Math team will start wth practice on Monday Sept 25, 2017.
You will need to return a signed permission slip (now available in office ) in order to take the bus to a meet.
Friendly Hills will meet as usual before school on Mondays 7:30 to 8:30 in room 121.
Heritage will meet as usual after school on Mondays from 3:20 until 4:30 in room 160.
Meets are on Mondays after school. The bus returns about 5:30.
Schedule
Sept 25 Practice
Oct 2 Practice
Oct 9 Practice
Oct 16 Meet1 at St Paul Academy. Bus returns HER 5:25, FH 5:40
Oct 23 Practice
Oct 30 Meet2 at St Paul Academy. Bus returns HER 5:25, FH 5:40
Nov 6 Practice
Nov 13 Practice
Nov 20 Meet3 at Heritage. Bus returns FH 5:20 (HER 5:00)
Nov 27 Practice
Dec 4 Practice
Dec 11 Meet4 at St Thomas Acad. Bus returns FH 5:10, HER 5:25
Dec 18 Practice
Jan 8 Meet5 at Friendly Hills. Bus returns HER 5:25 (FH 5:05)
Friday, September 1, 2017
Friday, January 20, 2017
Math Team Party
Heritage - party on Friday Jan 20 from 3:30 - 4:30 or 5:00 in the Room 121
Ice cream sundaes, awards, puzzles and games.
Friendly Hills - party on Monday Jan 23 from 3:30 - 4:30 or 5:00 in the Cafetorium
Ice cream sundaes, awards, puzzles and games.
Ice cream sundaes, awards, puzzles and games.
Friendly Hills - party on Monday Jan 23 from 3:30 - 4:30 or 5:00 in the Cafetorium
Ice cream sundaes, awards, puzzles and games.
Monday, January 9, 2017
Year End Results
Team Results
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Individual Results
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HERITAGE SCORES
FRIENDLY HILL SCORES
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SPA Gold | 182 | 122 | 122 | 126 | 118 | 670 |
SPA-Blue | 110 | 100 | 44 | 32 | 38 | 324 |
St Thomas Acad | 46 | 50 | 28 | 24 | 0 | 148 |
Individual Results
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HERITAGE SCORES
7 | Kathryn Lewis | 0 | 31 | ||
5 | Quinn Hendel | 6 | 8 | 14 | 31 |
7 | Meg Noggle | 0 | 25 | ||
5 | Lauren Noggle | 0 | 2 | 2 | 21 |
6 | Ruby Lipschultz | 0 | 2 | 2 | 14 |
7 | Gianna Heil | 0 | 13 | ||
5 | Giovanni Lombardi | 0 | 10 | ||
5 | Gray Gallant | 0 | 2 | 2 | 4 |
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Name | M5A | M5B | M5T | Yr Tot |
Oscar Halverson | 6 | 6 | 12 | 68 |
Thomas Hoffman | 4 | 8 | 12 | 56 |
Sophia Schomer | 4 | 6 | 10 | 52 |
Cyrus Martin Risch | 0 | 8 | 8 | 51 |
Stella Warwick | 2 | 6 | 8 | 40 |
Nina Kessler | 4 | 6 | 10 | 40 |
Will Gannon | 2 | 8 | 10 | 38 |
Miles Dunn | 0 | 2 | 2 | 37 |
Kallie Frett | 2 | 8 | 10 | 36 |
Nick Wendt | 0 | 6 | 6 | 32 |
Ben Sikkink | 0 | 30 | ||
Katherine Meyers | 0 | 6 | 6 | 30 |
Victoria Dzurilla | 4 | 2 | 6 | 28 |
Erik Essen | 2 | 2 | 4 | 28 |
Duy Hoang | 2 | 6 | 8 | 28 |
Amario Sisakda | 2 | 6 | 8 | 26 |
Luke Reisig | 2 | 6 | 8 | 26 |
Shakti Gurung | 0 | 8 | 8 | 23 |
Isaiah Walker | 2 | 2 | 4 | 20 |
Charles Cheesebrough | 0 | 6 | 6 | 19 |
Marcell Booth | 2 | 2 | 4 | 16 |
Claire Newmark | 0 | 4 | 4 | 14 |
Lucia Schomer | 0 | 6 | 6 | 14 |
Mia Cheesebrough | 0 | 2 | 2 | 13 |
T J Kronschnabel | 0 | 2 | 2 | 13 |
Pierce Moberg | 0 | 11 | ||
Sophia Kanavati | 0 | 6 | 6 | 8 |
Achyuth Sujith Kasthoori | 0 | 2 | 2 | 4 |
Maria Axinia | 0 | 4 | ||
Honey Adewuyi | 0 | 0 | ||
Mikayla Stebbing | 0 | 0 | ||
Daniella Sanchez | 0 | 0 | ||
Brittany Sanchez | 0 | 0 |
Monday, December 19, 2016
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 (3x^{2}+4x-7)-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}+4x-7) + ^{-}2 (x^{2}-2x+3) Now let's distribute that negative two.
(3x^{2}+4x-7) + (^{-}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 = (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 x^{3 }= 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 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
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Heritage Scores and Team Assignments
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 (3x^{2}+4x-7)-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}+4x-7) + ^{-}2 (x^{2}-2x+3) Now let's distribute that negative two.
(3x^{2}+4x-7) + (^{-}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 = (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 x^{3 }= 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 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
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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 |
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