and 
. Define 
for 
as 
, 
. Prove that 
exists and find it's value .
be a function that satisfies the conditions:
has Darboux’s property
at 
exist.
![\[ \int_{1}^{2}e^{x( x+\sqrt{x^2-1} )}dx \]](http://latex.artofproblemsolving.com/b/1/d/b1d1f5525bb9cb5fe1f1a7491ff5209cad5438d3.png)
such that
.
is an increasing function on 
,
.
,
for 
using the same 
and 
axes.
.
.
;
be an algebraic extension of 
,
be an algebraic closure of 
is an embedding such that 
for all 
,
extends to an automorphism of 
is an embedding. Then, there exists an embedding 
that extends 
,
is an isomorphism.
and 
that extends 
is also an algebraically closed field.
,
be the field of algebraic numbers, 
.
defined by
Then, according to the exercise above, 
How should we interpret 
?
be a prime number. A flea is staying at point 
of the real line. At each minute,
to the left or to the right.
minutes, it wants to be at 
the number of its strategies to do this
:
be a continuous function such that for all 
and for all 
,
Show that 
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Hi!
This is a Pre-Algebra Marathon and I would like to ask you to enter your answers with
Sorry if it is had to understand cause this is my first time posting.
Also, I would be more than glad if someone messaged me on how to use LaTex.
Here's the first problem:
This is a Pre-Algebra Marathon and I would like to ask you to enter your answers with
S[#ofproblem]P[#ofproblem]Sorry if it is had to understand cause this is my first time posting.
Also, I would be more than glad if someone messaged me on how to use LaTex.
Here's the first problem:
P1This post has been edited 1 time. Last edited by StarGalaxy457, Aug 15, 2015, 10:06 AM
Reason: I edited the post because I was fixing some minor things
Reason: I edited the post because I was fixing some minor things
,
,
?
.
are between 
?
and 196 is 
,
multiples of 7.
,
so, 
non-negative odd numbers?
![[asy] pair O; for(int i = 0; i < 5; ++i){
draw(O--((2/sqrt(3))^i)*dir(30*i));
}
for(int g = 0; g < 4; ++g){
draw( ((2/sqrt(3))^g)*dir(30*g)-- ((2/sqrt(3))^(g+1))*dir(30*g+30));
}
label("10 cm", O--(16/9)*dir(120), W);
label("$30^{\circ}$",.4*dir(0),dir(90));
label("$30^{\circ}$",.4*dir(25),dir(115));
label("$30^{\circ}$",.4*dir(50),dir(140));
label("$30^{\circ}$",.4*dir(85),dir(175));
real t = (2/(sqrt(3)));
draw(rightanglemark((1,.1),(1,0),(.9,0),s=3));
draw(rightanglemark(rotate(30)*(0,t**4),rotate(0)*(0,t**3),O,s=3));
draw(rightanglemark(rotate(0)*(0,t**3),rotate(-30)*(0,t**2),O,s=3));
draw(rightanglemark(rotate(-30)*(0,t**2),rotate(-60)*(0,t**1),O,s=3));
[/asy]](http://latex.artofproblemsolving.com/3/d/1/3d1ca59de5ec31242d2b311c96ced27d77211649.png)
![\[e^{W(2i\pi n)}=(e^{i\pi})^{W(2n)}=((-1)^2)^{W\cdot n}=1^{W\cdot n}=\boxed{1}\]](http://latex.artofproblemsolving.com/7/f/3/7f3119fda5230d7812683eaffb5309cfa1eef7ed.png)
and the exponent multiplication property.
?
.
for 
and 
?
and it is within this interval so this is one solution.
?
.
?
and obviously it's not in this interval so it's another no.
?
which is roughly slightly bigger than 5 and so yes this is the other solution.
.
