PartialMathematician

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

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

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

a = 9, b = 8, c = 7, and d = 1

9.87 - 0.1 = 9.77. The largest possible value is \(\boxed{9.77}\).

- PM

So the answer is \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).

- PM

Note that in this equation, \(x_1 + x_2 = 13\) and \(x_1x_2 = 4\).

But yes, a good solution would include the use of Vieta's formula. 

In the quadratic polynomial  \(P(x) = ax^2 + bx + c\), the roots \(x_1\) and \(x_2\) satisfy \(x_1 + x_2 = -\dfrac{b}{a}\) and \(x_1 x_2 = \dfrac{c}{a}\).

Using the quadratic formula, we have the two solutions \(\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}\) and \(\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}\). The reciprocals would be \(\dfrac{1}{\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}}\)  and \(\dfrac{1}{\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}}\). Adding them together, we have \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).

- PM

Oh yeah, I just realized that I never used it. Let me remove it. Thanks, CPhill! 

Thanks, CPhill!