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$$\\-8+\frac{3}{4}*3\\\\
-8+\frac{9}{4}\\\\
-8+2+\frac{1}{4}\\\\
-6+\frac{1}{4}\\\\
-5\frac{3}{4}\\\\$$

the button that you are describing anon looks like this

$${{\mathtt{x}}}^{{\mathtt{y}}}$$

$$3\times 10\;\; [x^y] \;\;8\;=\\\\$$

on the web2 calc you can just type in  

3*10^8                           (Just like mathematician said)  

$$\\f(x) = \frac{x^2}{5} + 2x\\\\
f'(x) = \frac{2x}{5} + 2\\\\
f"(x) = \frac{2}{5}\\\\$$

If a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

$$\\\small{\text{
$
x_{1,2}
=\dfrac{-4\pm\sqrt{16-4*(-k)}}{2}
=\dfrac{-4\pm\sqrt{16+4k}}{2}
=\dfrac{-4\pm\sqrt{4(4+k)}}{2}
$
}}\\\\
\small{\text{
$
x_{1,2}=\dfrac{-4\pm 2\sqrt{4+k}}{2}
=-2\pm \sqrt{4+k}
$
}}\\\\
\small{\text{
$
a=-2+ \sqrt{4+k} \quad | \quad a > b
$
}}\\
\small{\text{
$
b=-2- \sqrt{4+k}
$
}}\\
\small{\text{
$
a-b=-2+ \sqrt{4+k} -( -2- \sqrt{4+k} )=-2+ \sqrt{4+k} + 2 + \sqrt{4+k} )=2\sqrt{4+k}
$
}}\\
\small{\text{
$
a-b =2\sqrt{4+k}
$
}}$$

41=16/3% of X

$$\\41=\frac{16}{3}\div 100 \;\; * \;\; X\\\\
41=\frac{16}{3}\times \frac{1}{100} \;\; * \;\; X\\\\
41=\frac{16}{300} \;\; * \;\; X\\\\
41\times\frac{300}{16} =\frac{16}{300}\times\frac{300}{16} \;\; * \;\; X\\\\
41\times\frac{300}{16} =\; X\\\\
X=41\times\frac{300}{16} \\\\$$

$${\mathtt{41}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{300}}}{{\mathtt{16}}}}\right) = {\frac{{\mathtt{3\,075}}}{{\mathtt{4}}}} = {\mathtt{768.75}}$$

That is what inverse sin or arcsin is for

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.866\: \!025\: \!403\: \!784}}\right)} = {\mathtt{59.999\: \!999\: \!999\: \!95^{\circ}}}$$

$$Y(Y)^5=Y^1*Y^5=Y^{1+5}=Y^6$$

$$\\\frac{x+3}{10}=\frac{2}{5}\\\\
10\times\frac{x+3}{10}=10\times\frac{2}{5}\\\\
x+3=2\times\frac{2}{1}\\\\
x+3=4\\\\
x=1$$

OR

$$\\x+\frac{3}{10}=\frac{2}{5}\\\\
x+0.3=0.4\\
x=0.4-0.3\\
x=0.1\\$$

$$\left({\frac{{\mathtt{25}}}{{\mathtt{100}}}}\right){\mathtt{\,\times\,}}{\mathtt{450}} = {\frac{{\mathtt{225}}}{{\mathtt{2}}}} = {\mathtt{112.5}}$$

$$\\.5=(.85)^t\\\\
log(0.5)=log(0.85)^t\\\\
log(0.5)=t*log(0.85)\\\\
t=\frac{log(0.5)}{log(0.85)}\\\\$$

$${\frac{{log}_{10}\left({\mathtt{0.5}}\right)}{{log}_{10}\left({\mathtt{0.85}}\right)}} = {\mathtt{4.265\: \!024\: \!281\: \!798\: \!725\: \!7}}$$

I have not seen mathematician's method before either but I do like it!!

Rosala, I think this question could be rephrased as follows:

Nikola can complete N units of work on her own in 15 days.

Nikola and Ilia together can complete N units of work in 10 days.

In how many days could Ilia complete N units of work on her own?

Note that Nikola works at a rate of N/15 units of work per day.

Assume Ilia works at a rate of N/T units of work a day.  The idea is to find the value of T.

(N is unknown of course.)

.

.

Do you mean:

a)  $$(\ln(x))^2$$

or 

b) $$\ln{(x^2)}$$

and

c) $$\frac{1}{x(1+4etc.)}$$

or

d) $$\frac{(1+4etc.)}{x}$$

.

dont know about the rhymes...but i have a webpage for if you learn it...you'll never forget it!

http://web2.0calc.com/formulary/math/algebra

this question is kind of confusing me ...was their any other info given?

This would be written out as 

(a-5) X 2

a = "your number"

everything between the words "The product" and the next "and" should be grouped together in brackets as they are above.

The answer is 61.

In detail it is 61. not 61

You need to put the decimal place after the 61 or it could be considered, by some, to be ambiguous.

First lets find how many 15 minutes are in 1 hour

1 hour = 60 minutes

so 60/15 = 4 

at this point there are 4 "15 minutes" in 1 hour

Now lets find out how many are in 8 hours

take the 4 and multiply it by the total hours(8)

so 4 x 8  which = 32

There are 32 "15 minutes" in an 8 hour period

alrighty, so to make this easier forget about the denominator for now

here we start by multiplying the -5 by -8

so -5 x -8 = 40

then we attach that denominator back on the bottom

so  40/4

simplify divding the 40 by 4

so 40/4 = 10

and you are done.     Your answer is 10.

The length of a leg  = 6    ... proof....

Since the lengths of the two legs are equal, we have

√[6^2 + 6^2 ] = r

√[2 *6^2] = 6√2  = r = the hypoteneuse

We need to use the chain rule, here

We have two fucnctions...the "outside' function," arctanh(cosx) 

And the "inside' function," cos x

Let cos x = u  and     u' = -sinx

The derivative of arctanh(u)  is just      1 /( 1 - u^2)

And the derivative of u is just u' 

And we can think of Y' as being the "derivative of the outside function" times the "derivative of the inside function"....so we have...making the substitutions

Y' = 1/[1 - cos^2(x)] * (-sin(x))   = -sin(x)/ [sin^2(x)] = -1/sin(x) =-csc(x)

And we can just "tack" the "5" back on....so we end up with

Y'[5arctanh(cosx)] = -5csc(x)

Hope that helps...!!!

(x^2 -1) / 3  * 2/(x^2 - x)   let's "swap' denominators

(x^2 -1) / (x^2 -x) * 2/3    factoring the fraction on the left on top and bottom, we have

[ (x -1)(x + 1)] / [x (x -1)] *2/3       note.... the (x-1) terms "cancel"...and we're left with

(x + 1)/x  *  2/3 =

[2(x+ 1)] / [3x)  =  (2x + 2) / (3x)

@Mathematician

Can you explain it more simple?

Sorry 

Whenever you multiply fractions, just multiply the denominator by the denominator and the numerator by the numerator: (x^2 - 1) * (2) = 2x^2 - 2 and 3 * (x^2 - x) = 2x^2 - 2x, so therefore:

(x^2-1)/3 * (2)/(x^2-x) = (2x^2 - 2)/(2x^2 - 2x)

The 2x^2's cancel eachother out and so do the -2's. So, your simplified expression is 1/x.

Which trig function