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y  =3x + 8

So

x^2  + [ 3x +8 -1]^2   = 10

x^2 + [3x + 7]^2   =10

x^2 + 9x^2 + 42x + 49  =10

10x^2 + 42x + 39  = 0

The sum of the two solutions to this   =   -42 /10   = -4.2

The x  midpoint  of  these two solutions =  [sum of two solutions / 2 ] =  (-4.2 / 2 )  =   -2.1

And the y coordinate of this midpoint is  3(-2.1) + 8 =  1.7

(x,y) =  ( -2.1 , 1.7)   

Since EAG   =x

Then EGF  = x   {since  EG is a transversal cutting parallel lines}

But since  EF  = FG  then angle FEG  also  = x 

So

BEF  + FEG  + AEG   =180

130 + x  + x +  x  =  180

130 + 3x  = 180

3x  =  180 - 130

3x  = 50

x = 50 / 3

See here  :

https://web2.0calc.com/questions/pyramids_4

12 ft  =  12 (.3048)  = 3.6576 m

3.6576  =  -4.9t^2 + 42t + 18.9

-4.9t^2 + 42t + 15.2424  = 0

Using the quadratic formula, the ball reaches 12 ft  at

-42 -  sqrt ( 42^2 - 4 (-4.9)(15.2424) )                -42  - sqrt (2062.75104)

__________________________  =              ___________________   ≈   8.92 sec

                 2 * -4.9                                                          -9.8

Very nice, qjin  !!!

https://web2.0calc.com/questions/please-help-me_126

We can write   these two equations

2x - 9y   = 14     (1)

8x - 42  = y        (2)

Sub (2)  into (1)

2x - 9(8x - 42)  = 14

2x - 72x + 378  = 14

-70x  = 14 - 378

-70x = -364

x = -364 / -70  =  26/5 =  5.2

y = 8 (5.2) -42  =  -0.4

xy  =  (5.2) (-0.4)   =  -2.08

We have a binomial probability

We want to choose any 2 of the 5 to show a one and  a two

Each has a probability of (1/6) of being rolled

And the other 3 die can each show any value except one or two  (4 outcomes for each of the other three die)

The probability  is

C (5,2) * (1/6)^2 * (4/6)^3   =   20 / 243 

Since AB and CD are parallel, BC is a transversal making angles B and C  supplementary

So

B + C  = 180

B + 8B  =  180

9B  = 180

B =  180  / 9    = 20°

C = 45°

By the Law of Sines

8 /sin 45  =  BC / sin 60

8 sin 60 / sin 45  = BC =  4sqrt 6

And

8 /sin 45  = AC /sin75

8sin75 / sin 45 = AC  = 4 ( 1 + sqrt 3)

Area of triangle = A  =   (1/2)AB *BC *sin75  =  (1/2) (  8) ( 4sqrt 6) sin 75   =  24 + 8sqrt (3)

AB * BC * AC  =   8 * [ 4sqrt (6) ]  * [ 4 (1 + sqrt (3) ) ]  =   128 sqrt (6)  ( 1 + sqrt (3) )

Circumradius  =      [ AB * BC * AC ]  / [  4 A ]  =

[   128 sqrt (6)  ( 1 + sqrt (3) ) ]  / [ 4 * ( 24 + 8sqrt (3)) ]    =   4sqrt (2)

7/4

2(−5)^2−5p−15=0 which solves to p=7

Then we get 7x^2+7x+k=0

b2−4ac has to = to 0 so k = 7/4

So  we have

5  =   [ x^3 -2x^2 - 8x   ] / [ x + 2]  +  5 / [x + 2]  -  [11x]  / [x + 2]

Multiply through by  x + 2

5 (x + 2)  = [x^3 -2x^2 -8x ]   + 5  - 11x

5x + 10  = x^3 -2x^2 - 19x + 5

x^3 - 2x^2 - 24x - 5   =   0

We have the form

ax^3  + bx^2 + cx  + d = 0

By Vieta......the sum of the x values  that  satisfy  this  =  (-b) / a  =     - (-2) / 1  =  2