All Answers 359395 Answers

Are you sure the question is complete?  The graph below shows that all values in the region shaded in red (which extends up to infinity!) satisfies the condition ≥sin(x).

If tg(x) is restricted to the range 0 to 2pi as well, then it's all the red values that are shown above zero, stretching up to 2pi.

The answer is 132

What's the question?!!!

2^3= 8  

7^3= 343

8^3= 512

3^3 = 9

12^3 = 1728

9^3 = 729

28^3 = 21952

the serie will be then 2-7-8-3-12-9-28-29-52........

we take two terminal numbers if there is a repition 2 , 12 and 52

8 and 28

Not sure if this is according to Hoyle, but the most obvious??

56 - 34 = 22

You are quite right Anonymous!  I've now corrected my original.

wrong mistake made when multiplying out the brackets

x= -9/8

Another possibility:

 $$\frac{6}{3}+4*5=22$$

You haven't provided enough information for a unique answer!

6*4=24

5-3=2

24-2= 22

Because 3.14 is just a rounded version of pi (to 2 decimal places).

So pi is actually 3.14159265...etc. when 3.14 is just pi rounded to 2 d.p.

Congratulations Zegroes!   

I'm sorry blaster0 but I don't think it makes sense to me either.  Sorry.

Monday 18/5/14

This one is up for discussion:

You may find this talk by Adam Spancer interesting.

all you need to know is that sin-1 is exactly the same as asin.

That is inverse sine = arc sine

now just press 2nd sin like with any calc.

or

if you just leave your finger on the sin button a little longer than usual it will turn into arc sin.

I got it   

Melody and CPhill thank you very much and sorry for that thumbs 

There is an interesting novel about the Goldbach conjecture, called Uncle Petros and Goldbach's Conjecture by  Apostolos Doxiadis.  The publishers (Faber) offered a $1 000 000 prize to anyone who proved the theorem within 2 years of its initial publication (in 2000).  Needless to say, no-one claimed the prize!

I'm afraid this question reads like gobbledygook to me blaster0!! Have you typed it in correctly?

If this is log to the base 10, then log y = 3.8 means that y = 10^3.8

$${\mathtt{y}} = {{\mathtt{10}}}^{{\mathtt{3.8}}} \Rightarrow {\mathtt{y}} = {\mathtt{6\,309.573\: \!444\: \!801\: \!932\: \!494\: \!3}}$$

If it is log to a different base, say, b, then y = b^3.8.

Thanks

The final value here is given by

$${\mathtt{finalvalue}} = {\mathtt{6\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.09}}\right)}^{{\mathtt{4}}} \Rightarrow {\mathtt{finalvalue}} = {\mathtt{8\,469.489\: \!66}}$$

So final value = $8469.49

So the interest earned is 

$${\mathtt{8\,469.49}}{\mathtt{\,-\,}}{\mathtt{6\,000}} = {\mathtt{2\,469.49}}$$ dollars

Assuming the scores obey a normal distribution the following cumulative probability graph should help:

This is the cumulative probability for a normal distribution with mean 0 and standard deviation of 1.

Here we have z = (Joelie's test score - mean test score)/standard deviation

We can see from the graph that to get into the top 20% (ie reach a cumulative probability of 0.8) z must be at least 0.842.  So:

0.842 = (0.75 - 0.62)/stddev

$${\mathtt{stdev}} = {\frac{\left({\mathtt{0.75}}{\mathtt{\,-\,}}{\mathtt{0.62}}\right)}{{\mathtt{0.842}}}} \Rightarrow {\mathtt{stdev}} = {\mathtt{0.154\: \!394\: \!299\: \!287\: \!410\: \!9}}$$

So the standard deviation is 0.154 or 15.4%.

At least, I think this is what you are after!

20% is 60km

100% is 300km

80% is 240km

240km at 20km/hr is 12hours

12+2 :45= 14hours 45minutes.

let x=1:   $${\frac{{\mathtt{40}}}{{\mathtt{1}}}} = {\mathtt{40}}$$

let x=$$\frac{1}{10}$$=0.1: $${\frac{{\mathtt{40}}}{\left({\frac{{\mathtt{1}}}{{\mathtt{10}}}}\right)}} = {\mathtt{400}}$$

latex code: \frac{1}{10}

let x=$$\frac{1}{1000}$$=0.001: $${\frac{{\mathtt{40}}}{\left({\frac{{\mathtt{1}}}{{\mathtt{1\,000}}}}\right)}} = {\mathtt{40\,000}}$$

latex code: \frac{1}{1000}

let x=$$\frac{1}{10000000000}$$=0.0000000001: $${\frac{{\mathtt{40}}}{\left({\frac{{\mathtt{1}}}{{\mathtt{10\,000\,000\,000}}}}\right)}} = {\mathtt{400\,000\,000\,000}}$$

latex code: \frac{1}{10000000000}

let x=$$\frac{1}{100000000000000000000}$$=0.00000000000000000001:$${\frac{{\mathtt{40}}}{\left({\frac{{\mathtt{1}}}{{\mathtt{100\,000\,000\,000\,000\,000\,000}}}}\right)}} = {\mathtt{4\,000\,000\,000\,000\,000\,000\,000}}$$

latex code: \frac{1}{100000000000000000000}

...

let x=0=0.0000000000000000...: $$\boxed{\frac{40}{0}=\infty}$$

latex code: \boxed{\frac{40}{0}=\infty}

40/x  if x=0  is infinite!