[/quote]
,
be reals such that 
. Prove that
Let 
. Prove that
with circumcenter 
.
and 
intersect the altitude from 
to 
at points 
and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
s.t 
.
such that 
Lines 
and 
intersect line 
respectively at 
and 
.
intersects 
at 
intersects 
at 
and 
and 
.
be a point on 
.
.
.
Write only the answers to the questions given.
to be more exact:
in years 2002-08 time was 90' for part A and 120' for part B
red balls and 
white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...
unit. The area of the hexagon is ...
and 
.
is ...
so that 
and 
is ...
,
,
,
has an average of 
and a median of 
.
are integers greater than 
which are four consecutive quarters of an arithmetic row with 
.
and 
are squares of two consecutive natural numbers, then the smallest value of 
is ...
,
and 
.
and 
.
is ...
meet 
for each natural number 
.
is ....
provided that the difference of any two numbers at least 
is ...
which satisfies 
are as many as ...
and 
.
.
is a point on the side extension 
is perpendicular to 
and 
.
is ...
consists of 
with 
positive reals is ....
with 
is ....
square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of 
coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of 
distinct points in a plane such the circumradius of any 
,
lie on a line in this order and satisfy 
.
and 
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
,
and 
and 
and 
such that for all 
such that 
.
Y by
Hello everyone,
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!Attachments:
This post has been edited 3 times. Last edited by Vulch, Apr 11, 2025, 8:07 PM
distinct inputs, they cannot give the same output.
is a function. 
.
is one-one then for any choice of 
such that 
,
cannot be equal to 
,
.
and 
.
,
is not a correct example to contradict injectivity of the function.
such that 
.
.
are not 
.
be any nonzero real number, for surjectivity we need to show that there is a 
with 
.
,
and 
.
.
.
