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FGCU Invitational
Algebra 2 Individual
Solutions
1)
B
y=6x-3 x=6y-3  (x+3)/6 =f-1(x), evaluate for x=a-1 (a-1 +3)/6 (a+2)/6
2)
D
Factor to get 9x(1+1+1)=9x(3)= (32)x(3)=32x(3)=32x+1
3)
C
Looking for xy; One method is to multiply each equation by xy  y+2x=10xy and 3y-5x=3xy Using either substitution or elimination solve the system getting y= 1/3, plug into
an equation to solve for x , getting x=1/4, so xy=1/12
4)
D
Get common bases; (24)x+ 24x-2 =40 rewrite to be able to factor out an x term 24x +
24x/22=4024x(1+1/4)=4024x(5/4)=4024x=40(4/5)24x=3224x=254x=5x=5/4
5)
A
Let u =x1/3 and solve 2u2+u-6=0 u=-2 or u=3/2 so x1/3=-2 x=-8 or x1/3=3/2x=27/8
6)
C
Solve two cases(neg and pos): neg case 2<= -(5x-3)<7-2>=5x-3>-7 1/5>=x>-4/5; pos
case 2<=(5x-3)<7 1<=x<2
7)
A
f(g(h(x)))=f(g(2-x))=f(4-4x+x2)=3(4-4x+x2)-2=3x2-12x+10
8)
A,B,C,D Since the line can be perpendicular to any radius all slopes are possible.
9)
D
10)
B
Factor by grouping: a(a-p)+q(a-p)= (a+q)(a-p)
3x3 – 2x2 - 3x – 2
-4x3 + 0x2
- 6x2 + 5x
-4x - 7
-13
11)
D
A vertical line other than the line x=0, so the slope is undefined
12)
A
Expand (x-y)2= x2 – 2xy + y2 = 16, replace x2 + y2 with 124124-2xy=16  xy=54
13)
E
1/(3+4i)  1/(3+4i) *(3-4i)/(3-4i) (3-4i)/(9-16i2) (3-4i)/25
14)
A
with a hole at x=-1, undefined at x=-3 and an xintercept at x=5 (root)
15)
D
, the critical points are x=5 and x=7, check
the intervals : x<5, 5<x<7, and x>7 (recall 5 is not in the domain for x), select the
intervals <0
16)
B
4 x+y = 4 4x+y=4,and 4 x-y =4 -1x-y=-1; solve the system x=3/2 and y=5/2
17)
D
Using y-y1 = m(x-x1) and m=(3- -1)/(-2 – 5)=-4/7 then y+1=-4/7(x-5)
7y+7=-4x+204x+7y=13 2x+ 7/2y=13/2 so a+b = 20/2=10
18)
B
Expand (3-i)3=(3)3 +3(3)2(-i)+3(3)(-i)2+(-i)3=+ 27 -27i - 9 +i=18-26i so a=18 and b=-26
19)
B
The x-intercept is 4 and the y-intercept is 3 so b=4 and h=3 then Area = 1/2bh
FGCU Invitational
Algebra 2 Individual
Solutions
20)
D
ln53x = ln123xln5=ln12 x=ln12/(3ln5)
21)
C
Absolute value is > 0 so the smallest the function value can be is 2
22)
A
Absolute value is the distance |a+bi|=sqrt(a2 + b2) sqrt(72 + (5)2)
23)
A
Solving the first equation for x by squaring both sides (first rearrange to make less work)

value to get 2(5/4)-8(25/16)+2=-8

x=25/16 then plug in the
24)
E
Note the word MUST…since y=±5 and x =7, then xy=±35
25)
B
Joe was not driving each way the same amount of time. Since the distance really doesn’t
matter say Joe drove 5 miles to work and D=rt so on the way to work 5=55t where
t=1/11 hour. On the way home 5=45t and took 1/9hour. That means Joe drove 10 miles
in 20/99 hour so using D=rt 10=r20/99 r=99/2=49.5mph
26)
C
Original mix
water
New mix
Am’t drained(x)
quarts
15
x
15
15 -x
%
.30
0
.25
.30
Antifreeze
4.5
0
3.75
3.75
You know that you must
drain enough mixture to
reduce the antifreeze to
3.75 quarts so solve the
last row.
(15 –x).30 = 3.75 15-x=12.5x=2.5qt
27)
D
Using synthetic division 2) 2 0 -K 2 -12
4
8 16-2K 36-4K
2 4 8-K 18-2K 0
Solve -12 +(36-4k)=0
28)
B
Draw a triangle and use Pythagorean Thm: 92 + x2 = (x+3)2
29)
A
Let c = the current of the stream
rate
*time
=Distance So
4x+4c=72x+c=18
X+c
4
72
6x-6c=72-x+c=-122c=6c=3
X-c
6
72
30)
B
4y2 +8y
-25x2 + 50x
=121
2
4(y +2y ) -25(x2 - 2x
) =121
4(y2 +2y +1 ) -25(x2 - 2x+ 1 ) =121 + 4 -25
4(y + 1) 2 -25(x - 1) 2 =100