Problems solved, Scores and Team Assignments

Don't forget to see the previous post for on-line 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 = |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