All Answers 358098 Answers

$${{\mathtt{1\,000}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$  =  $${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{{\mathtt{1\,000}}}^{{\mathtt{2}}}}} = {\mathtt{100}}$$

1000² = 1 000 000

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{1\,000\,000}}}} = {\mathtt{100}}$$

Gruß radix !

$$\frac{-11}{4}\div 14=\frac{-11}{4}\times \frac{1}{14}=\frac{-11}{56}$$

Good thinking Chris  

More practice

Part B

11A+8B=94

Largest common factor of 11 and 8 is 1.  One is a factor of 94.  So this does have integral solution.

2 elevens and 9 eights confusions is the only solution.  

Thanks Chris and anon - I think we might need change too :((

Mellie this isn't really for you - I am just practicing doing diophantine equations  :))

Part C

8A+11B=69

The FCF of 8 and 11 is 1    and 1 goes into 69 so this does have integral solutions  :/

8*-276+11*207=69

8A+11B=69

So 

A=-276+11N          B=207-8N

276/11=25 and a bit

A=-276+11*26=10         B=207-8*26=-1

Mmm       There are not any solutions!

Thanks Chris - I am going to put my 2c worth in now  

$$\\\frac{9}{44} + \frac{17}{36} \\\\
=\frac{9}{11*4} + \frac{17}{9*4} \\\\
=\frac{9*9}{11*4*9} + \frac{17*11}{9*4*11} \\\\
=\frac{9*9+17*11}{11*4*9} \\\\
=\frac{81+187}{11*4*9} \\\\
=\frac{268}{11*4*9} \\\\
=\frac{268\div 4}{11*4*9\div 4} \\\\
=\frac{67}{11*9} \\\\
=\frac{67}{99} \\\\$$

Ok KidDanger BUT

if you use order of operation correctly  (that's  PEDMAS)  then

303.92*10^9 /11.2*10^6

actually means

$${\frac{{\mathtt{303.92}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{9}}}}{{\mathtt{11.2}}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}} = {\mathtt{27\,135\,714\,285\,714\,285.714\: \!285\: \!714\: \!285\: \!714\: \!3}}$$

$$\small{\text{$
\begin{array}{lcl}
$Let $ a, b, c, $ and $ n $ be positive integers. If $
a + b + c = 19 \cdot 97 $ and $ \\
\left[a + n = b - n = \dfrac{c}{n}\right],
$ compute the value of $ a .
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{lrrrcl}
& a+b+c=19\cdot 97 \\
\\
\hline
\\
(1)& a+n=k \\
(2) & b-n=k &\qquad \qquad (1)+(2): & a+b&=& 2\cdot k\\
(3) & \dfrac{c}{n}=k &\qquad \qquad $so$ & c &=& k\cdot n\\\\
& & & a+b+c &=& 2\cdot k + k\cdot n=19\cdot 97
\\
\hline
\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rrclrcl}
& 2\cdot k + k\cdot n &=& 19\cdot 97\\\\
& k\cdot(2+n)&=& 19\cdot 97\\\\
I. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{19}\cdot \textcolor[rgb]{0,0,1}{97}\\\\
& \underline{k=19} && \underline{2+n = 97 }& \qquad \Rightarrow
\qquad n&=& 95\\
& && & a+n&=& k\\
& && & a+95&=& 19\\
& && & a&=& -76 ~$ negative! $\\ \\
II. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{97}\cdot \textcolor[rgb]{0,0,1}{19}\\\\
& \underline{k=97} && \underline{2+n = 19 }& \qquad \Rightarrow
\qquad n&=& 17\\
& && & a+n&=& k\\
& && & a+17&=& 97\\
& && & a&=& 80 ~$ okay! $\\
\end{array}
$}}$$

Good thinking KidDanger :)

The modal value is the one that occurs most often.  If the 66 is truly 66 then the modal value is 1 because there are more 1's than any other number.  However, if you meant to write 6, 6 instead of 66 then the modal value would be 6 because there would now be more 6's than any other number.

. 

Divide 2121831 by 127016000 and multiply by 100 to find the percentage

.

The Smith family has 13 children, all born exactly 2 years apart. How old will the second-youngest child be when the oldest child is five times the age of the youngest child?

$$\small{\text{$
\begin{array}{rcl}
$arithmetical sequence $
~
\underbrace{c_1}_{child ~1} ~~
\underbrace{(c_1+1\cdot 2)}_{child ~2} ~~
\underbrace{(c_1+2\cdot 2)}_{child ~3} ~~
\underbrace{(c_1+3\cdot 2)}_{child ~4} ~~
\cdots\\\\
c_n=c_1+(n-1)\cdot2\\\\
c_{13}=c_1+12\cdot2\\
c_{13}=c_1+24\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
c_{13}=c_1+24 = 5\cdot c_1\\
c_1+24 = 5\cdot c_1\\
4\cdot c_1 = 24 \\
\mathbf{c_1 = 6} \\\\
c_2 = c_1+2\\
c_2 = 6+2\\
\mathbf{c_2 = 8}
\end{array}
$}}$$

The second-youngest child is 8 years old

Thanks for that great answer Heureka :)

A circle centre O and radius 2cm is inscribed in an equilateral triangle ABC and touches the side BC at N. What is AN?

$$\\
\small{\text{when~ $ h = \overline{AN} $~ and one side is a}}\\\\
\qquad
\tan{(60\ensurement{^{\circ}})}
= \dfrac{h}{\frac{a}{2}} \qquad
\tan{(30\ensurement{^{\circ}})}
= \dfrac{r}{\frac{a}{2}} \\ \\
\dfrac{a}{2} = \dfrac{h}{ \tan{(60\ensurement{^{\circ}})}
} = \dfrac{r}{ \tan{(30\ensurement{^{\circ}})} }\\\\\\
h= r\cdot \dfrac{ \tan{(60\ensurement{^{\circ}})} }{ \tan{(30\ensurement{^{\circ}})} }
\small{\text{$
\qquad \tan{(60\ensurement{^{\circ}})} = \sqrt{3}
\qquad \tan{(30\ensurement{^{\circ}})} = \dfrac{\sqrt{3} }{3}
$}} \\\\\\
h= r\cdot \dfrac{ \sqrt{3}}{ \frac{\sqrt{3} }{3} }\\\\\\
h= 3\cdot r\\\\
h= 3\cdot 2~\text{cm}\\\\
h=6~\text{cm}\\\\
\mathbf{\overline{AN} = 6~\text{cm}}$$

3^3=3*3*3=9*3=27

Tshirt=$10

pair of shoes=$10

candy=$1

$${\mathtt{dollar10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{dollar10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{dollar1}} = {\mathtt{dollar21}}$$

Overall, she does not have the money to buy what she wants. Not for the products, and if the state government she lives in is greedy, she also cannot pay the tax.

cuz $21(the total subtotal or total cost) is greater than $20 (the amount Holly has)

And for once, anon and I agree

Holly, youre one greedy girl!

$${\mathtt{0.05}}{\mathtt{\,\times\,}}{\mathtt{9\,000\,000\,000}} = {\mathtt{450\,000\,000}}$$

Overall, we can conclude that 5% of 9 billion is 450 million

9/44   + 17/36       we  need to find a common denominator between 44 and 36

Note that 44 factors as 4 X 11

And 36 factors as 4 X 9     

So.....taking each different factor found in both numbers and multiplying them together, we get  4 x 9 x 11  = 36 x 11 = 396....and this is the common denominator

So  9/44  =  [9 x 9]/[11 x 4 x 9] =  81/396

And   17/36  =  [17 x 11]/ [11 x 4 x 9]  =  187/396

So we have ........     81/396 + 187/396  =  268/396    .... and  since both numerator and denominator are even, we can divide them by 2 ....and we get....

134/198     and we can divde both by 2, again....and we get

67/99

And that's it!!!!

Just multiply 847 x .59.......your answer will be in dollars and cents

Anonymous is correct....here's a pic.....

NB = 2√3   AB = 4√3  .......and by the Pythagorean Theorem.........

AN = √[(AB)^2 - (NB)^2]  =   √[(4√3)^2  -  (2√3)^2 ]  = √[48 - 12] = √36  = 6

Holly is greedy!

.