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Since we know the length of the square's side is 12 cm, we can calculate the area of the square:

Area of square ABCD=12 cm×12 cm=144 cm2

To determine the area of the shaded region, we need to calculate the areas of the semicircles and subtract them from the area of the square.

First, let's find the area of one semicircle:

Area of semicircle = 21​πr2

We know the diameter of the semicircle is equal to the length of the square's side, which is 12 cm. Therefore, the radius of the semicircle is 6 cm.

Area of semicircle = 21​π×62 cm2=18π cm2

Since there are two semicircles, we need to double the area of one semicircle:

Total area of semicircles = 2×18π cm2=36π cm2

Finally, we can subtract the area of the semicircles from the area of the square to find the area of the shaded region:

Area of shaded region = Area of square ABCD - Total area of semicircles

Area of shaded region = 144 cm2−36π cm2

The quadratic equation $x^2-5x+5=-4x+7$ has two roots. Find the absolute difference of these roots.  

                                         x^2-5x+5=-4x+7  

Combine like terms           x² – x – 2  =  0  

Factor the quadratic         (x – 2)(x + 1)  =  0  

So, the roots are                 x  =  2  

                                           x  =  –1   

Can you take it from there?  

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The quadratic equation x² – 5x + t = 3x has only positive integer roots. Find the average of all distinct possible values of t.  

Subtract 3x from both sides        x² – 8x + t = 0    

Note that the resulting   

equation will factor.                     (x – 1)(x – 7) = 0   gives us t =   7  

                                                   (x – 2)(x – 6) = 0   gives us t = 12  

                                                   (x – 3)(x – 5) = 0   gives us t = 15  

                                                   (x – 4)(x – 4) = 0   gives us t = 16  

                                            7 + 12 + 15 + 16          50

Average of possible t's  =  –––––––––––––––  =  ––––  =  12.5  

                                                       4                        4  

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The real numbers x and y satisfy x (x+y) = 27 and (x+y) = 54. What is the value of (x+y)^2?  

If  (x+y) = 54  then  (x+y)² = 54²  =  2916   

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Combining like terms, we have the inequality 2x2+nx−135<0. We can use the quadratic formula to solve for the roots of 2x2+nx−135=0.

The quadratic formula gives us:

x=2⋅2−n±n2−4⋅2⋅−135​​

x=4−n±n2+1080​​

Since we are given that the inequality has no real solutions, the discriminant n2+1080 must be negative. This gives us the inequality n2<−1080.

Since n is an integer, the only possible values of n are −34, −33, −32, ..., 32, 33, and 34. Therefore, there are 67​ possible values of n.