We have: 145+176+147+167+198+132+150+141+138+143+146=1683

She have $1485 so: 1683-1485=198

So we remove the Birthday (Katty Perry): 198$

Chúng tôi có tất cả: 145 + 176 + 147 + 167 + 198 + 132 + 150 + 141 + 138 + 143 + 146 = 1683 $

Cô ấy có $ 1485 như vậy: 1683-1485 = 198 $

Vì vậy, chúng tôi loại bỏ các sinh nhật (Katty Perry): 198 $

For a = 15 and b = 6.

=> a # \(b=15.6+15^2+6^2=90+225+36=375\)

(a#b)@a = \(\dfrac{375}{3}\)- a = 125 - 15 = 110 

Thank you very much 

 Let O₁ and O₂
be the centers of C₁ and C₂

respectively (We are assuming C has radius 1 and C
has radius 3). Then the desired locus is an annulus centered at the midpoint of O₁O₂

with inner radius 1 and outer radius 2.  For a fixed point Q on C₂

, the locus of the midpoints of the segments PQ for P lying on C₁ is the image of C₁ under a homothety centered at Q of radius 1/2,
which is a circle of radius 1/2. As Q varies
, the center of this smaller circle traces out a circle C₃ of radius (again by homothety). By considering the two positions of Q on the line of centers of the circles
, one sees that C₃ is centered at the midpoint of O₁O₂

and the locus is now clearly the specified annulus

what is "The product" ?, i don't understand ! 

Suppose \(\dfrac{x}{x-4}\Rightarrow a=\dfrac{x}{\overline{x-4|x}}\)

a, 

a = \(-\dfrac{3}{13}=\dfrac{-3}{13}=\dfrac{3}{-13}\)

\(\Rightarrow\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Easy see x = 3 => a \(=-\dfrac{3}{13}\)

So x = 3

b,

a = 1/6 

=> a - 4 \(\ge0\)

=> a > 4

\(\Rightarrow5\le x\le9\)

\(\Rightarrow1\le x-4\le5\)

\(\Rightarrow a\in\left\{\dfrac{5}{15};\dfrac{6}{26};\dfrac{7}{37};\dfrac{8}{48};\dfrac{9}{59}\right\}\)

Have 8/48 = 1/6

So a = 8/48

=> x = 8

Dao Trong Luan's solution isn't logical at all. I need another one.

Let a, b be the month and the date in which he was born

\(\left(a,b\in Z^+;a< 13;b< 32\right)\). We have : \(31a+12b=170\)

\(12b\ge12\Rightarrow31a\le158\Rightarrow a\le5\). Moreover :

\(170⋮2;12b⋮2\Rightarrow31a⋮2\Rightarrow a⋮2\). So, a = 2 ; 4

If a = 2, \(b=\dfrac{170-31.2}{12}=9\)

If a = 4, \(b=\dfrac{170-31.4}{12}=\dfrac{23}{6}\)(asburd)

So, he was born on 9/2

Cause \(2\cdot\left(3+4\right)>2\cdot3+4\)

So the positive difference between the value of 2 × (3 + 4) and the value of 2 × 3 + 4 is :

\(2\cdot\left(3+4\right)-\left(2\cdot3+4\right)=14-10=4\)

Answer : 4.

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