holds inequality 
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
such that
diverges for 
(harmonic sequence) but converges for 
because 
,
converges
![$f(x)\in R[x]$](http://latex.artofproblemsolving.com/a/1/5/a15dc97c051e81ae9ff0c060eaa4367bc723ccc9.png)
show that 
has at least one root of multiplicity one
be a metric space and 
be a function such that 
.
Prove that for all 
,
exists, where 
is 
.
Prove that the amount of the limit does not depend on choosing 
.
![\[\iint_{R} \sin(xy) \,dx\,dy, \quad R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]\]](http://latex.artofproblemsolving.com/a/2/0/a205868dd99f4794de53ad1db50f34b17c70a923.png)
Y by Rounak_iitr
The coordinates of 
are 
, 
, 
, with 
. The area of is 
. What is the value of 
?
![[asy]
size(10cm);
draw((5,7)--(11,7)--(3,11)--cycle);
label("$A(5,7)$", (5,7),S);
label("$B(11,7)$", (11,7),S);
label("$C(3,y)$", (3,11),W);
[/asy]](//latex.artofproblemsolving.com/d/5/e/d5ed8caf884979a843424f5f8958763603b9776c.png)
are 
, 
, 
, with 
. The area of is 
. What is the value of 
?![[asy]
size(10cm);
draw((5,7)--(11,7)--(3,11)--cycle);
label("$A(5,7)$", (5,7),S);
label("$B(11,7)$", (11,7),S);
label("$C(3,y)$", (3,11),W);
[/asy]](http://latex.artofproblemsolving.com/d/5/e/d5ed8caf884979a843424f5f8958763603b9776c.png)
This post has been edited 2 times. Last edited by rnatog337, Jan 25, 2024, 5:45 PM
,
.
![[asy]
size(10cm);
draw((5,7)--(11,7)--(3,11)--cycle);
draw((3,11)--(3,7)--(5,7),red);
draw((3,7.5)--(3.5,7.5)--(3.5,7));
label("$A(5,7)$", (5,7),S);
label("$B(11,7)$", (11,7),S);
label("$C(3,y)$", (3,11),W);
label("$D(3,7)$", (3,7),SW);
[/asy]](http://latex.artofproblemsolving.com/6/c/3/6c3d5e8cefbc0744689ea5ca762f90f5209701e9.png)
as the point at which 
.![$[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$](http://latex.artofproblemsolving.com/f/3/0/f308cc469812756561fc3e1e8503d82e6b1c9e6a.png)
,
are 
and 
.
are 
Now,![\begin{align*}
[\triangle ABC] = 12 &= \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 \\
&= 3(y-7).
\end{align*}](http://latex.artofproblemsolving.com/4/a/3/4a37bae26ebbd9a4b33446698be5d3b5e83e9ab4.png)
Dividing by 
gives 
so 
because 
and keeps on going until it touches the same ground that 
and 
rest on.
so the diff is in the y intercept so it can be 4 units more or less, clearly by the graph and the equation 
it needs to be 
is 
thus, 
equivalent to 
