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$$5^3$$

means

$$5\times5\times5=125$$

infinity x infinity  =  infinity

A tangent is also a conversional form that leads from and to unrelated subjects and ideas.

For example: someone starts talking about apples then that leads to pears then to bananas then to monkeys then to Chimps then to primates then to missing links then to certain members on this forum.

Each may just be touched on before moving on to the next.

Next time a conversional cotangent.

Find the cube root of both sides:

 (k^3)^(1/3)  =  (1/216)^(1/3)

                k      =  (1)^(1/3) / (216)^(1/3)  

                k      =  1 / 6

Since you have no x-term on the right side, just simplify the right side:  f(-2)  =  5.

If your function had been  f(-2)  =  -2x + 7,

then you would have replaced x with -2 to get:  f(-2)  =  -2(-2) + 7  =  4 + 7  =  11.

24m - 135m/(-3)*3

Do multiplications and divisions before additions and subtractions (negative divided by a negative is positive):

24m + 45m*3

24m + 135m

159m

1/x  =  0.049

Multiply both sides times x:

1  =  0.049x

Divide both sides by 0.049:

1/0.049  =  x

x  =  1000/49  or  20.408...

5(0.2 + 0.8)^10                 (do what is inside parentheses)

= 5(1)^10                          (do exponents before multiplication  1^10 = 1)

= 5 x 1

= 5

10-11d>-5d-4    add 5d to both sides

10 - 11d + 5d > -4        subtract 10 from both sides

-11d + 5d > -14           simplify

-6d > -14                    divide both sides by -6 and reverse the inequality sign

d < -14/-6

d < 7/3

After simplifying all the multiplications and divisions, you get:

Numerator:  (20) * log(5) * log(e) * log(2) * log(100)

Denominator:  log(20) * log(25) * log (16)

The change of base formula says that ln(20) = log(20) / log(e)

         so log(e) / log(20)  =  1 / ln(20)

Now, the numerator that we have left is:  20 *  log(5) *  log(2) * log(100)

the denominator is:                                   ln(20) *  log(25) * log(16)

Now, log(100)  =  2   and    log(25)  =  log(5^2)  =  2log(5)    and    log(16)  =  log(2^4)  =  4log(2)

Making these substitutions, we have:

Numerator:    20 * log(5) * log(2) * 2

Denominator:  ln(20) * 2 * log(5) * 4 * log(2)

Cancelling, we end with:  5/ln(20)

It means

1.0476e-9

$$1.0476\times 10^{-9}$$

The second last column.

Are you blind?

Which bit is sqare rooted?  Just the 2 or more than that?

anyway, just use the logarithms formulas on the formulary page - It's one of the tabs at the top of the page.

Fri 24/10/14

1) This simultaneous equations question was different - I liked it 

ok lets think about this

$$\\a^2+ab = 15 \qquad (1a)\\
a(a+b)=15 \qquad (1b)\\
a=\frac{15}{a+b} \qquad\qquad (1c)\\\\\\
2b^2 = 20 - 2ab\qquad (2a)\\
b^2=10-ab\qquad (2b)\\
b^2+ab=10\qquad (2c)\\
b(b+a)=10\qquad (2d)\\
b=\frac{10}{b+a}\qquad\qquad(2e)\\$$

So if

$$\\If\;\;a+b=35\;\;then\;\;a=15/35\;\;and\;\; b=10/35\\\\
If\;\;a+b=25\;\;then\;\;a=15/25\;\;and\;\; b=10/25\\\\
If\;\;a+b=-25\;\;then\;\;a=-15/25\;\;and\;\; b=-10/25\\\\
If\;\;a+b=-5\;\;then\;\;a=-15/5\;\;and\;\; b=-10/5\\\\
simplifying\;\; each \;\;of \;\;these\\\\\\
If\;\;a+b=35\;\;then\;\;a=3/7\;\;and\;\; b=2/7\\
BUT\qquad 3/7+2/7\ne35\;\; So\;\; no\;\; good\\\\
If\;\;a+b=25\;\;then\;\;a=3/5\;\;and\;\; b=2/5\\
BUT\qquad 3/5+2/5\ne25\;\; So\;\; no\;\; good\\\\
If\;\;a+b=-25\;\;then\;\;a=-3/5\;\;and\;\; b=-2/5\\
BUT\qquad -3/5+-2/5\ne-25\;\; So\;\; no\;\; good\\\\
If\;\;a+b=-5\;\;then\;\;a=-3\;\;and\;\; b=-2\\
AND\qquad -3+-2=-5\;\; So\;\; this\;\;answer\;\;is\;\; good\\\\$$

So a+b could be worth -5

-5 [5-(-9)] = -5[5 + 9] = -5[14] = -70   ......remember....a negative outside parentheses or brackets just changes the signs on the terms inside

1 mol silver contains 6.022*10^23 atoms

So 7.3 moles contain

7.3 x (6.022 * 10^23) atoms = 4396060000000000000000000 atoms = 4.39606 x 10¹⁹ atoms

2x^2+12x+3=0

This polynomial can't be factored.....it does have "real" solutions, though...usng the on-site solver, we have

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{5.738\: \!612\: \!787\: \!525\: \!830\: \!6}}\\
{\mathtt{x}} = -{\mathtt{0.261\: \!387\: \!212\: \!474\: \!169\: \!4}}\\
\end{array} \right\}$$

yes is he walking or traveling by jet?

55 candies would be left because 700-645=55

wow correct anwser (Y) u r genius :)

Please present you question better - I cannot be bothered with all that scrolling. 

since your question has not been answered i assume I am notthe only one that feels this way.

(70x²)/(100x)  =  (70/100)(x²/x)  =  (7/10)(x)  =  (7x)/10

$$16+3x(-5)=16-15x$$

If this represents constant or smooth acceleration, then:

             acceleration = (final velocity - beginning velocity) / time

                               a  =  ( 15 m/sec - 24 m/sec ) / 12 sec

                              a  =   (-9 m/sec)/12 sec  =  -0.75 m/sec²