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934,650 rounded to nearest ten=934,650

934,650 rounded to nearest hundred=934,700

934,650 rounded to nearest thousands=935,000

hope that this helps!

Yes they are thank you so much

129,999 rounded to the nearest ten=130,000

129,999 rounded to the nearest hundred=130,000

129,999 rounded to nearest thousand=130,000

i am not sure if they are correct. This got me confused.

3^2 = 3 x 3 = 9

9.

D**n, multiplication seemed logical to me, nevertheless this should be able to be figured out as well.

The last number is relatively easy since it's the result of $${{\mathtt{2}}}^{{\mathtt{n}}}$$, which means the last digit regularly skips between this array of numbers: 2,4,8,6.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}\right)}$$ the last digit is 6.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}$$ the last digit is 2.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ the last digit is 4.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}$$ the last digit is 8.

We know the last digit is multiplied 1122 times, $${\frac{{\mathtt{1\,122}}}{{\mathtt{4}}}} = {\frac{{\mathtt{561}}}{{\mathtt{2}}}} = {\mathtt{280.5}}$$ so it is of the form $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ which means the last digit is 4.

While doing this I figured out a better way to solve this problem, we know this of the exponent:

$${\mathtt{1\,122}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{17}}$$

Another thing we know for sure is that while multiplying two numbers only the last two digits of each number affect the last two digits of the answer. Hence all digits save for the last two are irrelevant for further multiplication.

Since only the last two digits matter of the 1122 being multiplied we can drop the first two digits:

$${{\mathtt{22}}}^{{\mathtt{1\,122}}} = \left({\left({\left({\left({{\mathtt{22}}}^{{\mathtt{2}}}\right)}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}\right)$$

$${\left({\left({{\mathtt{484}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${\left({\left({{\mathtt{84}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

$${\left({{\mathtt{592\,704}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}} = {\left({\mathtt{4\,194\,304}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${{\mathtt{4}}}^{{\mathtt{17}}}$$

$${{\mathtt{4}}}^{{\mathtt{17}}} = {{\mathtt{4}}}^{{\mathtt{11}}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {\mathtt{4\,194\,304}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}}$$

Only last two digits matter

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {{\mathtt{4}}}^{{\mathtt{7}}} = {\mathtt{16\,384}}$$

Last two digits of $${{\mathtt{1\,122}}}^{{\mathtt{1\,122}}}$$ are 84, case closed.

3 s squared?

that is how 3s^2 is read.

if you rationalise the denominator (multiply by multiplying by √7 + √3) . Thats gives 10 + 2 √21 / 4.

4 3/.75 =4

12.00x 1 1/2= 6$$\mathrm{\ }$$

  .55 is fifty five/100.   Divide both by 5 =      11/20

You'll get a raise of $ 4800, meaning you'll make $16,800 yearly.  If you can't figure something as easy as this, I'd CUT your pay.

.36 is 36 one hondreths, or   9/25

13  1/3

 the answer is 290

Not necessarily, the question mentions that the weight is supported by the two wires which indicates it's at rest.

The lens formula is:

$${\frac{{\mathtt{1}}}{{\mathtt{f}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{d1}}}} = {\frac{{\mathtt{1}}}{{\mathtt{d2}}}}$$

In which

f: focal length

d1:distance to source

d2:distance to image

Since the sun has an almost infinite distance to the lens d1=infinity, which means:

 $${\frac{{\mathtt{1}}}{{\mathtt{\infty}}}} = {\mathtt{0}}$$

so

 $${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{d2}}}}$$

Which means answer A is correct.

Please show me the full proof, the main difficulty with this problem is the danger of falling for circular reasoning and I must be sure this is not the case in your reasoning. On another note the existence of the bisector in this triangle is completely irrelevant. The  only thing that constitutes this particular shape is the relationship of the median to the altitude. 

Tues 2/9/14

*1) Finding the last 2 digits of a large power or a multiple.   Some very impressive number work here (Thanks Honga)

@@ End of Day Wrap:    Mon 1/9/14       Sydney, Australia      Time 1:30am (Really Tuesday morning)     ♬

Hi all,

Today our great answers were supplied by CPhill, AzizHusain, Bertie, Honga, Heureka and TakahiroMaeda.  A very big thank you to each of you.  

Now straight to our interest posts:

1) This one looks interesting.

Thanks, Melody, for that one !!!

Thanks Chris 

I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.   

$$\begin{array}{rlll}
\frac{12}{sinB}&=&\frac{11}{sin30}\\\\
\frac{sinB}{12}&=&\frac{sin30}{11}\\\\
SinB&=&\frac{12sin30}{11}\\\\
B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\
B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\
B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\
\end{array}$$

$$If B is $33^0$ then \\
$C=180-30-33=117^0$\\\\
If B is $147^0$ then \\
$C=180-30-147=3^0$\\\\
So the 2 possible solutions are $

I assume by "length," you mean "height"

The volume of a cone is (1/3) of that of its associated cylinder (a cylinder of the same radius and height)

So

V = (1/3)(r^2)(h) = (1/3)(4^2)(6) = 32 cu in.

I think that means "how much more will i get paid"

40% of 12000=40/100*12000 

You can finish it on the site calc on the home page 

I am not sure if you want me to do this with calculus or with physics formulas.

I'll do it with physics formula.

$$\\u=80km/hour\;=\frac{80000}{60*60}m/sec = \frac{200}{3*3}m/sec= \frac{200}{9}m/sec\\\\
v=20km/hour\;=\frac{20000}{60*60}m/sec = \frac{50}{9}m/sec\\\\
a=-2m/s^2\\\\
find\;t\\\\
v=u+at\\\\
\frac{50}{9}=\frac{200}{9}+-2t\\\\
\frac{-250}{9}=-2t\\\\
\frac{250}{18}=t\\\\
t=13.9\;sec \qquad \mbox{to nearest one tenth of a second}\\\\$$