(1) Rewrite in standard form:

`3x^2-18x-y^2-2y+20=0`

`3(x^2-6x)-(y^2+2y)=-20`

Complete the square on the expressions in the parantheses:

`3(x^2-6x+9)-(y^2+2y+1)=-20+27-1`

`(x-3)^2/2-(y+1)^2/6=1`

This is a hyperbola opening left/right. If the equation is in the form `(x-h)^2/a^2-(y-k)^2/b^2=1` then the center of the hyperbola is at (h,k) and the slopes of teh asymptotes are `+-...

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(1) Rewrite in standard form:

`3x^2-18x-y^2-2y+20=0`

`3(x^2-6x)-(y^2+2y)=-20`

Complete the square on the expressions in the parantheses:

`3(x^2-6x+9)-(y^2+2y+1)=-20+27-1`

`(x-3)^2/2-(y+1)^2/6=1`

This is a hyperbola opening left/right. If the equation is in the form `(x-h)^2/a^2-(y-k)^2/b^2=1` then the center of the hyperbola is at (h,k) and the slopes of teh asymptotes are `+- b/a`

**Here we have `m=+-sqrt(6)/sqrt(2)=+-sqrt(3)` **

The graph:

(2) Use `P(n,r)=(n!)/((n-r)!)` :

`(P(8,5))/(P(12,4)P(8,4))=((8!)/(3!))/((12!)/(8!)*(8!)/(4!))`

`=(8*7*6*5*4)/((12*11*10*9*8!*8*7*6*5*4!)/(8!*4!))`

`=6720/(12*11*10*9*8*7*6*5)`

`=6720/19958400=1/2970`

(3) Solve `sin^2x-sinx+1=cos^2x`

Rewrite the right side in terms of sinx:

`sin^2x-sinx+1=1-sin^2x`

`2sin^2x-sinx=0`

`sinx(2sinx-1)=0`

`sinx=0 ==> x=0,pi`

`sinx=1/2 ==> x=pi/6,(5pi)/6`

The graphs:

(4) Solve 6x+2y=0 and 3x+7y=24 simultaneously by elimination:

6x+2y=0

3x+7y=24

---------------------Multiply the second equation by -2:

6x+2y=0

-6x-14y=-48

----------------

-12y=-48 ==> y=4 so `x=4/3`

The solution is `(4/3,4)`

(5) ----

(6) ----

(7) Find the points that are in the solution set of y>-2x-1.

First graph the line -- no points on the line are solutions to the inequality. Next choose a point not on the line; say (0,0). Checking the inequality we see that 0>-1 is true -- so the half-plane defined by the line containing the point (0,0) has all of the solutions. If this is a multiple choice question, just see which points are on that side of the line. The graph:

Points "above" or to the "right" of the line are solutions.

(8) Solve the system 5x-2y=25; 5x+3y=5

5x-2y=25

5x+3y=5

---------- Subtract the first equation from the second equation

5y=-20 ==> y=-4 then `x=17/5`

The solution is `(17/5,-4)`

(9) ---

(10) We are given three sides of a triangle: a=22,b=39,c=19 and we are asked to find the measure of angle B (assumed to be opposite the side b).

Use the Law of Cosines: `b^2=a^2+c^2-2acCosB`

`1521=484+361-2(22)(19)cosB`

`676=-836cosB`

`cosB~~-.8086124402`

`B~~cos^(-1)(-.8086124402)~~143.961^@`

The measure of angle B is approximately 144 degrees. (This makes sense since b is the longest side.)