geno3141

If the ratio is 4:3, there are 7 parts.

The x-distance from (-2,-5) to (5,-3)  is  5 - -2  =  7     --->     (1/7) x 7  =  1     --->     move 4 x 1 spaces to the right of -2, you end at 2

The y-distance from (-2,-5) to (5,-3)  is  -3 - -5  =  2    --->     (1/7) x 2  =  2/7  --->     move 4 x (2/7)  space up from -5, you end at -3 6/7

The point is (2, -3 6/7)

Let  n  =  number of terms

Let  t₁  =  the first term

Let  t_n  =  the n^th term

Let  d  =  the common difference

Let  S  =  the sum of the terms

Two formulas:  S  =  n · (t₁ + t_n) / 2                        t_n  =  t₁ + (n - 1) · d

t_n  =  t₁ + (n - 1) · d     --->      t_n  =  18 + (20 - 1) · 11     --->     227

S  =  n · (t₁ + t_n) / 2     --->     S  =  20 · (18 + 227) / 2     --->     2450

Since all the cards are dealt out, x must evenly divide into 54.

The numbers that are factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.

So, when x = 1, y = 54; x = 2, y = 27; x = 3, y = 18; x = 6, y = 9; x = 9, y = 6; x = 18, y = 3; x = 27, y = 2; and x = 54, y = 1.

For x to be at least 2 and y to be at least 5, the possibilities are (2, 27), (3,18), (6,9), (9,6).

I believe that the answer is b.

iv won't be correct if there is a constant in the remainder.

hit x hit x miss  =  .80 x .80 x .20  =  .128

hit x miss x hit  =  .80 x .20 x .80  =  .128

miss x hit x hit  =  .20 x .80 x .80  =  .128

Adding these together:                     .384

You will need 3 6's and 2 not-6's.

The probability of getting a 6 is 1/6; the probability of getting a not-6 is 5/6.

(1/6)(1/6)(1/6)(5/6)(5/6)  =  25/7776.

You will need to multiply this answer by the number of ways that 3 6's can be chosen:  ₅C₃ = 10.

Answer =  10 x 25 / 7776  =  0.0322  or  125/3888.

Since angle(YOW) = 100^o, arc(YZW) = 100^o.

Since angle(ZYW) = 10^o, arc(ZW) = 20^o.

Subtracting, arc(YZ) = 80^o.

Angle(YWZ) = ½Arc(YZ) = 40^o.

If you graph these, you can notice that these form a right triangle; therefore these three points are all found on one circle. 

The endpoints of a diameter of this circle are (2,6) and (8,10). 

The midpoint between these two points is (5,8).

This point is the center of the circle; the circumcenter.

First:  notice that triangle(DCE) is similar to triangle(BCF) by AA.

Since BC = 3·DC:  if DE = x, then BF = 3x 

                        and if CE = y, then CF = 3y.

Since triangle(ABE) is a right triangle:  (21 + 3x)²  =  (27)² + (30 + y)²

                                 --->            441 + 126x + 9x²  =  729 + 900 + 60y + y²

                                --->        9x² - y² + 126x - 60y  =  1188                                          (Equation #1)

Since triangle(ADF) is a right triangle:  (27 + x)²  =  (21)² + (3y + 10)²

                                 --->              729 + 54x + x²  =  441 + 9y² + 60y + 100

                                 --->      x² - 9y² + 54x - 60y  =  -188                                               (Equation #2)

Subtracting Equation #2 from Equation #1:     8x² + 8y² + 72x  =  1376                         (Equation #3)

We need to get rid of the y²-term; let's look at triangle(CDE):  x² + y²  =  100     --->     y²  =  100 - x²

Substituting this value into Equation #3:  8x² + 8y² + 72x  =  1376     --->     8x² + 8(100 - x²) + 72x  =  1376      

                   --->     8x² + 800 - 8x² + 72x  =  1376     --->     800 + 72x  =  1376     --->     72x  =  576     --->     x  =  8

Since  x² + y²  =  100  and  x  =  8     --->     y  =  6.

You can now determine all the length of all the sides; divide the region into non-overlapping triangles and find the appropriate areas.

Since they are consecutive integers, they can be labeled:  a, a + 1, and a + 2

Product of the smallest and largest:  a(a + 2)

17 more than three times the median integer:  17 + 3(a + 1)

The equation:     a(a + 2)  =  17 + 3(a + 1)

                            a² + 2a  =  17 + 3a + 3

                        a² - a - 20  =  0

                    (a - 5)(a + 4)  =  0

                Either  a = 5  or  a = -4

Since they must be positive, we can only use  a = 5     --->   a + 1  =  6     --->     a + 2  =  7