Don't forget to see the previous post for online lessons on this stuff.
SOLVING THE PRACTICE TEST PROBLEMS
1. What is the intersection of y = 3x + 4 and y = x  2
Since y = x  2, let's substitute (x  2) for y in the first equation:
x  2 = 3x + 4 (Now let's add 2 to both sides)
x = 3x + 6 (Let's get all the x's on the left by adding 3x to both sides)
2x = 6 (To get x all alone on the left, divide both sides by 2)
x = 3 (So we know x. To find y, we can use either equation. Let's use the simplest one y = x  2)
y = x  2 (But we know x = 3, so substitute 3 for x)
y = 3  2 or y = 5. So the lines intersect at (3, 5)
Let's check! (3, 5) should make both equations true
(5) = 3(3)+ 4
5 = 9 + 4 True!
(5) = (3) 2 Yes, that's true too!
2. Find the two points where y = x4 and y = ⅔ x  2 intersect.
This is a little tricky because we have to remember that x4 could =y or x4 could = y
y = x4 y = x4
y = x + 4 (I multiplied both sides by 1 so I'd have just y on the left)
Next I take that second equation and substitute each of these expressions for y.
x4 = ⅔ x  2 and x + 4 = ⅔ x  2 (Let's multiply both sides of both equations by 3)
3x  12 = 2x  6 and 3x + 12 = 2x  6 (Next subtract 2x from both sides)
x  12 = 6 and 5x + 12 = 6 (and now we will solve for x)
x = 6 and 5x = 18
x = 18/5
x = 18/5 = 3
⅗
So we have the two x values (6, y) and (3
⅗, y). Let's find y by using our two equations.
y = x  4 and y = x  4
y = 64 and y = x + 4
y = 2 and y = 4  ( 3
⅗ ) = ⅖
(6,2) and (3
⅗, ⅖)
Check these answers by plugging them into the original 2 equations and you'll be convinced!
3. Simplify (3x^{2}+4x7)2(x^{2}2x+3)
Where most of you went wrong was losing that minus sign in the middle.
If you remember that subtracting is the same as adding a negative, you won't lose the sign.
(3x^{2}+4x7) + ^{}2 (x^{2}2x+3) Now let's distribute that negative two.
(3x^{2}+4x7) + (^{}2x^{2 }+ 4x  6) Did you remember that ^{}2 x ^{}2 = 4 ?
Combine like terms: x^{2 }+ 8x + 1 Done!
4. Simplify
x^{2} + 2x  24
x  4
Looks hard, but here's an easy way to think about this one. 2 x 3 can be represented by two rows of three squares. We want to know what two binomials (a fancy word for a mathematical expression having two terms  like x+1) have a product of x^{2} + 2x  24. (x+a)(x+b) Draw box and let's fill in the four products.
x + a
x  x^{2}  ax 
+ 
b _bx__ab__
Now it can be seen that the product of axb=24 and the sum of a+b=2. 24 = 2x12 or 3x8 or 4x6.
Aha 4 and 6 have a product of 24 and a sum of 2. If you had trouble coming up with this, look at the denominator for a hint  one of the numbers is 4.
x^{2} + 2x  24 = (x4)(x+6) = x + 6
x  4 x4
5.What is the area of a parallelogram formed by y = 1, y = x+3, y = 5, and y = x3.
We could graph this, but we don't even have to. A = h x w. The bottom and top of our parallelogram are y = 1 and y = 5. The difference of 4 is the height. The other two equations have a difference of 6. w x h = 6 x 4 = 24 sq units
Test two:
1. Line segment MH is rotated 180 degrees about the origin to create segment M'H'.
M = (1,4) H = (5,2) Where is M'? Where is H'?
This means M' is going to 180 degrees, or exactly opposite the origin. M'=(1,4) and H'=(5,2)
2. A cube has a volume of 729 cubic inches. What is the length in inches of each edge?
V = l x w x h, but on a cube l=w=h. So x^{3 }= 729. 5x5x5125 so that's too low. 10x10x10 = 1000 so thats a little high. Try 9x9x9 81x9=729 aha! Answer: 9 inches
3. A rectangle has a width of 5 units and an area of 75. What's the length of the diagonal?
Rectangle: A = L x W Hence 75 = L x 5. L must be 15. With a little help from Pythagorus, we know that a^{2 }+ b^{2 }= c^{2}
So 5^{2 }+ 15^{2 }= c^{2}
25 +225 = c^{2}
250 = c^{2}
c=√250
=√5x5x10
=5√10
4. Sanjay's basketball has Volume = 256/3 π. How big must the rim be? Express as an inequality in terms of C (circumference) and π.
Volume of a sphere is 4/3πr^{3 }so 256/3 π = 4/3πr^{3 }Hence 256 = 4r^{3 }or 64 = r^{3 }Thus r = 4.
C = 2πr so C =2π4 = 8π. So that's if the ball exactly fits the rim, but the rim needs to be slightly bigger to allow the ball to fall through. Hence C > 8π inches.
5. Sanjay's backboard is 6ft by 3.5 ft. He will use tape to mark out the rectangular "sweet spot." It's height is ¾ its length. The ratio of the area of the "sweet spot"to the area of the backboard is 1:7. How many feet of tape will Sanjay need?
Let's calculate the size of the sweet spot. The back board is 6x3.5 = 21 sq ft. The sweet spot is one seventh or hence 3 sq ft. So height x width=3 but also h=¾w, so substituting we get
¾w x w = 3 ( Now multiply both sides by 4 to get...)
3w x w =12 ( Now divide both sides by 3 to get...)
w x w = 4 So w = 2 Width of the "sweet spot" is 2ft. Its height is ¾ that or ¾x2 =1.5ft
So he'll need 2 + 1.5 + 2 +1.5 to go all the way around the "sweet spot". That 7 ft.
Friendly Hills Scores and Team Assignments
>
Oscar Halverson 
10 

42.0 
Thomas Hoffman 
6 
6 
34.7 
Kallie Frett 
7 
6 
32.3 
Nick Wendt 
2 
4 
26.0 
Cyrus Martin Risch 
0 
6 
24.3 
Nina Kessler 
5 
2 
18.3 
Sophia Schomer 
2 
2 
18.0 
Victoria Dzurilla 
2 
4 
16.0 
Miles Dunn 
0 
2 
15.0 
Katherine Meyers 
0 
0 
14.7 
Stella Warwick 
0 
0 
14.7 
Will Gannon 
0 
0 
13.3 
Erik Essen 
0 
2 
12.7 
Ben Sikkink 


12.3 
Amario Sisakda 
0 
2 
10.7 
Duy Hoang 
0 
0 
9.3 
Isaiah Walker 
0 
2 
8.7 
Luke Reisig 
0 
0 
7.3 
Marcell Booth 
0 
2 
6.0 
Shakti Gurung 


5.0 
Charles Cheesebrough 
0 
0 
4.3 
Sophia Kanavati 
0 
0 
4.0 
Mia Cheesebrough 
0 
0 
3.7 
T J Kronschnabel 


3.7 
Claire Newmark 
0 
0 
3.3 
Lucia Schomer 
0 
0 
2.7 
Daniella Sanchez 
0 
0 
1.3 
Brittany Sanchez 
0 
0 
1.3 
Achyuth Sujith Kasthoori 
0 
0 
0.7 
Heritage Scores and Team Assignments
Kathryn
Lewis 
8 
26.3 
Quinn Hendel 
0 
6.3 
Lauren Noggle 
0 
6.3 
Gianna Heil 

4.3 
Ruby Lipschultz 

4.0 
Gray Gallant 
0 
2.0 