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Complete the square  on the  first equation                   Complete the square  on the  second equation 

x^2 + ax + a^2/4   =   -1 + a^2/4                                        x^2 -  x  + 1/4  =  a + 1/4

(x + a/2)^2  =  (a^2  - 4)  /4                                               ( x - 1/2)^2  = ( 4a + 1) / 4

                                                    Take  both roots  

( x + a/2)  =  ± √(a^2 - 4)  /2                                             ( x  -1/2)   = ± √(a + 1/4) /2

x = ± √(a^2  - 4) / 2   - a/2                                                  x   =  ± √(4a + 1) / 2  + 1/2

Note  that   when  a   = 2      

x  =   √ (2^2 - 4) / 2   -  2/2  =  -1                                      x  =  -  √  ( 9)/ 2  + 1/2 = -3/2 + 1/2  =  -1

So

x^2  + 2x  + 1  =  0      has the double  root of  -1

And

x^2  - x - 2  = 0           has the  roots  2  and  -1 

So....they share  the  root  of  -1      when  a  =  2

The only solution is x = 1.

So would the length that the catcher throws be the diagonal of the diamond + 3?

10 c 2 = 45

Using the formula (a+b)/2*n

we have $a$ as the first term

and $b$ as the last term

$(3N+1)/2*N=80*N$

Since it is average we have to multiply by the amount of terms

Solving we get $N=53$

That's a nice solution method, guest   !!!!!

The diagonal of  the  square  will  be  the  distance  from  home plate to  second base

The diagonal of a square =  side * sqrt (2)

So....This distance  is    90 * sqrt (2)  ≈  127.28 ft  = 127 ft

So .....from 3  ft behind home plate  the distance =    127 +  3  =   130 ft

8 times          8 x 7 = 56      

             NINE x 7 = 63  <==== OVER

Angle DAC   = 180    - 85  =   95° 

And since ABCD  is a cyclic quadtilateral, then angle   DCB =  180  -95  =  85°

Then angle BCF  =180   -  85  = 95°

So angle FBC   =180  - 50  -95   = 35°

And this  =  (1/2)  measure of  minor arc BC    =   2 (35)    =  70°

But angle CAB    is an inscribed angle itercepting this arc....so its measure = (1/2) (70) =  35°

   x = 2 pointers     then 

  52-x = 1 pointers

2x     +  1(52-x) = 89

x = 89-52 = 37  2 point shots      then    52-37 = 15   1-pointers

( I guess this could be considered a 'system' of equations !)

Let the number of  1 point shots =   x

Let the numberof 2 point shots   =  y

So

x  +   y     =    52

1x  + 2y   =    89              subtract the frist equation  from the second

y =  37  = the number of 2 point shots

x +  y  = 52

x + 37  =52

x = 52   - 37   =  15   = the  number of  1 point shots

In the first case    r  =sqrt ( 3^2 + 4^2)   = sqrt (25)   = 5

So sin (a)  =  3/5         cos ( a )   = 4/5

In the second case   r =  sqrt ( 5^2 + 12^2)  = sqrt (169)   =13

So   sin (b)  =  5/13       cos (b)  =  12/17

sin (a + b)   =   

sin a   cos  b     +     sin b    cos  a   =

(3/5)  ( 12/17)   +   (5/13) (4/5)   =

( 36 / 85 )  + (20 / 65)   =

808 /  1105

The product  of   Wilma's  roots  = c/1 =  c

So   1 * 8   =  8    = c

The sum of Greg's roots  =    -b/1  = -b

So    1 +  -4  =  -3

So b =  3

So  the polynomial is

x^2   + 3x   +  8   =   0

Using the  quadratic  formula  the roots are

( -3   ± √  (3)^2  - 4*8 ))  / 2   =   ( -3 ± √-23 )  / 2  =     (-3 ±  i √23) / 2

We compute the complement: we'll count the number of losing tickets.

We saw in part (a) that there are 1200 total possibilities.

To have a losing ticket, you must have at most one correct white ball, and miss the SuperBall.

You miss all 3 white balls if your ticket contains 3 of the 7 white numbers that were not drawn, so there are  possibilities.

You hit 1 white ball and miss the others if your ticket contains 1 of the 3 white numbers that were drawn and 2 of the 7 white numbers that were not drawn, so there are  possibilities.

You miss the SuperBall if you have one of the 9 red numbers that were not drawn.

Therefore, there are  losing tickets.

Hence, there are  winning tickets, and your probability of winning a super prize is

Note: We can approach this problem using direct counting, but there are a number of cases: 1) Matching 3 white balls with any super ball. 2) Matching 2 white balls with any super ball. 3) Matching 1 white ball and matching the super ball. 4) Matching 0 white balls and matching the super ball.

Given the numerous cases here, a complementary counting approach is a faster approach.

We compute the complement: we'll count the number of losing tickets.

We saw in part (a) that there are 1200 total possibilities.

To have a losing ticket, you must have at most one correct white ball, and miss the SuperBall.

You miss all 3 white balls if your ticket contains 3 of the 7 white numbers that were not drawn, so there are  possibilities.

You hit 1 white ball and miss the others if your ticket contains 1 of the 3 white numbers that were drawn and 2 of the 7 white numbers that were not drawn, so there are  possibilities.

You miss the SuperBall if you have one of the 9 red numbers that were not drawn.

Therefore, there are  losing tickets.

Hence, there are  winning tickets, and your probability of winning a super prize is

Note: We can approach this problem using direct counting, but there are a number of cases: 1) Matching 3 white balls with any super ball. 2) Matching 2 white balls with any super ball. 3) Matching 1 white ball and matching the super ball. 4) Matching 0 white balls and matching the super ball.

Given the numerous cases here, a complementary counting approach is a faster approach.

The answer is 4.

When the ball hits the ground, h will be zero

sooooo    0 = -16t^2 + 60t + 54      Use quadratic formula to solve for t  (throuw out negative value)  

                       a = -16      b=60     c = 54

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

\(t = {-60 \pm \sqrt{60^2-4(-16)(54)} \over 2(-16)}\)

12.5 % in decimal form is .125

.125 * 2000 = $ ........

3x^2 + 10x + 8 = (3x+  4   )(x+  2  )     Take it from here......

If we start listing out the gcds, you'll notice that they repeat, 1, 2, 1, 4. 

I'm pretty sure this is due to the +4 in 3n + 4. 

So the possible values are 1, 2, and 4. 

1 + 2 + 4 = 7

I hope this helped. :)))

=^._.^=

If you do it enough times in mod 5, you'll notive a pattern of 2, 4, 1, 2, 4, 1. 

100/3 = 33 remainder 1. 

So our answer is 2. 

I hope this helped. :)))

=^._.^=

xxxxxxxxxxx

Could you clarify please, re the square root signs and the missing f(z) at the end.

6*5/2 = 15 (total ways to paint 2 faces)

2 and 4 are the only way to have the product of 2 numbers 1-6 equal 8.

15 - 1 = 14

14 is our answer.

=^._.^= 

f(-7) = 2(-7)  + 9

f(-7) = -14 + 9

f(-7) = -5

=^._.^=