All Answers 356032 Answers

3!!!

=6!!

= 720!

= a number too big for the calc to disply but it is not infinity.  

I wonder if there is a tricky way to present this in scientific notation?

2010 - 2003 = 7 years elapsed

7/3 =2 1/3 number of doublings

2^(2 1/3) =5.04 x 100 =504% increase in price of gold in 7 years.

504% ^(1/7) =5.04^(1/7) =~26% per year - increase in gold price.

Next year's price =This year's price x 1.26

Constant of decay, when half-life is known, is always:

ln(1/2) / half-life =ln(1/2) / 8 =-0.08664......[constant of decay is always negative].

Yes it is a nice clip :)

It's possible this question is looking for the cube root of the expression in brackets, in which case we have:

.

4(Z-4)/6 + 3Z-(1/6)=3/6

\(\frac{4(Z-4)}{6} + 3Z-\frac{1}{6}=\frac{3}{6}\\ \frac{4(Z-4)}{6} + \frac{18Z}{6}-\frac{1}{6}=\frac{3}{6}\\ \frac{4(Z-4)+18Z-1}{6} =\frac{3}{6}\\ 4(Z-4)+18Z-1=3\\ 4Z-16+18Z-1=3\\ 22Z-17=3\\ 22Z=20\\ 11Z=10\\ Z=\frac{10}{11}\)​

simplfy this expression. sqrt3(((8*x^3)*yx^6)*z^3)

\(\begin{array}{|rcll|} \hline && \sqrt{3}\cdot \{~[(8\cdot x^3)\cdot yx^6]\cdot z^3~\} \\ &=& \sqrt{3}\cdot 8 \cdot x^9 \cdot y \cdot z^3 \\ \hline \end{array} \)

Light of wavelength 619 nm passes through a double slit and makes its first max (m=1) at ab abgke if 0.226 deg. What is the separation of the slits, d, IN MILLIMETERS? Rearrange and solve, please. 

\(d\sin(\theta)= m\lambda\)

\(\begin{array}{rcll} Let\ \lambda &=& 619\ nm \\ \lambda &=& 619 \cdot 10^{-9}\ m \\ \lambda &=& 619 \cdot 10^{-6}\ mm \\\\ Let\ \theta &=& 0.226^{\circ} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline d\sin(\theta) &=& m \lambda \\ d &=& m\cdot \frac{\lambda}{\sin(\theta)} \quad & | \quad m=1 \\ d &=& 1\cdot \frac{619 \cdot 10^{-6}\ mm }{\sin(0.226^{\circ})} \\ d &=& 156929.997726 \cdot 10^{-6} \ mm \\ d &=& 0.156929997726 \cdot 10^{6}\cdot 10^{-6} \ mm \\ d &=& 0.156929997726 \ mm \\ \hline \end{array} \)

580 nm light shines on a double slit with d=0.000125 m.
What is the angle of the third dark interference minimum(m=3)?

\(\begin{array}{rcll} Let\ d &=& 0.000125\ m \\ d &=& 125 \cdot 10^{-6}\ m \\\\ Let\ \lambda &=& 580\ nm \\ \lambda &=& 580 \cdot 10^{-9}\ m \\ \end{array} \)

\(\begin{array}{|rcll|} \hline d\sin(\theta) &=& m \lambda \\ \sin(\theta) &=& m\cdot \frac{\lambda}{d} \quad & | \quad m=3 \\ \sin(\theta) &=& 3\cdot \frac{580 \cdot 10^{-9}\ m }{125 \cdot 10^{-6}\ m} \\ \sin(\theta) &=& 3\cdot \frac{580}{125} \cdot 10^{-9+6} \\ \sin(\theta) &=& \frac{1740}{125} \cdot 10^{-3} \\ \sin(\theta) &=& 13.92 \cdot 10^{-3} \\ \sin(\theta) &=& 0.01392 \\ \theta &=& \arcsin(0.01392) \\ \theta &=& 0.79758300970^{\circ} \\ \hline \end{array} \)

Call A the original amount of the substance....and we have this equation

.5A = Ae^(-k*t)     where k is the decay factor and t is the time in days = 8

........divide both sides by A

.5 =  e^(-k*8)     take the ln of both sides

ln (.5)   = ln e^(-k*8)     and we can write

ln (.5)  =( -k*8)       and ln e   = 1   so we can ignore this

ln(.5)  = -k * 8        divide both sides by -8

-ln(.5) / 8  = k ≈ 0.09 [rounded ]

if x+ 1/x= 2, what is x2 + 1/x2

if  \(x+ \frac{1}{x}= 2\) , what is \( x^2 + \frac{1}{x^2}\)

\(\begin{array}{|rcll|} \hline x+ \frac{1}{x} &=& 2 \quad & | \quad \text{square both sides} \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 2^2 \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 4 \\ x^2 +2\cdot x\cdot \frac{1}{x} + \frac{1}{x^2} &=& 4 \\ x^2 +2 + \frac{1}{x^2} &=& 4 \quad & | \quad -2 \\ x^2 + \frac{1}{x^2} &=& 4-2 \\ \mathbf{x^2 + \frac{1}{x^2}} & \mathbf{=} & \mathbf{2} \\ \hline \end{array} \)

At what point do the graph of the equations shown below intersect?

-8x+2y=4

6x+3y=-9

Intersection at  point (-0.83333333333,-1.33333333333) or (\( -\frac{5}{6},-\frac{4}{3}\) )

l remember reading "The Adventures of Tom Sawyer"... in third grade. lt's amazing how much literature can open your eyes.

Thanks, HSC.......Twain is one of my favorite writers........

Hi Rarinstraw,

You have tried to do it on simple interest but your answer is too little becasue you have only charged interest for 1 year, not for three.

a car costing 5995. financed at 9% reducing interest for 36 mos will cost what per month?

C=??

PV=5995

n=36

i=0.09/12 = 0.0075

\(5995=C*\frac{1-1.0075^{-36}}{0.0075}\\ C=5995\div (\frac{1-1.0075^{-36}}{0.0075})\)

5995/((1-1.0075^-36)/0.0075) = $190.64   per month

l mean if you look up the library of Babel... you'll find that number of pages based on any combination of letters, symbols, and punctuatution. Anything and everything stated is not original.

Neon Green.

Oh, right. About 9. Or 11.

Thinking about it now I have finished I think guests answer is a better interpretation of your meaning. But you could mean either.  :)

3(-2(-1)-3)to the second power

\([3(-2(-1)-3)]^2\\ =[3*(-2*-1-3)]^2\\ =[3*(2-3)]^2\\ =[3 *-1]^2\\ =[-3]^2\\ =9\)

Thanks for the tip!

Here's a tip for these types of problems that may prevail later on. Set the denominator eual to zero and try to factor the denominator.

(5995+(5995*9%))/36 = 181.5152777777777777777777777777777777777777777777777777777777777778

OR, in shorter numbers

~$181.52

100

Reason: 10^{log(100, 10)} 

10 to the power of log base 10 cancels

100 is all you are left with.