All Answers 358156 Answers

$$v\ne0$$

$$0.1=\frac{6}{v^2}+0.0005v^2-0.033\\\\
0.133=\frac{6}{v^2}+0.0005v^2\\\\
0.133v^2=6+0.0005v^4\\\\$$

substitute x for v²

$$0.0005x^2-0.133x+6=0\\$$

$${\mathtt{0.000\: \!5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.133}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{133}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5\,689}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{5\,689}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{133}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{57.574\: \!540\: \!107\: \!467\: \!690\: \!7}}\\
{\mathtt{x}} = {\mathtt{208.425\: \!459\: \!892\: \!532\: \!309\: \!3}}\\
\end{array} \right\}$$

 $$v^2=57.574540\:\:\rightarrow v=\pm7.58779\:\: \mbox{approx}$$   or

$$v^2=208.42545989\:\:\rightarrow\:\:v=\pm14.4369\:\:\mbox{approx}$$

I think that is all okay.

$${\mathtt{0.12}}{\mathtt{\,\times\,}}{\mathtt{4\,115\,267}} = {\mathtt{493\,832.04}}$$

A really food answer.

but the final answer should have the x first.

x=-3

huh??????????

You need to provide some numbers

PLUS

The SA above is not quite right.

$${\mathtt{SA}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,\times\,}}{\mathtt{h}}$$

should be

$${\mathtt{SA}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,\times\,}}{\mathtt{h}}$$

$$$$

Melody, the second one is essentially the same as the first and the fourth is the same as the third.  The minus sign would be part of the constant that you need to turn the proportionality into an equality.

x^2-11x+30 

to factorise this you need two numbers that multiply to +30 (they must have the same sign)

and add to -11

ok -5 and -6 work

(x-5)(x-6)=(x+a)(x-6)

a=-5

$$x\:\:\alpha\:\: y\:\:\mbox{direct variation}\\

x\:\:\alpha\:-y\\

x\:\:\alpha\:\: \frac{1}{y}\:\:\mbox{inverse variation}\\

x\:\:\alpha\:\: \frac{-1}{y}$$

I get confused with variation - I think the 1st and 3rd are right but I am not too sure about the others.

Does someone want to give me a lesson please?

I have an error to my answer. The 2 should be changed from the coefficent but it should be multiplyed to the existing exponent. The ln and e would still cancel, but your answer would be six.

Thanks Alan,

You have used 150 instead of 1500 Melody.

Maybe it is a sequence in a different base (not10) 

Where does that online book of sequences hang out.  Yes there really is such a thing.

Can anyone find it easily?

We need some numbers here!

Thank you anonymous.  This has been a really interesting question.

Thank you to Chris and Kitty as well!

$$\frac{3}{150}=\frac{x}{11721722}$$

$${\frac{{\mathtt{3}}}{{\mathtt{150}}}}{\mathtt{\,\times\,}}{\mathtt{11\,721\,722}} = {\mathtt{234\,434.44}}$$

$$x=$$234434.44

Our answers appear to be different by a power of 10.

The question asker can decide which answer is correct.  One of us has made a silly mistake i guess.

THIS IS WRONG - SEE ALAN'S COMMENT.

n=3: the number of diagonals = 0 + 1 + 2   -  3

n=4: the number of diagonals = 0 + 1 + 2 + 3  -  4

n=5: the number of diagonals = 0 + 1 + 2 + 3 + 4  -  5

n=6: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 -  6

...

n: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + (n-1)  -  n

the number of diagonals = $$\frac{(1+(n-1))*(\;((n-1)-1)\;+\;1\;)}{2} - n = \frac{n(n-1)}{2} - n=\binom{n}{2}-n$$

the number of diagonals = $$\boxed{\binom{n}{2}-n=\frac{n(n-3)}{2}}\quad n\ge3$$

S.

If you are talking about a function then the range is the set of all  possible y values.

If you are talking about statistics then it is the highest score - the lowest score.

If the number in question is 39^{(46252+73)*4733} then, since the unit value of powers of 39 alternate between 1, if it's an even power, and 9, if it's an odd power we have: (46252+73)*4733 =  219256225 is odd, so the value in the units position of 39^{(46252+73)*4733} is 9.

If the number in question is 39⁴⁶²⁵²+73⁴⁷³³ then the unit digit of the first term will be 1 (because of the even power) and we need to find the unit digit of the second term.

The unit digit of powers of 73 run through the sequence 1, 3, 9, 7 depending on the remainder of the power after dividing by 4 (0,1,2, 3 respectively).  Since the remainder of 4733 when divided by 4 is 1, the unit digit place of 73⁴⁷³³ is 3.  

When the two terms are added together therefore, the value in the units position of 39⁴⁶²⁵²+73⁴⁷³³  is 1+3 = 4.

Or, you could just put it straight into the web2 calc using the "math (input=Result)" tab.

$${\mathtt{6}}{\mathtt{\,\times\,}}{\left({\mathtt{9.36}}{\mathtt{\,-\,}}{\mathtt{8.974}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{7.65}}{\mathtt{\,-\,}}{\mathtt{8.974}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{9.1}}{\mathtt{\,-\,}}{\mathtt{8.974}}\right)}^{{\mathtt{2}}} = {\mathtt{6.200\: \!532}}$$

Just type this, exactly as written, into the calculator on the Home page here.

I take it C is the angle, and you are talking about a triangle in which case you can use the Cosine rule in the form

$$c^2=a^2+b^2-2abcos(C)$$  to find c.

$${\mathtt{c}} = {\sqrt{{{\mathtt{192}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{291}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{192}}{\mathtt{\,\times\,}}{\mathtt{291}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{41.3}}^\circ\right)}}} \Rightarrow {\mathtt{c}} = {\mathtt{193.896\: \!211\: \!065\: \!943\: \!792\: \!5}}$$

or c ≈193.9

I'll leave you to use the sine rule $$\frac{sin(A)}{a}=\frac{sin(C)}{c}$$ to find A, and similarly for B (or just use the fact that the three angles must add to 90° to get the last one). 

Good try Anonymous, but, unfortunately, in your answer you forgot the warning about being careful with the units!  P is in millions, so we have:

1103.4*e^0.0149t = 1500

Divide both sides by 1103.4

e^0.0149t = 1500/1103.4

Take log to base e of both sides

0.0149t = ln(1500/1103.4)

Divide both sides by 0.0149

t = ln(1500/1103.4)/0.0149

$${\mathtt{t}} = {\frac{{ln}{\left({\frac{{\mathtt{1\,500}}}{{\mathtt{1\,103.4}}}}\right)}}{{\mathtt{0.014\: \!9}}}} \Rightarrow {\mathtt{t}} = {\mathtt{20.608\: \!643\: \!372\: \!615\: \!506}}$$  years

This is 21 years to the nearest year, so it will be 2005+21 = 2026 when India's population reaches 1500million.

$$We have $$\sqrt{(1-cos(x))(1+cos(x))} = \sqrt{1-cos^2(x)}=\sqrt{sin^2(x)}=sin(x)$$

There are an infinite number of solutions to this as you can probably infer from the graph below:

However, if you mean within the range 0 to 360° then use the calculator in the form x = asec(2.5) to find the first one:

$${\mathtt{x}} = {asec}{\left({\mathtt{2.5}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{66.421\: \!821\: \!521\: \!798^{\circ}}}$$

If you look carefully at the graph above you will see that the next one is obtained by subtracting this from 360°

$${\mathtt{22}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}} = {\mathtt{41}}$$