or physics formulas?

Calculus:

Initially  

 $$\\\ddot y=-9.8\;m/s^2, \qquad \dot y=0\;m/s, \qquad y=78.4\;m \\\\
\mbox{Find t when y=0}\\\\$$

Ongoing:

$$\\\ddot y=-9.8\;m/s^2\\\\
\ddot y=-9.8t+C_1 \\\\
c_1 =0 \qquad \mbox{because of initial conditions therefore} \\\\
\dot y=-9.8t\\\\
y=\frac{-9.8t^2}{2}+C_2\\\\
C_2 =78.4 \qquad \mbox{because of initial conditions therefore} \\\\
y=-4.9t^2+78.4$$

find t when y=0

$$\\0=-4.9t^2+78.4\\\\
-78.4=-4.9t^2\\\\
16=t^2\\\\
t=4\;seconds$$

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

If you want to do it by physics formula you will use         $$s=ut+\frac{1}{2}at^2$$

and you will get the same result.    

What İs The Question ? 

Not Problem Melody

I just tell of Answer

:)

Thanks Amoonstar, 

You could just do it as 24/100 * 215    

The calculator is on the home page.  

First of all we must find %1

For This

İf %24 of 215, %1=?

$${\frac{{\mathtt{215}}}{{\mathtt{24}}}} = {\mathtt{8.958\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

İf %1=8.958333, What is the %100

Question of us wants it, What is the %100

İf 1 piece is 8.958333, What is the 100 piece

For This

$${\mathtt{8.958\: \!333}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{895.833\: \!3}}$$

The Result is 895.83¯

Wed 10/9/14      ♬  

*1) limits (Thanks Alan)

@@ End of Day Wrap:      Tues 9/9/14        Sydney, Australia      Time 10:45pm     ♬

Hi all,

Our great answerers today were CPhill, HaleyWoodall17, AzizHusain, Lola200, Alan, DragonSlayer554, Nba2k14, Pumpkinguy, NinjaDevo, ThAnswerGuy, Kitty<3, Milkshake, Heureka and Demogorgon.  I even answered a few myself LOL.  Thanks, you are all great!     

Our new answerers today were ThAnswerGuy, Nba2k14 and Milkshake.  Artorias is new too.  He/she didn't answer any questions but he/she introduced themselves on the forum in a special post.  I thought that was nice so I have included this new member too.   Anyway we are very pleased that you have all joined us and we hope that all your experiences here are really good ones.   

Now for some interest posts.  I pulled out quite a few today.   

*1) Using the web 2 calc to find the equation of a quadratic through 3 given points. (Thanks Alan)

İf you divided by 3 to 4

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}} = {\mathtt{0.75}}$$

The Result is 0,75

Great answers Answer Guy!

I really should look at some of these tutorials as well. !  thanks Chris.    

Hey AnswerGuy, you can't just leave it at that.  That doesn't educate me or anyone else!

Please tell us what a Hardy Ramanujan number is!    Thank you.

Just in case my tone comes across wrong, I am not cranky LOL.      I am just curious.      

1729 is the smallest Hardy Ramanujan number

what is the domain and range of $$\sqrt{81-x^2}$$

   $$\\81-x^2\ge0\\
81\ge x^2\\
x^2 \le 81\\
-9\le x \le 9 \qquad \mbox{The the domain is [-9,9]} \\$$

The square root of anything is greater or equal to 0  therefore  range does not include less than 0.

$$\sqrt{81-x^2} = \sqrt{81-positive\; number} \;\le \sqrt{81}\le9$$

$$Therefore the range is [0,9]$$

Parallel lines have the same slope.

So that,

for Ax + By = C, we have y = (-A/B)x + C/B, where the slope is -A/B

for Ax + By = D, we have y = (-A/B)x + D/B, where the slope is -A/B

Hence, the two lines are parallel.

Note that:

if A = B = 0, y = 0/0 + C/0 (or D/0). The lines are undefined.

if C = D, it is a single line crossing the ordinate axis at the same point.

well for the pearl one to get the most you cna you draw 2500 of each one with one white pearl left over.

does that count?

as an exponent it would be, 10^13 km  does that help you?

If we assume the density of asphalt is about 2000kg/m³, and 1 ton = 907kg then

volume = 907/2000 m³

For a thickness of 166mm (=0.166m) the area covered is

$${\mathtt{area}} = {\frac{{\mathtt{907}}}{\left({\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{0.166}}\right)}} \Rightarrow {\mathtt{area}} = {\mathtt{2.731\: \!927\: \!710\: \!843\: \!373\: \!5}}$$

or area ≈ 2.7 m²