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That is a really cool technique.  Thanks Tiggsy.

You are very welcome. :)

Multiplication is reapeated addition so just add $3$ to itself untill you get $12$.. Then you will se that we have to add $4$ times so blank is equal to $4$.

The 3 congruences can be re-written as follows:

x mod 2 =1,
x mod 9  =6,
x mod 25   =17, solve for x

Using CRT plus MMI

i=0;j=0;m=0;t=0;a=(2, 9, 25);r= (1, 6, 17);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

x =450m + 267, where m=0, 1, 2, 3.........etc.

x =(450 x 2) + 267

x =1167 - This is the least 4-digit solution.

Is the LaTeX not working?

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

(Test)

If ΔABC ≅ ΔDEF, 

DE = 17

EF = 13

DF= 9

And BC = 2x-5

Then which of the following is the value of x? None of the ''above''

BC = EF       2x - 5 = 13      x = 4   

Trapezoid ABCD has base AB = 20 units and base CD = 30 units. Diagonals AC and BD intersect at X. If the area of trapezoid ABCD is 400 square units, what is the area of triangle BXC?

area of triangle BXC = 96 squared units

(10 / 3) / (4 / 1) = 10*1 / 3*4 = 10 / 12

My tips: That's all.

 ABCDEFGH is a regular octagon of side 12cm. Find the area in square centimeters of trapezoid BCEF.

Area of trapezoid BCEF is 296.7350647 cm²

10/12

An octagon is inscribed in a square so that the vertices of the octagon trisect the sides of the square. The perimeter of the square is 60 centimeters. What is the number of square centimeters in the area of the octagon?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The square side is 15 cm, and the side of an octagon is 5 cm.

Octagon area = 8(2.5 * 7.5) = 150 squared centimeters.

Thanks! Both of your answers are great!

Use pythagorean theorem to find AC, then multiply by height AE to get 30

The chance of drawing an ace or a king ... I think 86% seems high. 

I would have said the first card has an 8/52 chance. 

If that card isn't an ace or a king, then the second card has an 8/51 chance. 

Added together is 0.3107 ... 31% still seems high, but whatever. 

  .75 x   <= 30.75

See numerous solutions here:

https://www.wolframalpha.com/input/?i=%282*a*b%2Bb*c%2Bc*a%29+mod13%3D%3D0%2C+%28a*b%2B2*b*c%2Bc*a%29mod13%3D%3D6*a*b*c%2C+%28a*b%2Bb*c%2B2*c*a%29mod13%3D%3D8*a*b*c%2C+integer+solution

By computer program:

for [a = 1, a <= 12, a = a + 1]

for [b = 1, b <= 12, b = b + 1]

for [c = 1, c <= 12, c = c + 1]

  if (2*a*b + b*c + c*a % 13 == 0)

  if (a*b + 2*bb*c + c*a % 13 == 6*a*b*c % 13)

  if (a*b + b*c + 2*c*a % 13 == 8*a*b*c)

  output(a,b,c)

output: a = 7, b = 11, c = 3, so the answer is 8.

can someone help me ASAP i need this last question to submit my assignment.

If it is: 22^(22*(11^11)) mod 10^20 =81,356,903,237,166,301,184 - These are the last 20 digits.

If it is: 22^(22^(11^11)) mod 10^20 =76,852,909,220,939,431,936 - These are the last 20 digits.

The line 2y=3x intersects the ellipse (x²/8)+(y²/18)=1 at the points Q and R. What is the length of segment QR?

The mean score will always be a z score of 0

Lower than the mean will have negative z scores and higher will have positive z scores.

99.7% of all the scores will have z scores of between   -3 and  +3.

Both Ann and Bill did badly because their z score is negative (less than 0)

The one that is more unusual will be the one whose score is furthest from 0. 

In this case that worst one.  Which one is that?