All Answers 358511 Answers

Sorry Chloe, I think you messed up a little.  

$$\\(3xy^2z^2)(-3x^3y^3z)^2\\\\
(3xy^2z^2)(9x^6y^6z^2)\\\\
27x^7y^8z^4\\\\$$

When you square root a negative number you get an imaginary number.

You probably don't know about those so you cannot square root a negative number - there are no REAL solutions.

Thanks Kitty :)

Not quite Chloe,

 $$\sqrt{9}=3$$

because 3*3=9

Really?  I didn't know that!

Thanks anon,

I'm not sure, is 4/9 a perfect square?  Perhaps someone would like to google that. 

Are you multiplying adding or what?

Multiply both sides by 7

I'll start you off

write it as a fraction and then mult top and bottom by 100 the simplify.

Thanks Alan.

Use the compound interest formula

FV=P(1+r)^n

It is the 2nd button at the top  

Hi chilledz3non,

Well the exact instantaneous velocity is given by h'(t) = 2t    so when t=2  velocity=2*2=4 units per time.

Now perhaps you are meant to do it from first principles so lets think about that.

Average velocity between t=some value, I'll call it t₁ and t=2 would be     $$\frac{f(t_1)-f(2)}{t_1-2}$$

The instantaneous velocity would be given by this as t₁ tends to 2

That is 

$$\\\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
f(t_1)=(t_1)^2-1\\
f(2)=2^2-1=3\\\\
$Instantaneous velocity at t=2 is $\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-1-3}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-4}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{((t_1)-2)((t_1)+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1-2)(t_1+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;(t_1+2)\\\\
=4$$

Maybe Alan or Heureka could tell me why my limits are not displaying properly.  

I'll answer by drawing some graphs:

Perhaps this graph might help:

Alan would you fancy answering this one?

108 = 2²*3³ so a/(108b) = 2⁴*3⁴*5/⁽2²*3³*b) = 2*2*3*5/b

This is only an integer if b is 1 or 2 or 3 or 4 or 5 or 6 or 10 or 12 or 15 or 20 or 30 or 60

f(x)=x^2e^-x/2, [-1,8]

$$\\f(x)=x^2e^{-\frac{x}{2}}, [-1,8]\\\\
f'(x)=2xe^{-\frac{x}{2}}+\frac{-1}{2}e^{-\frac{x}{2}}x^2\\\\
f'(x)=2xe^{-\frac{x}{2}}-0.5x^2e^{-\frac{x}{2}}\\\\
f'(x)=xe^{-\frac{x}{2}}(2-0.5x)\\\\
$stationary points when $ f'(x)=0\\\\
x=0\;\;or\;\;2=0.5x\\
x=0\;\;or\;\;x=4\\$$

So you now need to find the y values for 

$$x=0,\;x=4,\;x=-1\;\; and\;\; x=8$$

Then you will have your minimum and your maximum values for the given region.

c² - 9c + 8 = 0 can be factored as (c - 1)*(c - 8) = 0

So the two solutions are c = 1 and c = 8

nanometer = 10^-9 meters

angstrom = 10^-10 meters

Write it as y=1/x

f(x) = 1/x  x≠0

Angle of elevation = angle of depression.

sin(angle) = 30/40 

$${\mathtt{angle}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{30}}}{{\mathtt{40}}}}\right)} \Rightarrow {\mathtt{angle}} = {\mathtt{48.590\: \!377\: \!890\: \!729^{\circ}}}$$

First ask what's the probability that none of them have a birthday in common.  Put the individuals in order from first to 23rd.

The probability that the 2nd person doesn't have the same birthday as the first is 364/365.

The probability that the 3rd doesn't have the same birthday as either of the first two is 363/365.

The probability that the 4th doesn't have the same birthday as any of the first three is 362/365.

...

...

The probability that the 23rd doesn't have the same birthday as any of the others is 343/365.

Therefore the overall probability that no one has a birthday in common is obtained by multiplying all the above probabilities together.

This is 364*363*362*...*343/365²² = 364!/((365-23)!*365²²) = 365!/((365-23)!*365²³)

The probability that at least two people have the same birthday is therefore 1 - 365!/((365-23)!*365²³) as Melody has noted.