All Answers 358173 Answers

raise both sides to the power of (1/73035)

1.0000157^(1/73035) = b

ive just cured a headache and woah im getting one again!

whosoever invented the calc , shouldnt have invented it before inventing a common sense dose!

Thanks Heureka  

I was thinking that I could go live on a mountain to lose some weight (Mt Everest is 29000 feet) but Heurika has determined that the diffenence will be 0.4 of a pound so it is probably not going to be worth the moving expense. How sad.           

$$\sqrt{-2}=\sqrt{(-1)(2)}=\sqrt{-1}\sqrt{2}=\imath\sqrt{2}.$$

$$\sqrt{-35}=\imath\sqrt{35},$$    etc.

Here is one wire but there  are 3 of these placed around each tree to hold it straight.

And there are 6 trees.

Can you work it out now?

i is the base unit of the imaginary number system.

 $$i=\sqrt{-1}$$

infinity 

that is the answer I do believe.

$$\\log_{10}10^9 = 9\\
log_{10}10^{30}=30\\
etc$$

I like your graph Alan.   :)

what alan has done is  

$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\
$you can take this another step$\\\\
$Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\
^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$

What about the curved bit anon?

find the perimeter of semi circular plate of radius 3.85cm

$$\\P=0.5* 2\pi r\;\; +\;\; 2r$$

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{3.85}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3.85}} = {\mathtt{19.795\: \!131\: \!716\: \!320\: \!704}}$$

P = 19.8  cm

do you mean

$$\\(a^3)^2\qquad if\;\; a = 7?\\\\\\
(7^3)^2=7^6$$

$${{\mathtt{7}}}^{{\mathtt{6}}} = {\mathtt{117\,649}}$$

do you get 11% each year or 11% at the end of 10 years. ?

The question is not clear.   

One universal symbol for exponent is ^

5^6

looks like this

$${{\mathtt{5}}}^{{\mathtt{6}}} = {\mathtt{15\,625}}$$

what is 460sqm as a percentage of 174,320sqm?

460/174320 *100   then put a percent sign.  

On the web2 calc it usually comes out as both. 

I don't know what tital is :P
If you mean total volume them it's as simple as:

$${\mathtt{120}}{\mathtt{\,\times\,}}{\mathtt{120}}{\mathtt{\,\times\,}}{\mathtt{80}} = {\mathtt{1\,152\,000}}$$

$${\mathtt{1\,152\,000}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{11\,520\,000}}$$cm3

11 520 000cm3 = 11 520 dm3 = 11,52 m3

It's the same as asin.

Press shift and and click sin

it does not matter what the original population is.

$$\\1.03=1(1+r)^{10}\\\\
1.03=(1+r)^{10}\\\\
log(1.03)=log(1+r)^{10}\\\\
log(1.03)=10log(1+r)}\\\\
10^{\left(\frac{log(1.03)}{10}\right)}=10^{log(1+r)}}\\\\
10^{\left(\frac{log(1.03)}{10}\right)}=1+r\\\\
r=10^{\left(\frac{log(1.03)}{10}\right)}-1\\\\$$

$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\mathtt{1.03}}\right)}{{\mathtt{10}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.002\: \!960\: \!253\: \!145\: \!653\: \!4}}$$

so the yearly rate of increase is approximately    0.296%

I've drawn the graph below - see if it helps (I've assumed you meant L to be 4x-3y+60=0).

.

sorry, I made a mistake on "have you learnt these: i=1, i^2=-1,i^3=-1,i^4=1??? " IT SHOULD BE" i=i, i^2=-1,i^3=-i,i^4=1

i is stande for a negative square root of a number 

such as 2i=$${\sqrt{-{\mathtt{2}}}}$$ , 35i=$${\sqrt{-{\mathtt{35}}}}$$ 135i=$${\sqrt{-{\mathtt{135}}}}$$e.t.c, coz when you type negative square root of something to the calculator , "error" will be shown , the mathematician design a symbol of negative square root of a number, 

have you learnt these: i=1, i^2=-1,i^3=-1,i^4=1??? 

Just type 2^31+1 in the calculator

$${{\mathtt{2}}}^{{\mathtt{31}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{2\,147\,483\,649}}$$

.

perimeter of semi circular plate =2r (r=radius)=2*3.85=7.7cm 

the diameter doesn't change no matter the circular plate is half or not!

Melody's answer assumes the probability is of the number appearing at least once.

To calculate the probability of it appearing exactly once, use (n/6)(5/6)^n-1.

The two probabilities are compared as a function of n in the graph below:

(The final sentence in the question suggests that Melody's answer is what was being sought.)

.

If I weigh 141.2lbs at sea level what will I weigh 29,000ft above sea level ?

29000 ft = 8839.2 m

Only Free-air correction: 

Accounts for the 1/r^2 decrease in gravity with the distance from the center of the Earth:

$$g_1=\boxed{g_{(\text{ see level })}=\dfrac{GM_E}{R_E^2}}$$ 

and in 8839.2 m

$$g_2=\boxed{g_{(\ 29000\ ft \text{ above sea level})}=\dfrac{GM_E}{( R_E +8.8392\ km )^2}}$$

$$F_2 = F_1\cdot \dfrac{g_2}{g_1}\cdot \dfrac{\not{m}}{\not{m}}$$

$$\small{\text{
$
F_2 = F_1\cdot \dfrac{R_E^2} {( R_E +8.8392\ km )^2}
=F_1\cdot \dfrac{6371^2} {( 6371 +8.8392\ km )^2}=F_1\cdot 0.99723094065
$
}}$$

So we have 141.2lbs * 0.99723094065 = 140.809008820 lbs