I have no idea what you are talking about.
Can you explain it better?
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.
The area of triangle ABX is 24*sqrt(2).
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
Nice one! I think you're right :D
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) ]
yes!
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?
Yes.