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.

## Friday, January 20, 2017

## 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

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

(3x

Combine like terms: x

4. Simplify

x

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

x | x

+ |-----|-------|

b |_

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

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. 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

So 5

25 +225 = c

250 = c

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

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 - 24x - 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 + 6x - 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 inches3. 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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