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G9x) has to cancel the -7x^4    (and the 3x^3)    so g(x) has to be of degree 4

360 = b-d   + b    + b +d    + b + 2d

 360 = 4b + 2d

        = 288+ 2d

d = 36                       Angle D =    b + 2d = 72 + 2(36) = 144 degrees

Let

\[f(x) = \sum_{k = 0}^\infty x^k = \frac{1}{1 - x}.\]

Taking the derivative, we get

\[f'(x) = \sum_{k = 1}^\infty kx^{k - 1} = \frac{x}{(1 - x)^2}.\]

Then

\[\sum_{k = 1}^\infty kx^k = \frac{x^2}{(1 - x)^2}.\]

Therefore,

\[\sum_{k = 1}^\infty kr^k = \frac{r^2}{(1 - r)^2}.\]

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7       --->                      -4x + 3y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z = 3        --->         z  =  3 - kx

Substituting these:        -6x - (3 - kx)  =  -15

                                         (k - 6)x  = -18

If  k - 6  =  0,  there is no value for x that will result in a product of -18

so  k = 6  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of -18.

So the answer is k = 6.

If the slopes are the same then there are no solutions because theya re parallel and do not cross

Arrange each into y = mx+ b form      where m = slope       find the value of 'a' that make them the same slope

   the first equation is already in y = mx + b form         m = 11

2y-4 = ax +5y+3

3y = -ax -7      

y = - a/3  x   - 7/3       so     -a/3 = 11    solve for 'a'

thanks so much

√(x + 24) / x = 7

√(x + 24) = 7x

square both sides

x + 24 = 49x^2

49x^2 - x - 24 = 0

x = (1 + √(1^2 + 4*24*49)) / 2*49 

Let me correct the question first (red ink).

In a triangle, ABC, points P and Q lie on the side BC. It is given that BP = 1, PQ = 3, and QC = 1. The line passing through P and parallel to AC intersects AB in K and the line passing through Q and parallel to AB intersects AC in L. Show that KL || BC by proving that triangles AKL and ABC are similar. 

Find another way to prove that KL || BC, by showing that a certain quadrilateral in the diagram is a parallelogram.

Hint:    ∠A = 90º         AB = 3        AC = 4

Now you can solve it. 

1a + 2   =   2b + 1    I'll switch to  use x and y

2y  = x+1

y = 1/2x + 1/2

    now solve for integers 

       when x = 3   y = 2    too small

       when x = 5   y = 3  

base 3     21₃ = 7

base 5    12₅  = 7                 7

h(t)=-16t^2 +198

100 = -16t^2 + 198 

-16t^2 + 98 = 0

16t^2 = 98

t^2 = 98/16

t = 2.475  sec

You get the maximum when a = b = c = 6/7.  Then min{1/2*ab, ac, 2bc} = 18/49.

Graphical soln to #2

Graphical soln to #1

n = 2501

(4-5i) (-5-5i)      expand

-20 -20i +25i + 25i²         simplify    (remember i²  = - 1 )

-20 -25     -20i+25i

    -45 + 5i

Ways to arrange the math books

1x1xxx       4 ways x 2 =8

1xx1xx        3 ways x2 6

1xxx1x        2 ways x 2 =4

1xxxx1         2 ways

                   total   20 ways      

                 the other three slots can be arranged amonsth the remaining 4 books in  4 * 3 * 2 *1 = 24 ways for each math book arrangement

20 * 24 = 480 ways

Binary number in base 10 =  2 + 4 + 8 64 + 256 + 1024  = 1358

Convert 1358 to base 8

     2   x 512

     5  x 64 

     1 x 8

      6 x 1                        2516₈

As follows:

Disregard.......re-read the Q .....

ab = 25    is stated in your question.....

There is only one     x = -2.   sum = -2

f = 9s

In three years :

   f+3  = 6 (s+3)      Sub in the red equation

  9s +3  = 6 (s+3)

   9s + 3 = 6s + 18

   3s      = 15

    s = 5  y/o     then f = 9s = 45 y/o

The sum of the squares of the diagonal =  sum of sides squared

    40^2 +d^2  =  30^2 + 30^2 + 20^2 + 20^2

         d^2 = 1000

          d = sqrt (1000)     simplify as needed ......     ~EP