All Answers 358087 Answers

It is easier if you change all the units to metres first.

0.5*0.5   m^2 each tile

Total = 0.5*0.5*108

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{108}} = {\mathtt{27}}$$

27 m^2

how many 4-digit numbers are there such that the thousands digit is equal to the sum of the other three digits?

 $$\small{\text{4-digit-number:~$abcd = a10^3+b10^2+c10+d\qquad a = b+c+d$}}
\\
\small{\text{$\Rightarrow (b+c+d)10^3+bcd_{10} < 10000$}}
\\
\small{\text{$\Rightarrow (b+c+d)10^3< 10000-bcd_{10}$}}
\\
\small{\text{$\Rightarrow (b+c+d)< \dfrac{10000-bcd_{10}}{ 10^3 }$}}
\\
\small{\text{$\Rightarrow (b+c+d)< 10 -
\underbrace{ \dfrac{ bcd_{10} }{ 1000 } }_{ 0\dots 0.999}
$}} \\\\
\small{\text{$\Rightarrow \boxed{\mathbf{(b+c+d)< 10} }
$}} \\\\$$

$$\small{\text{$b+c+d < 10 \qquad
\begin{array}{|r|c|l|}
\hline
(b=0) & c & d \\
\hline
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 2 \\
0 & 0 & 3 \\
0 & 0 & 4 \\
0 & 0 & 5 \\
0 & 0 & 6 \\
0 & 0 & 7 \\
0 & 0 & 8 \\
0 & 0 & 9 \\
\hline
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 1 & 2 \\
0 & 1 & 3 \\
0 & 1 & 4 \\
0 & 1 & 5 \\
0 & 1 & 6 \\
0 & 1 & 7 \\
0 & 1 & 8 \\
\hline
0 & 2 & 0 \\
0 & 2 & 1 \\
0 & 2 & 2 \\
0 & 2 & 3 \\
0 & 2 & 4 \\
0 & 2 & 5 \\
0 & 2 & 6 \\
0 & 2 & 7 \\
\hline
0 & 3 & 0 \\
0 & 3 & 1 \\
0 & 3 & 2 \\
0 & 3 & 3 \\
0 & 3 & 4 \\
0 & 3 & 5 \\
0 & 3 & 6 \\
\hline
0 & 4 & 0 \\
0 & 4 & 1 \\
0 & 4 & 2 \\
0 & 4 & 3 \\
0 & 4 & 4 \\
0 & 4 & 5 \\
\hline
0 & 5 & 0 \\
0 & 5 & 1 \\
0 & 5 & 2 \\
0 & 5 & 3 \\
0 & 5 & 4 \\
\hline
0 & 6 & 0 \\
0 & 6 & 1 \\
0 & 6 & 2 \\
0 & 6 & 3 \\
\hline
0 & 7 & 0 \\
0 & 7 & 1 \\
0 & 7 & 2 \\
\hline
0 & 8 & 0 \\
0 & 8 & 1 \\
\hline
0 & 9 & 0 \\
\hline
\end{array}
\qquad sum_{b=0} = 1+2+3+4+\dots +10
$}}$$

$$\small{\text{$
\begin{array}{lclcr}
b &\qquad & sum \\
\hline
0 & & 1+2+3+4+5+6+7+8+9 +10 & = & 55\\
1 & & 1+2+3+4+5+6+7+8 + 9 & = & 45\\
2 & & 1+2+3+4+5+6+7+ 8 & = & 36\\
3 & & 1+2+3+4+5+6+7 & = & 28\\
4 & & 1+2+3+4+5+ 6 & = & 21\\
5 & & 1+2+3+4+ 5 & = & 15\\
6 & & 1+2+3+4 & = & 10\\
7 & & 1+2+3 & = & 6\\
8 & & 1+2 & = & 3\\
9 & & 1 & = & 1\\
\hline
& & sum &=& 220
\end{array}
$}}$$

 There are 220 - (abcd = 0000) = 219  4-digit numbers

Hallo Melody, i think add for the first digit 5 the following digits 221 with 3 numbers of ways and you get

216+3 also 219  4-digit numbers

I am not really sure about the notation that is needed.

but

P(N rooms empty) = $$^5C_N * 0.4^N * 0.6^{5-N}$$           where     $$0\le N\le5\;\;\;and\;\;\;N\in Z$$ 

Thanks Radix,      

I'll just take a look   

$$\\\frac{x^2-1}{x^2-3x+2}>0\\\\
\frac{(x-1)(x+1)}{(x-1)(x-2)}>0\\\\
$NOTE: x cannot be 1 or 2$\\\\
\frac{(x+1)}{(x-2)}>0\\\\
\frac{(x-2)(x+1)}{1}>0\\\
(x-2)(x+1)>0\\\\
consider\;\;y=(x-2)(x+1)\\\\
$This is a concave up parabola with roots at x=2 and x=-1$\\\\
$It will be above the x axis (positive) at the two ends$\\\\
$roughly sketch it to convince yourself$\\\\
$therefore true for$\\\\
x<-1\;\;\;and \;\;\;x>2$$

thanks but the answer which is written n my book is:x<-1 or x>2

x^2-1 > x^2-3x+2      | -x^2

     -1 > -3x +2

     -3 > -3x                  |   * (-1)

  3 < 3x                        | : 3        =>   1 < x      =>       x > 1

I hope it is right !

Gruß radix !

Look here:

cos(w) = 2/15

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2}}}{{\mathtt{15}}}}\right)} = {\mathtt{82.337\: \!744\: \!339\: \!234^{\circ}}}$$

The angle is 82.34°

Gruß radix !

$$\\\frac{(e^x+7)}{(3e^{-x}+5)}=3\\\\
(e^x+7)\div (3e^{-x}+5)=3\\\\
(e^x+7)\div (\frac{3}{e^{x}}+5)=3\\\\
(e^x+7)\div (\frac{3+5e^{x}}{e^{x}})=3\\\\
(e^x+7)\times (\frac{e^{x}}{3+5e^{x}})=3\\\\
(e^x+7)\times e^{x}=3(3+5e^{x})\\\\
(e^x)^2+7e^{x}=9+15e^{x}\\\\
(e^x)^2-8e^{x}-9=0\\\\$$

$$\\let\;y=e^x\\\\
y^2-8y-9=0\\\\
(y-9)(y+1)=0\\\\
y=9\;\;or\;\;y=-1\\\\
e^x=9\;\;or\;\;e^x=-1\\\\
e^x>0\;\;$for all real x so $ \\\\
e^x=9\\\\
x=ln9\\\\
x=ln3^2\\\\
x=2ln3$$

There are 6 numbers after the point so the 1with 6 zeros will be on the bottom.

0.000023 = 23/1000000

HAPPY BIRTHDAY JILLYBEAN  

  HAVE A REALLY GREAT DAY !!   

There are a lot of us so I thought I would make some cupcakes for everyone to share :))

Hi Sabi,

Yes i guess I missed a couple of steps, :)

$$\frac{(1-x)}{(x-2)}>0$$

FIRST:   I must state that x-2 cannot be zero so  x cannot equal 2

Now I am allowed to multiply both sides by whatever I want and in so doing I will cancel the denpominator out.

If this was an equation (not an inequality) then I would multiply both sides by (x-2)  BUT that causes some problems here because x-2 could be negative or positive.

It would make life easier if we multiply by something that HAS to be positive.

So I multiplied by   $$(x-2)^2$$

$$\\\frac{(1-x)}{(x-2)}\times \frac{(x-2)^2}{1}>0\times \frac{(x-2)^2}{1}\\\\
\frac{(1-x)}{1}\times \frac{(x-2)}{1}>0\\\\
\frac{(1-x)(x-2)}{1}>0\\\\
(1-x)(x-2)>0$$

Now  the rest is the same as before :)

If you need more explanation - just ask :)

$$\\(1-x)(x-2)>0\\
$If you let $ y=(1-x)(x-2)\\
$then the statement will be true when $y>0\\
y=(1-x)(x-2)\\
y=-1x^2+3x-2\\
$this is a parabola- the coefficient of $ x^2$ is -1$\\
$So it is concave down.$\\
$The middle bit will be above the x axis where y is positive.$\\
$The roots are x=1 and x=2$\\
SO\\
(1-x)(x-2)>0\quad when\;\;1

Go TR  

Thanks Chris :))

When working in base ten, the first place after the decimal point is  tenths, the second is tenths squared,  the third is tenths ^3  etc

When working in base eight, the first place after the decimal point is  eigths, the second is eigths squared,  the third is eigths ^3  etc

So I needed to change 59/10^2   to something over eight to the power of some positive integer.

I just chose to do it to three decimal places so I used 8^3.

Does that make sense?

You are very welcome    

Thank you so much!!

with this calc put the first one enter it likie this

asin(25/30)

and this will be the output :)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{25}}}{{\mathtt{30}}}}\right)} = {\mathtt{56.442\: \!690\: \!238\: \!079^{\circ}}}$$

Geno is right     but it is undefined.  YOU CANNOT divide by 0.   NOT EVER!

High Sweaty,

Start by multiplying both sides by 3  Then see if you can do it.  

Let me know how you get on :)

Anything divided by 1 stays the same so

X=890/450 = 89/45 = 1 and 44/45

67.4 and 34.4

do

674/344 = $${\frac{{\mathtt{674}}}{{\mathtt{344}}}} = {\frac{{\mathtt{337}}}{{\mathtt{172}}}} = {\mathtt{1.959\: \!302\: \!325\: \!581\: \!395\: \!3}}$$

the ratio is        337:172

First, divide by 2:  72 = 2 x 36

Since 36 is even, divide again by 2:  72 = 2 x 2 x 18

Since 18 is even, divide agian by 2:  72 = 2 x 2 x 2 x 9

Since 9 = 3 x 3, finish:  72 = 2 x 2 x 2 x 3 x 3

So, the prime factors are 2 and 3; three 2s and two 3s.

You can google that for descriptions, from (relatively) easy to (relatively) complete.

-2(1 + p) + p(p - 1)

=  -2 - 2p + p² - p

=  -2 - 3p + p²

or:  p² - 3p - 2

Because 0.9999 ... (repeating forever) 

IS  1.0000 ... (repeating forever)