\(\dfrac{2^{10}\cdot5^6\cdot30^5}{4^2\cdot5^9\cdot60^2}=\dfrac{4^5\cdot5^6\cdot30^5}{4^2\cdot5^9\cdot2^2\cdot30^2}=\dfrac{4^5\cdot5^6\cdot30^5}{4^3\cdot5^9\cdot30^2}=4^2\cdot5^{-3}\cdot30^3=3456\)

x+2y = 8 and 5x+2y = 48

=> \(\left(5x+2y\right)-\left(x+2y\right)=48-8=40\)

=> 4x = 40 => x = 10

=> 10 - 2y = 0 => 2y = 10 => y = 5

So the answer is: \(\dfrac{10+5}{2}=\dfrac{15}{2}\)

average = 1 --> sum 1st,3rd add to 2 --> 1 + 1, 2 + 0 --> 2 numbers
average = 2 --> sum add to 4 --> 1 + 3, 2 + 2, 3 + 1, 4 + 0 --> 4 numbers
average = 3 --> sum add to 6 --> 1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1, 6 + 0 --> 6
average = 4 --> sum add to 8 --> 1 + 7, 2 + 6, 3 + 5, 4 + 4, 5 + 3, 6 + 2,............................ 7 + 1, 8 + 0 --> 8 numbers
average = 5 --> sum add to 10 --> 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4,............................ 7 + 3, 8 + 2, 9 + 1 --> 9
ave = 6 --> sum add to 12 --> 3 + 9, 4 + 8, 5 + 7, 6 + 6, 7 + 5, 8 + 4,
............................ 9 + 3 --> 7
ave = 7 --> sum add to 14 --> 5 + 9, 6 + 8, 7 + 7, 8 + 6, 9 + 5 --> 5

ave = 8 --> sum add to 16 --> 7 + 9, 8 + 8, 9 + 7 --> 3 numbers
ave = 9 --> sum add to 18 --> 9 + 9 --> 1 number
2 + 4 + 6 + 8 + 9 + 7 + 5 + 3 + 1
appears to be the sum of the digits from 1 to 9
sum digits = n(n+1)/2 = 9(10)/2 = 90/2 = 45 numbers
111, 210
123, 222, 321, 420
135, 234, 333, 432, 531, 630
147, 246, 345, 444, 543, 642, 741, 840
159, 258, 357, 456, 555, 654, 753, 852, 951
369, 468, 567, 666, 765, 864, 963
579, 678, 777, 876, 975
789, 888, 987
999

Answer: 45 numbers
 

Put H is intersection of AB and CD

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}AC^2=HC^2+AH^2\\AD^2=HD^2+AH^2\end{matrix}\right.\Rightarrow AC^2-AD^2=\left(HC^2+AH^2\right)-\left(HD^2+AH^2\right)=HC^2-HD^2\\\left\{{}\begin{matrix}BC^2=HC^2+BH^2\\BD^2=HD^2+BH^2\end{matrix}\right.\Rightarrow BC^2-BD^2=\left(HC^2+BH^2\right)-\left(HD^2+BH^2\right)=HC^2-HD^2\end{matrix}\right.\)

=> AC² - AD² = BC² - BD²

1/

We have:

\(\dfrac{2011}{1.11}+\dfrac{2011}{2.12}+...+\dfrac{2011}{100.110}=2011\left(\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{100.110}\right)\)

\(=2011\cdot\dfrac{1}{10}\left(\dfrac{11-1}{1.11}+\dfrac{12-2}{2.12}+...+\dfrac{110-100}{100.110}\right)\)

\(=\dfrac{2011}{10}\left(1-\dfrac{1}{11}+\dfrac{1}{2}-\dfrac{1}{12}+...+\dfrac{1}{100}-\dfrac{1}{110}\right)\)

\(=\dfrac{2011}{10}\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}-\dfrac{1}{11}-\dfrac{1}{12}-...-\dfrac{1}{110}\right)\)

\(=\dfrac{2011}{10}\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)\)

\(\dfrac{2012}{1.101}+\dfrac{2012}{2.102}+...+\dfrac{2012}{10.110}=2012\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right)\)

\(=2012\cdot\dfrac{1}{100}\left(\dfrac{101-1}{1.101}+\dfrac{102-2}{2.102}+...+\dfrac{110-10}{10.110}\right)\)

\(=\dfrac{2012}{100}\left(1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{10}-\dfrac{1}{110}\right)\)

\(=\dfrac{2012}{100}\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)\)

\(\Rightarrow\dfrac{2011}{10}\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)x=\dfrac{2012}{100}\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)\)

\(\Rightarrow\dfrac{2011}{10}x=\dfrac{2012}{100}\Leftrightarrow x=\dfrac{1006}{10055}\)

The distance the trip fly is:

450.2,5 = 1125 [mi]

The speed of the original headwind is:

1125 : 3 = 375 [mi/h]

Answer: 375 miles/h

So the answer is:

|15-2| = |2-15| = 13

||8-3| = |3-8| = 5

So the answer is 13 or 5

Tell those numbers are a,b

=> a+b = 11 and ab = 24

=> ab - a - b = 24-11 = 13

=> a(b-1) - b + 1 = 14

=> a(b-1) - (b-1) = 14

=> (a-1)(b-1) = 14

We have table:

So (a,b) = (2,15); (15,2); (3,8); (8,3)

1st = 1.3

2nd = 2.3

3rd = 3.3

............

=> 99th = 99.3 = 297

The sum time to Nathan ran 2,5 miles is:

2,5.7'36'' = \(2,5\cdot\dfrac{19}{150}=19'\)

The sum time to Nathan run 5 miles is:

5.7'24'' = \(5\cdot\dfrac{37}{300}=37'\)

So the last time to Nathan run last 2,5 miles is:

\(37'-19'=18'\)

So his pace be for the next 2,5 miles is:

\(\dfrac{18'}{2,5}=7'12''\)

So his pace be for the next 2,5 miles is 7 minutes 12 seconds

\(\widehat{CAB}=180^o-90^o-30^o=60^o\)

\(tan\widehat{CAB}=\dfrac{BC}{CA}\)

\(\Leftrightarrow tan60^o=\dfrac{9}{CA}\Leftrightarrow CA=\dfrac{9}{\sqrt{3}}=3\sqrt{3}\) cm

\(\angle CAD=\dfrac{60^o}{2}=30^o\Leftrightarrow\angle ADC=180^o-90^o-30^o=60^o\)

\(tan\angle ADC=\dfrac{CA}{CD}\)

\(\Leftrightarrow tan60^o=\dfrac{3\sqrt{3}}{CD}\)

\(\Leftrightarrow\sqrt{3}=\dfrac{3\sqrt{3}}{CD}\Leftrightarrow CD=3cm\)

\(\left\{{}\begin{matrix}\dfrac{x}{y}=2\\x=3z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\x=3z\end{matrix}\right.\Leftrightarrow x=2y=3z\)

\(\Rightarrow4x=8y=12z=10\)

\(\Rightarrow x+y+z=\dfrac{10}{4}+\dfrac{10}{8}+\dfrac{10}{12}=\dfrac{55}{12}\)

\(3^{17}\cdot7^{23}=\left(3^4\right)^4\cdot3\cdot\left(7^4\right)^5\cdot7^3=...1\cdot3\cdot...1\cdot343=...9\)

So the digits in the units of 3¹⁷ x 7²³ is 9

Draw \(CH\perp AB\)

Connected AC

\(\left\{{}\begin{matrix}DA\perp AB\\CH\perp AB\end{matrix}\right.\)=> DA // CH

=> \(\angle HCD+\angle ADC=180^o\) [2 angles in the same side]

But \(\angle ADC=120^o\Rightarrow\angle HCD=60^o\Leftrightarrow\angle BCH=120^o-60^o=60^o\)

\(\angle BCH=60^o\Rightarrow\angle HBC=90^o-60^o=30^o\)

We have:

tan \(\angle HBC=\dfrac{HC}{BC}\)

\(\Rightarrow HC=tan30^o\cdot8=\dfrac{\sqrt{3}}{3}\cdot8=\dfrac{8\sqrt{3}}{3}\left(units\right)\)

The same, we have:

tan \(\angle HCB=\dfrac{HB}{BC}\)

\(\Rightarrow HB=tan60^o\cdot8=\sqrt{3}\cdot8=8\sqrt{3}\left(units\right)\)

\(\Rightarrow S_{BHC}=\dfrac{\dfrac{8\sqrt{3}}{3}\cdot8\sqrt{3}}{2}=32\left(units^2\right)\)

We have: 

111³ = 1367631 have 7 digits, remove

222³ = 10941048 , remove

333³ = 36926037, satisfy

A > 3 => \(\left(AAA\right)^3=\overline{X...........}\) with:

* X > 7 when (AAA)³ have 7 digits

* X < 9 when (AAA)3 have 8 digits 

So A = 3

3 dozen oranges = 36 oranges

So the answer is: \(\dfrac{90}{4}\cdot36=810\) [cents]

\(f\left(g\left(-3\right)\right)=f\left(-3\cdot2+4\right)=f\left(-2\right)=\left(-2\right)^2-2=2\)

The number of marbles of Mac is:

25.20% = 5 [red marbles]

The number of marbles of Thayer is:

\(20.\left(100\%-75\%\right)=5\) [marbles]

So the answer is: 5 - 5 = 0 [marble]

I never have studied inequality Co-si yet. So can you give me different cases? 

P/s: I am grade 7 

Tell the commission of Mr. Jones and Mr. Smith  is a and b

a = 3% widgets

b = 5% widgets

=> a/b = 3/5

=> The answer is: 500/5.3 = 300 widgets