Call Alice's age is a

        Betty's age is b

 and  Clara is c 

Follow the threads , we have :

a + b + c = 29  (1)

b = a + 4  (2)

c = b + 6   (3)

Change (3) , (2) to (1)

a + b + c = a + (a + 4) + (b + 6)

<=> a + (a + 4) + (a + 10) = 29

=> 3a + 14 = 29

=> 3a = 15

=> a = 5

So , Alice's age is 5

\(F=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3}}}=\dfrac{1}{1+\dfrac{1}{\dfrac{6}{3}+\dfrac{1}{3}}}=\dfrac{1}{1+\dfrac{1}{\dfrac{7}{3}}}=\dfrac{1}{1+\dfrac{3}{7}}=\dfrac{1}{\dfrac{7}{7}+\dfrac{3}{7}}=\dfrac{1}{\dfrac{10}{7}}=\dfrac{7}{10}=0,7\)

We have the following summary : 

1 mile <=> 1p12' = 72' = \(\dfrac{1}{50}h\)

So , in 1 hour , the car can travel :

1mile x 50  <=>  \(\dfrac{1}{50}.50h=1h\)

=> x = 50 (mile)

There are : (200 - 1) : 1 + 1 = 200 numbers from 1 to 200

We have a sequence of two-digit numbers are as follows :

2 ; 12 ; 22 ; 32 ; ....... ; 192

Apply a formula to count numbers , we have :

(192 - 2) : 10 + 1 = 20 

There are also numbers with 2 digits such as

20 ; 21 ; 23 ; 24 ; .... ; 29 ; 120 ; 121 ; 122 ; 123 ; ...... 129

And there are have 18 also numbers .

So , from these information , we have 38 numbers have digit 2

We have :

200 - 38 = 162 (numbers)

There are 162 numbers from 1 to 200 do not have the digit 2 

a) There are : 7 x 7 x 6 x 5 = 1470 numbers

b) There are : 4 x 3 x 2 x 1 = 24 old numbers 

a) There are :7 x 7 x 6 x 5 = 1470 numbers

b) There are :4 x 4 x 3 x 2 = 96 even numbers

The hole stands anywhere , is it ? Or in the middle ?

We have :

2²⁰⁰⁹ is even number 

3²⁰⁰⁹ = 3^4.502 . 3 = .....1 . 3 = ......3 (odd number)

7²⁰⁰⁹ = 7^4.502 . 7 = .....1 . 7 = .....7(odd number)

 = 9^4.502 . 9 = ......1 . 9 = .......9(odd number)

So : 2²⁰⁰⁹ + 3²⁰⁰⁹ + 7²⁰⁰⁹ + 9²⁰⁰⁹ =  .....2 + .....3 + ....7 + .....9 = ......21 (odd number)

Because the lowest common multiple of a and b is 50 . So all possible values of a and b is :

(a;b) = {1 ; 2 ; 5 ; 10 ; 25 ; 50}

b³ = 3375 

=> b = \(\sqrt[3]{3375}=15\)

Because a,b,c are three consecutive odd number  and b = 15 so

a = 13 ; b = 15 ; c = 17

ac = 15 . 17 = 221 

We have :

\(\left|x\right|\ge0\)

\(\left|x\right|+2004\ge2004\)

\(\dfrac{\left|x\right|+2004}{-2005}\le\dfrac{2004}{-2005}\)

So MAX_B = \(\dfrac{-2004}{2005}\)

\(\Leftrightarrow\dfrac{\left|x\right|+2004}{-2005}=\dfrac{-2004}{2005}\)

=> |x| + 2004 = 2004

=> |x| = 2004 - 2004 

=> |x| = 0

=> x = 0

We have :

a \(\le\) c + 2

b + 1 \(\le\) c + 2

c + 2 = c + 2

=> a + b + 1 + c + 2 \(\le\) c + 2 + c + 2 + c + 2

=> (a + b + c) + (1 + 2) \(\le\) 3c + 6

=> 1 + 3 \(\le\) 3c + 6

=> 4 \(\le\) 3c + 6

=> \(-\dfrac{2}{3}\) \(\le\) c

=> Min_c = \(-\dfrac{2}{3}\)

=> a + b +( \(-\dfrac{2}{3}\)) = 1 

=> a + b = \(\dfrac{5}{3}\)

With a = b + 1

=> a + b = \(\dfrac{5}{3}\)

=> b + 1 + b = \(\dfrac{5}{3}\)

=> b = \(\dfrac{1}{3}\)    =>    a = \(\dfrac{4}{3}\)  (1)

With a < b + 1

=> b + 1 + b < \(\dfrac{5}{3}\)

=> b < \(\dfrac{1}{3}\)    =>    a < \(\dfrac{4}{3}\)  (2) 

From (1) and (2) 

=> b \(\le\dfrac{1}{3}\) and \(a\le\dfrac{4}{3}\)

So the smallest value of c = \(-\dfrac{2}{3}\)

When b\(\le\dfrac{1}{3}\) and \(a\le\dfrac{4}{3}\)

Can i do this :

We have : \(\dfrac{4}{n}=\dfrac{1}{n}+\dfrac{3}{n}\)

With n = 3k (k is the natural number greater than 1) , i have :

\(\dfrac{4}{n}=\dfrac{1.\left(n+1\right)}{n.\left(n+1\right)}+\dfrac{3}{n}=\dfrac{1}{n+1}+\dfrac{1}{n.\left(n+1\right)}+\dfrac{3}{n}=\dfrac{1}{3k}+\dfrac{1}{3k.\left(3k+1\right)}+\dfrac{3}{3k}=\dfrac{1}{3k}+\dfrac{1}{9k^2+3k}+\dfrac{1}{k}\)

With n = 3k + 1

\(\dfrac{4}{n}=\dfrac{1}{n}+\dfrac{3}{n}=\dfrac{1}{n}+3\left(\dfrac{1}{n-1}-\dfrac{1}{n\left(n-1\right)}\right)=\dfrac{1}{n}+\dfrac{3}{n-1}-\dfrac{3}{n\left(n-1\right)}=\dfrac{1}{3k+1}+\dfrac{3}{3k}+\dfrac{-3}{3k\left(3k+1\right)}=\dfrac{1}{3k+1}+\dfrac{1}{k}+\dfrac{1}{-k\left(3k+1\right)}\)

With n = 3k + 2

\(\dfrac{4}{n}+\dfrac{1}{n}+\dfrac{3}{n}=\dfrac{1}{n}+3.\left(\dfrac{1}{n+1}+\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{n}+\dfrac{3}{n+1}+\dfrac{3}{n\left(n+1\right)}=\dfrac{1}{3k+2}+\dfrac{1}{k+1}+\dfrac{1}{\left(3k+2\right)\left(k+1\right)}\)

So with every n greater than 4 then 4/n can analytical 3 different fraction the form \(\dfrac{1}{n}\)

a) equal to \(\dfrac{3}{5}\) :

\(\dfrac{6}{10};\dfrac{9}{15};\dfrac{12}{20};\dfrac{15}{25};\dfrac{18}{30}\)

b) equal to \(\dfrac{5}{8}\) :

\(\dfrac{10}{16};\dfrac{15}{24};\dfrac{20}{32};\dfrac{25}{40};\dfrac{30}{48}\)

c) equal to \(\dfrac{36}{108}\):

\(\dfrac{1}{3};\dfrac{2}{6};\dfrac{3}{9};\dfrac{4}{12};\dfrac{5}{15}\)

a) \(19:14=\dfrac{19}{14}=1\dfrac{5}{14}\)

b) \(28:13=\dfrac{28}{13}=2\dfrac{2}{13}\)

c) \(27:21=\dfrac{27}{21}=\dfrac{9}{7}=1\dfrac{2}{7}\)

d) \(15:10=\dfrac{15}{10}=\dfrac{3}{2}=1\dfrac{1}{2}\)

e) \(45:6=\dfrac{45}{6}=\dfrac{15}{2}=7\dfrac{1}{2}\)

a) \(6:9=\dfrac{6}{9}=\dfrac{2}{3}\)

b) \(7:4=\dfrac{7}{4}\)​

c) \(12:5=\dfrac{12}{5}\)

d) \(9:19=\dfrac{9}{19}\)

e) \(7:13=\dfrac{7}{13}\)

The edge length of the cube is :

\(\sqrt[3]{216}\) = 6 (cm) because 6 x 6 x 6 = 216 

The area is :

6 x 6 = 36 (cm²)

The female students is :

\(60x\dfrac{3}{5}=36\)(students)

The male student is :

60 - 36 = 24(students)

Ohhh , these answer are very similar , are you copy them together ???

We have :

-20 = -20 

16 - 36 = 25 - 45