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1. The product of two positive integers minus their sum is 39. The integers are relatively prime, and each is less than 20. What is the sum of the two integers?

2. If a is a constant such that 4x^2-12x+a is the square of a binomial, then what is a?

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.

 Dec 23, 2018
 #1
avatar+700 
+2

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

 Dec 23, 2018
 #2
avatar+4322 
+4

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    \(11+5=\boxed{16}.\)

.
 Dec 23, 2018
edited by tertre  Dec 23, 2018
 #3
avatar+700 
+1

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

 Dec 23, 2018
 #4
avatar+103122 
+3

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......!!!

 

 

cool cool cool

 Dec 23, 2018
edited by CPhill  Dec 23, 2018
edited by CPhill  Dec 23, 2018
 #9
avatar+4322 
0

Oh, I get what PM means. We have \(\frac{\pi*r^2}{4},\frac{9\pi}{4}.\)

tertre  Dec 23, 2018
 #5
avatar+103122 
+2

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

 

 

cool cool cool

 Dec 23, 2018
 #6
avatar+4322 
+1

No problem! smiley

tertre  Dec 23, 2018
 #8
avatar+700 
0

No problem! laugh

PartialMathematician  Dec 23, 2018
 #7
avatar+700 
0

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

 Dec 23, 2018
edited by PartialMathematician  Dec 23, 2018
 #10
avatar+700 
+1

Here is an image representation. The entire circle is not in the square. Only a 1/4 of the circle is inside. That is why the answer is \(\boxed{\dfrac{9}{4}\pi}\) units^2, not \(9\) units^2.

 

Hope this helps, 

- PM

PartialMathematician  Dec 23, 2018
 #13
avatar+103122 
+1

Sorry, PM.....I forgot we had a quarter-circle.....your answer is correct!!!

 

I'll make an edit

 

Thanks for pointing out my silly error!!!!

 

 

cool cool cool

CPhill  Dec 23, 2018
 #11
avatar+4322 
0

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

 Dec 23, 2018
 #12
avatar+700 
+1

and also here.

PartialMathematician  Dec 23, 2018

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