If you want to lear more about one of these topics, do a goole search of the topic under kahn academy. For example, "kahn academy factors"
TOPICS FOR MEET 1 (No calculators allowed!)
common factors
primes
evaluating expressions (order of operations)
factorials
fractions and decimals
ratios
turning English sentences into expressions using variables
area and perimeter
coordinate plane
averages  mean, median, and mode
logic problems
Here's a cool formula (Heron's formula) for finding the area of a triangle when you know all three sides.
Say 3", 4", 5"
First calculate the semiperimeter (half the perimeter)
s=(a+b+c)/2 s=(3+4+5)/2 = 12/2 = 6
Now multiply s(sa)(sb)(sc) = 6(63)(64)(65) = 6 * 3 * 2 * 1 = 36
Now take the square root  what number times itself = 36? The answer is 6
The area of the triangle is 6 sq in. The nice thing is that this works for all triangles!
Friday, September 23, 2016
Monday, August 29, 2016
2016 2017 Schedule Updated
To sign up, pick up a permission form from your teacher or from the office. Have a parent sign and return it to math club or to the office by the morning of Monday October 10th.
PRACTICE
Friendly Hills Mondays 7:30 am  8:30 am Room 121
Heritage Mondays 3:25 pm  4:30 pm Room 160
SCHEDULE
Sept 26 Practice
Oct 3 No School
Oct 10 Meet 1 at SPA Pickup Her. 5:25 FHMS 5:40
Oct 17 Practice
Oct 24 Practice
Oct 31 Practice
Nov 7 Meet 2 at SPA Pickup Her. 5:25 FHMS 5:40
Nov 14 Practice
Nov 21 Meet 3 at Heritage Pickup Her. 5:00 FHMS 5:20
Nov 28 Practice
Dec 5 Practice
Dec 12 Meet 4 at St Thomas Academy Pickup Her. 5:10 FHMS 5:25
Dec 19 Practice
Dec 26 No school
Jan 2 No school
Jan 9 Meet 5 at Friendly Hills Pickup FHMS 5:05 Her. 5:25
PRACTICE
Friendly Hills Mondays 7:30 am  8:30 am Room 121
Heritage Mondays 3:25 pm  4:30 pm Room 160
SCHEDULE
Sept 26 Practice
Oct 3 No School
Oct 10 Meet 1 at SPA Pickup Her. 5:25 FHMS 5:40
Oct 17 Practice
Oct 24 Practice
Oct 31 Practice
Nov 7 Meet 2 at SPA Pickup Her. 5:25 FHMS 5:40
Nov 14 Practice
Nov 21 Meet 3 at Heritage Pickup Her. 5:00 FHMS 5:20
Nov 28 Practice
Dec 5 Practice
Dec 12 Meet 4 at St Thomas Academy Pickup Her. 5:10 FHMS 5:25
Dec 19 Practice
Dec 26 No school
Jan 2 No school
Jan 9 Meet 5 at Friendly Hills Pickup FHMS 5:05 Her. 5:25
Monday, January 11, 2016
Meet 5 Results
FINAL STANDINGS
INDIVIDUAL
TEAM STANDINGS
MEET 5 RESULTS
INDIVIDUAL

Team  Meet1  Meet2  Meet3  Meet4  Meet5  Total 
SPAGold  156  166  130  168  129  749 
SSP Maroon  134  116  102  114  71  537 
St Thomas Acad  134  122  102  104  70  532 
Heritage Gold  112  86  68  66  51  383 
SPABlue  102  98  68  74  38  380 
Friendly Hills Gold  92  104  66  66  27  355 
SSP White  120  68  58  52  36  334 
Friendly Hills Red  64  24  24  32  26  170 
Trinity River Ridge  0  0  72  56  33  161 
Heritage Red  48  14  0  14  8  84 
Humboldt  0  0  2  4  8  14 
MEET 5 RESULTS
EventA  EventB  Team  Total  
SPAGold  40  49  40  129 
SSP Maroon  23  20  28  71 
St Thomas Ac  16  30  24  70 
Heritage Gold  2  13  36  51 
SPABlue  8  6  24  38 
SSP White  6  6  24  36 
Trinity River Ridge  8  9  16  33 
Friendly Hills Gold  4  11  12  27 
Friendly Hills Red  0  6  20  26 
Heritage Red  0  0  8  8 
Humboldt  0  0  8  8 
Sunday, January 10, 2016
Final Meet
Final Math Meet of the Year at Friendly Hills on Monday Jan 11.
Heritage students will be dismissed at 2:50 to catch the bus.
Team trophies and individual medals will be awarded.
Top 6 scorers in each school will be given ribbons, but not Monday evening.
PARENTS: Pickup at FHMS is 5:05, at HERITAGE  5:25. Don't be late  it's cold out there!
Heritage students will be dismissed at 2:50 to catch the bus.
Team trophies and individual medals will be awarded.
Top 6 scorers in each school will be given ribbons, but not Monday evening.
PARENTS: Pickup at FHMS is 5:05, at HERITAGE  5:25. Don't be late  it's cold out there!
Monday, January 4, 2016
Jan 4 Results
Next week the final meet of the year is at Friendly Hills.
Team Practice Results
Only the White Team Remembered what Mode is  the most common number in the sample.
The length and width of the pyramid increase by 20% (1.20) but the height stayed the same. Sorry Red team  I think I misled you.
FHMS GOLD 16
FHMS RED 8
FHMS WHITE 6
Heritage GOLD 22
Solutions to 3 problems:
#2 What is the intersection of x3y=6 and y=8x2
Since y=8x2, let's substitute in the first equation:
x3(8x2)=6 Now change those minuses to plus a negative:
x+3(8x+2)=6 Next we'll distribute the 3 across the parentheses
x24x+6=6 Now let's combine the x's
23x+6=6 Subtract 6 from both sides
23x=0 Divide both sides by 23
x=0 Now substitute 0 for x in either equation
03y=6
3y=6
y=2 so (0,2) is the point we were looking for.
Doublecheck our answer by plugging (0,2) into the 2nd equation:
2=8(0)2 yup!
#5. What's the volume of the New Year Ball in Times Square? It's 12' in diameter.
r= 6' Vol=4/3 pi r3 = 4/3 x 6x6x6 pi = (4/3 x 6)(6x6)pi = 8x36pi=288pi
#9. The side lengths of a right square pyramid are each increased by 20%. If the
height stays the same, by what percentage does he volume of the pyramid increase?
Two dimensions increase by 20% the third remains the same. Therefor the volume
increases by 1.20 x 1.20 = 1.44 or a 44% increase in volume
Team Practice Results
Only the White Team Remembered what Mode is  the most common number in the sample.
The length and width of the pyramid increase by 20% (1.20) but the height stayed the same. Sorry Red team  I think I misled you.
FHMS GOLD 16
FHMS RED 8
FHMS WHITE 6
Heritage GOLD 22
Solutions to 3 problems:
#2 What is the intersection of x3y=6 and y=8x2
Since y=8x2, let's substitute in the first equation:
x3(8x2)=6 Now change those minuses to plus a negative:
x+3(8x+2)=6 Next we'll distribute the 3 across the parentheses
x24x+6=6 Now let's combine the x's
23x+6=6 Subtract 6 from both sides
23x=0 Divide both sides by 23
x=0 Now substitute 0 for x in either equation
03y=6
3y=6
y=2 so (0,2) is the point we were looking for.
Doublecheck our answer by plugging (0,2) into the 2nd equation:
2=8(0)2 yup!
#5. What's the volume of the New Year Ball in Times Square? It's 12' in diameter.
r= 6' Vol=4/3 pi r3 = 4/3 x 6x6x6 pi = (4/3 x 6)(6x6)pi = 8x36pi=288pi
#9. The side lengths of a right square pyramid are each increased by 20%. If the
height stays the same, by what percentage does he volume of the pyramid increase?
Two dimensions increase by 20% the third remains the same. Therefor the volume
increases by 1.20 x 1.20 = 1.44 or a 44% increase in volume
Monday, December 21, 2015
Team Assignments for Meet 5
Friendly Hills
Heritage
6  Oscar Halverson  6  6  43.3 
6  Cyrus Martin Risch  0  4  28.0 
6  Nick Wendt  0  0  20.7 
6  Nina Kessler  2  1  20.7 
5  Miles Dunn  0  4  19.3 
5  Claire Newmark  16.7  
6  Lucas Lindgren  15.3  
6  Dain Dolan  14.7  
6  Sophia Schomer  0  0  13.7 
5  Duy Houng  12.0  
5  Will Gannon  0  4  12.0 
7  Lily Pince  0  2  10.7 
5  Amario Sisakda  10.0  
5  Stella Warwick  8.0  
5  Charles Cheesebrough  0  4  7.3 
6  Victoria Dzurilla  0  0  7.3 
5  Ellyanna Lee  6.7  
5  Melissa Irwin  0  2  5.3 
5  Jackson Cercioglu  2.7 
8  Anthony Rocke  42.7  
8  Ben Nickson  0  6  25.3 
8  Steve Nickson  0  0  20.0 
5  Aidan Mallberg  6  2  18.7 
6  Kathryn Lewis  2  2  18.0 
5  Emma Lawrence  0  2  12.7 
5  Colten Bartlette  7.3  
5  Ruby Lipschultz  7.3  
5  Dia Balderramos  2.7  
5  Maraya Lucio  2.7  
5  Alycia Gonzales Lewis  2.7  
5  Sophie Todaro  1.3 
Monday, December 14, 2015
Simultaneous Equations
1. What is the point of intersection of 2x + 3y = 7 and 5y = 3x  1 ?
Here are 32 lessons on simultaeous equations (aka system of equations) from Kahn Academy
https://www.khanacademy.org/math/algebra/systemsoflinearequations
Unfortunately, you probably need to start at the beginning of Kahn Academy's algebra course to really understand this.
First of all, what is our answer going to be? A point, like (0, 0).
5th grade way: We need an x and y which make both equations
true. 2x is always going to be even. 3y will be odd only if y is odd.
We want the sum to be 7, an odd number, so y must be odd.
Let's try a 1. 2x + 3 x 1 = 7. A little arithmetic tells us x=2 since
2 x 2 + 3 x 1 = 7. So our point is (2, 1) but does this solve the
other equation: 5y = 3x 1 ?
5 x 1 = 3 x 2  1
5 = 6  1 Yes! Answer: (2, 1) If it hadn't
worked on our second equation, we would have tried y = 3 or y = 1.
If these hadn't worked we'd have to use the 7th grade way.
7th grade way: We have two equations with two variables. Let's
get rid of the y's as our first step. So let's get the same number of y's
in both equations. To do this let's multiply both sides of the first
equation by 5 and the second by 3:
2x + 3y = 7 5y = 3x  1
10x + 15y = 35 15y = 9x  3
15y = 10x + 35
Since both expressions on the right = 15y, they must = each other!
So 9x  3 = 10x + 35
(add 10x to each side) 19x  3 = 35
(Now add 3 to each side) 19x = 38
(now divide both by 19) x = 2
We have x. To get y, use either of our starting equations and
put in a 2 for x.
2x + 3y = 7
2 x 2 + 3y = 7
4 + 3y = 7
(take 4 from each side) 3y = 3
(divide each side by 3) y = 1.
So x= 2 and y = 1. Our point is (2, 1)
Third way: Graph both equations on a piece of graph paper and see
where the lines cross. To graph a line, simply pick a value for x or y
and figure out the missing one.
Example: 2x + 3y = 7.
When y = 3, 2x + 9 = 7 or 2x = 2. Hence x = 1 Plot (1, 3)
When y = 1, 2x + 3 = 7 or 2x = 10. Hence x = 5 Plot (5, 1)
Use a ruler on your graph paper to draw a straight line thru both pts.
Now do the same for the other equation. Look where they cross 
it should be at (2, 1).
Here are 32 lessons on simultaeous equations (aka system of equations) from Kahn Academy
https://www.khanacademy.org/math/algebra/systemsoflinearequations
Unfortunately, you probably need to start at the beginning of Kahn Academy's algebra course to really understand this.
First of all, what is our answer going to be? A point, like (0, 0).
5th grade way: We need an x and y which make both equations
true. 2x is always going to be even. 3y will be odd only if y is odd.
We want the sum to be 7, an odd number, so y must be odd.
Let's try a 1. 2x + 3 x 1 = 7. A little arithmetic tells us x=2 since
2 x 2 + 3 x 1 = 7. So our point is (2, 1) but does this solve the
other equation: 5y = 3x 1 ?
5 x 1 = 3 x 2  1
5 = 6  1 Yes! Answer: (2, 1) If it hadn't
worked on our second equation, we would have tried y = 3 or y = 1.
If these hadn't worked we'd have to use the 7th grade way.
7th grade way: We have two equations with two variables. Let's
get rid of the y's as our first step. So let's get the same number of y's
in both equations. To do this let's multiply both sides of the first
equation by 5 and the second by 3:
2x + 3y = 7 5y = 3x  1
10x + 15y = 35 15y = 9x  3
15y = 10x + 35
Since both expressions on the right = 15y, they must = each other!
So 9x  3 = 10x + 35
(add 10x to each side) 19x  3 = 35
(Now add 3 to each side) 19x = 38
(now divide both by 19) x = 2
We have x. To get y, use either of our starting equations and
put in a 2 for x.
2x + 3y = 7
2 x 2 + 3y = 7
4 + 3y = 7
(take 4 from each side) 3y = 3
(divide each side by 3) y = 1.
So x= 2 and y = 1. Our point is (2, 1)
Third way: Graph both equations on a piece of graph paper and see
where the lines cross. To graph a line, simply pick a value for x or y
and figure out the missing one.
Example: 2x + 3y = 7.
When y = 3, 2x + 9 = 7 or 2x = 2. Hence x = 1 Plot (1, 3)
When y = 1, 2x + 3 = 7 or 2x = 10. Hence x = 5 Plot (5, 1)
Use a ruler on your graph paper to draw a straight line thru both pts.
Now do the same for the other equation. Look where they cross 
it should be at (2, 1).
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