geno3141

Because ABCD is a parallelogram, angle(D)  =  60°.

Because AB || DC. the alternate interior angle below the letter A is also 60°.

Angle(A)  =  90° - x°.

Thus the three angles to the left of diagonal line through A are:  (90° - x°) + 60° + 2x°  =  180°.

Solving:     150° + x°  =  180°     --->      x  =  30°

Let     x  =  mom's age (now)

Then  y  =  my age (now)

Seven years ago:         (x - 7)  =  7(y - 7)       --->     x - 7  =  7y - 49      --->     x - 7y  =  -42

Eight years from now:  (x + 8)  =  2(y + 8)     --->     x + 8  =  2y + 16     --->     x - 2y  =  8

Combining:     x - 7y  =  -42                                         --->       x - 7y  =  -42

                       x - 2y  =  8          --->   multiply by -1     --->      -x + 2y  =  -8

Adding down the columns:                                                            -5y  =  -50     

Dividing:                                                                                --->     y  =  10     (my age now) 

Substituting:   x - 2y  =  8     --->     x - 2(10)  =  8     --->     x  =  28  (mom's age now)

Triangle(BPS) is similar to Triangle(BAC)     --->     Area(BPS) / Area(BAC)  =  BP² / BA²  =  2² / 3²  =  4/9

     Consequently, Area(APSC)  =  5/9 · Area(ABC)

Similarly, Area(DWT) / Area(DAC)  =  4/9     --->     Area(WACT)  =  5/9 · Area(DAC)

Therefore, Area(APSCTW)  =  5/9 · Area(ABCD)     --->     Area(APSCTW)  =  5/9 · 180  =  20

Since  a  and  d  are both positive, and these are sides of a right triangle,  a + 2d  must be the hypotenuse.

Pythagorean Theorem:  (a)²  +  (a + d)²  =  (a + 2d)²

--->     a² + a² + 2ad + d²  =  a² + 4ad + 4d²

--->     a² - 2ad - 3d²  =  0

--->     (a - 3d)(a + d)  =  0

--->     Either  a = 3d  or  a = -d

--->     Since  a  and  d  are both positive,  a  cannot equal  -d.

--->     When  a  =  3d     --->     a/d  =  3  

I think that the third problem is:  (9-x²) / (9x + 27)

This, when factored becomes:  [ (3-x)(3+x) ] / [ 9(x+3) ]

Since (3+x) cancels (x+3):           (3-x) / 9

The calculator at this site will multiply these two number for you.

However, you needn't multiply the entirety of the two numbers together to get your answer; all you need to do is multiply 89 x 21.

89 x 21  =  1869; so the last two digits of your answer will be 69; any digits before the 89 and the 21 will not affect the last two digits.

I am going to assume that the particle can move only either to the East (E) or to the North (N).

Going from (0,0) to (3,1) requires, in some order E x E x E x N.

Going from (0,0 to (1,3) requires, in some order E x N x N x N.

So:  E x E x E x N  =  16 x ( E x N x N x N)

--->   E³ x N  =  16 x E x N³

Dividing both sides by E x N:

--->   E²  =  16 N²

--->   E² / N²  =  16

--->   E / N  =  4

Since the probability of E is  p,  the probability of N is  1 - p:

--->   E / N  =  p / (1 - p)  =  4     --->     p  =  4(1 - p)     --->     p  =  4 - 4p     --->     5p  =  4     --->      p  =  4/5

I have a different way of looking at this problem (that results in a different answer):

Choose two of the 13 different ranks to get the two pairs:  ₁₃C₂  =  78

Choose two of the 4 different suits for the first pair:  ₄C₂  =  6

Choose two of the 4 different suits for the second pair:  ₄C₂  =  6

Choose one of the 44 cards that remain:  44

Multiply these answers together:  78 x 6 x 6 x 44  =  123 552

A is true.

B is false because the diagonals of a rhombus are also perpendicular.

C is true.

D is true.

E is true.

I have an answer for this, but it takes a lot of steps -- I welcome a quicker solution.

1) Find the length of the diameter AC: 

          Since AC is a diameter, I can use triangle ABC to find this length.

          Both AB and BC = 5 and triangle ABC is an isosceles right triangle, AC  =  5·sqrt(2).

2) Find the length PC:

          Triangle ABC is a right triangle (angle ABC is inscribed in a semicircle), so use the Pythagorean Theorem.

          PC² + PA²  =  AC²     --->     PC² + 4²  =  [5·sqrt(2)]²     --->     PC²  +  16  =  50     --->     PC²  =  34     --->     PC  =  sqrt(34).

3) Find the size of angle(PAC):

          Use the Law of Cosines for triangle PAC).

           PC²  =  PA² + AC²  -  2·PA·AC·cos(PAC)     --->     [sqrt(34)]²  =  (4)² + [5·sqrt(2)]² - 2·(4)·5(sqrt(2)·cos(PAC)

                   --->     34  =  16 + 50 - 40·sqrt(2)·cos(PAC)     --->     -32  =  -40·sqrt(2)·cos(PAC)     --->     PAC  =  55.5501^o

4) angle(PAD)  =  45^o + 55.5501^o  =  100.5501^o

5) Use the Law of Cosines on Triangle(APD):

          AP = 5; AD = 5; angle(PAD) = 100.501^o     --->     DP²  =  AP² + AD² - 2·AD·AP·cos(PAD)

                  --->     DP²  =  (4)² + (5)² - 2·(4)·(5)·cos(100.5501)      --->     DP²  =  48.3238      --->      DP  =  6.95

Two comments:

1)  I hope that I haven't made a mistake!

2)  I hope that someone can find an easier way!