Urgent

1) we have X and Y as our two integers. We have the equation \(X\cdot Y - (X+Y) = 39 \Rightarrow XY - X - Y = 39\). The two numbers are \(11\) and \(5\). The sum of these two numbers is \(\boxed{16}\).

- PM

We have \(xy-x+y=39\) . By inspection, that both are relatively prime, as stated in the problem.Thus, we have \(x=11, y=5\), so their sum is    \(11+5=\boxed{16}.\)

2) We can factor the binomial into \((2x - 3)(2x - 3)\). Factoring it out, we have \(4x^2 - 12x + 9\), so \(\boxed{a=9}\). The 4 in front of x^2 must be factoring into 2*2 to be a square, and 12 must turn into 6 and 6. 

- PM

3. Three of the four vertices of a rectangle are (5,11), (16,11) and (16,-2). What is the area of the intersection of this rectangular region and the region inside the graph of the equation (x-5)^2+(y+2)^2=9? Express your answer as a common fraction in terms of pi.

The remaiining vertex of the rectangle is (5, -2)

And this is the center of the circle, as well

Here's a graph :

The region is just the area of a quarter-circle with a radius of 3  =

pi * (3)^2  / 4 =

9/4  pi units^2

Thanks to PM for spotting my earlier mistake......!!!

Thanks, guys.....!!!!

I think this is the answer to 3) 

We can make out based on the vertices given that the 4th vertex of the rectangle is (-5, 2). By using the Pythagorean Theorem, we can determine the radius of the circle, which is (\(3\sqrt{9}\)).

We also determine the center of the circle by equating x - 5 = 0 and y + 2 = 0. We solve and get the coordinates (5, -2). We notice this is one of the vertices of the rectangle, and since each corner of a rectangle is 90 degrees, we find the area of the sector to be 1/4 of the total area of the circle.

We then, using the circle area formula, get \(\boxed{\dfrac{9}{4}\pi}\), the answer. The answer is not 9 pi units^2 because we need to multiply by the sector of the total area of the circle. 9 * 1/4 = 9/4.

- PM

Also, the last questions was posted here: https://web2.0calc.com/questions/help_81210