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 = |x-4| and y = ⅔ x - 2 intersect.
This is a little tricky because we have to remember that x-4 could =y or x-4 could = -y
y = x-4 -y = x-4
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.
x-4 = 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 = 3⅗
So we have the two x values (6, y) and (3⅗, y). Let's find y by using our two equations.
y = x - 4 and -y = x - 4
y = 6-4 and y = -x + 4
y = 2 and y = 4 - ( 3⅗ ) = ⅖
(6,2) and (3⅗, ⅖)
Check these answers by plugging them into the original 2 equations and you'll be convinced!
3. Simplify (3x2+4x-7)-2(x2-2x+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.
(3x2+4x-7) + -2 (x2-2x+3) Now let's distribute that negative two.
(3x2+4x-7) + (-2x2 + 4x - 6) Did you remember that -2 x -2 = 4 ?
Combine like terms: x2 + 8x + 1 Done!
4. Simplify
x2 + 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 x2 + 2x - 24. (x+a)(x+b) Draw box and let's fill in the four products.
x + a
x | x2 | 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.
x2 + 2x - 24 = (x-4)(x+6) = x + 6
x - 4 x-4
5.What is the area of a parallelogram formed by y = 1, y = x+3, y = 5, and y = x-3.
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 x3 = 729. 5x5x5-125 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 a2 + b2 = c2
So 52 + 152 = c2
25 +225 = c2
250 = c2
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 4/3πr3 so 256/3 π = 4/3πr3 Hence 256 = 4r3 or 64 = r3 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
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| 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 |
