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Angle DAN = arctan( 10/20 ) = 26.565º

AP = PS = PQ = cos (26.565º) * 10 = 8.94427191

The area of a square PQRS = PS * PQ = 80 cm²   

The length of  AC = sqrt( 6² + 2² ) = 6.32455532

The length of  BC = AC / 4 = 1.58113883

The coordinates of point B =>   B (3.5, 0.5)  

Without using a calculator, find the largest prime divisor of \(5^{12}\) minus 2 times \(10^6\) plus \(2^{12}\).

\(\begin{array}{|rcll|} \hline && 5^{12} - 2* 10^6 + 2^{12} \\ &=& 5^{12} - 2* \left(5*2\right)^6 + 2^{12} \\ &=& 5^{12} - 2* 5^6*2^6 + 2^{12} \\ &=& \left(5^{6}- 2^{6} \right)^2 \\ &=& \Big( (5^{3}- 2^{3}) (5^{3}+ 2^{3}) \Big)^2 \\ &=& \Big( (125-8) (125+8) \Big)^2 \\ &=& \left( 117*133 \right)^2 \\ &=& \Big( (3^2*13)*(7*19) \Big)^2 \\ \hline \end{array} \)

The largest prime divisor is \(\mathbf{19}\)

The sum of all angles in a pentagon is 540, so wecan use that to find out the missing angle. 98+98+125+108=429, 540-429=111.  

Now we have 2x+11=111

Subract 11 from both sides 2x=100

Divde both sides by 2 x=50

Now for y, since the line is straight the single side will measure 180 degrees.  Subtract 111 from 180 and you get 69.  

Now we have y+8=69

subtract 8 from btoh sides and you get y=61

So now we have X = 50 and Y = 61

If you have any questions feel free to ask

Haha very funny, and incorrect

The ratio is 7:8.

The answer is C.

alpha works out to 60 degrees.

Could you please explain how you found those sums? I don't quite understand.

After working on this problem for a while, I found that the sum of the first row is \(\frac{1}{1-a}\) because the formula for the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), where r=the common ratio. Similarly,  I found that the sum of the second row is \(\frac{b}{1-a}\), and the sum of the third row is \(\frac{b^2}{1-a}\). This led me to believe that the sum of each row could be multiplied by b to get the sum of the next row, and therefore, the sums of the rows form a geometric series. I then found the sum of that series in the same way. Since the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), I found that the sum of all the squares in the grid is \(\frac{\frac{1}{1-a}}{1-b}=\frac{1}{(1-a)(1-b)}\). I haven't worked on part b yet, but I assume the sum can be found the same way.

The difference in our solutions has me wondering if I made a mistake, and if so, where I made it. I would really appreciate if someone could help me out with this!

625  x   625  x  625 =244,140,625

244,140,625  - [2 x 1,000,000] =242,140,625

2^4  x  2^4  x  2^4 =16  x  16  x 16 =4,096

242,140,625  +  4,096 =242,144,721

242,144,721  = 7 x 7 x 9 x 9 x 13 x 13 x 19 x 19

The largest prime divisor = 19

( √32 + √2 )² / √8 - 2 =  a ( b + √2 )  =   25 ( 1 + √2 ) = 60.3553906  

A = 2 ( 1+√2 ) a²

your question evolved from non-question to fine question.

A = [12 * sin(30º)] * 12

angle QPR = 102º

angle NPR = 78º

4x + 78º = 180º      (solve for x )

A = 1/2 [ 6 ( 6 / tan30º) ]

No problem! Have a nice day!

Thanks, didn't realise that at first

I just proved that any number wouldn't work.  There shouldn't be a way to prove it does work if there is proof it doesn't

Oh! Got it!!

If one acute angle of a right triangle is, then the other is, so the triangle is a 45-45-90 triangle. The hypotenuse is times the length of each leg, so each leg has 6. Therefore, the area of the triangle is.

Is there a way to prove that any numbers would work?