On February first my group and I headed downstairs to finish off the rest of the trials and gather the data we needed. We had already finished the long jump for Miller, so Doyle was up. We decided where Doyle was going to jump from and measured from there where we thought the balloon should be. 156 inches away, we measured the height of the balloon, to set the height of the parabola at 78 inches. He sprinted, and jumped forward, barely touching the balloon, and we recorded the data. Next up was Ed, and we set the balloon first at 118 inches away, keeping the height the same, 78 inches. We had to bump the distance the balloon was from Ed's point of depature down to 98 when he couldn't quite make it. The second time trying he actually jumped off of his own two feet the and toppled over in mid air. We laughed, but continued on with the experiment until we acquired the appropriate data. Finally I was up, and I could make it to about 132 inches away from my point of jumping, with the balloon at the same height as before, 78 inches. Finally once we collected all our data, we headed back upstairs. I had Miller photocopy five copies of the data for everyone. There was a problem as to how to create a parabola from the data acquired, seeing as it wasn't a perfect parabola everytime we jumped. I consulted Mr. Rochester and he stated that he wants us to imagine it like we're starting from the negative of the distance we jumped, and the vertex of the parabola is zero and the height of the balloon, and then the other x-intercept is the positive of the distance we jumped. So if I were to make my graph of my long jump, the vertex would be (0,78) and the two x-intercepts would be (-132,0) and (132,0).
The next scribe will be... Maxx Kim.
Showing posts with label benchmark. Show all posts
Showing posts with label benchmark. Show all posts
Monday, February 4, 2008
Thursday, January 31, 2008
Scripe post for January 31st
Mr. Rochester demonstrated again what we had to do for the first part of our benchmark. Then after we collected our materials we all went out in the halls and set to work. With our balloons taped on to ours sticks and tape measures in hand the jumping began. Most groups started with the "broad jump" where you stood still and had to jump as high and as far as you could (you had to touch the balloon). then it was full speed ahead to the "long jump". Here you got to have a running start (still had to touch the balloon). Some groups even thought they were so good it was worth filming (oh wait Rochester said it would be a nice addition to our graphs... so never mind everyone should film at some point). So over all today SLA's recored for both the long jump and the broad jump was broken, or i guess set. We will have to find out though when benchmarks are due who holds it though.
For those of you who missed this sports filled class you should talk to your group members and set up a time when they can help you collect your data.
next time on Scribe Post we will have Sam reporting from the right side of the room. Thanks for tuning in to another Scribe Post, until next time.
For those of you who missed this sports filled class you should talk to your group members and set up a time when they can help you collect your data.
next time on Scribe Post we will have Sam reporting from the right side of the room. Thanks for tuning in to another Scribe Post, until next time.
Wednesday, January 23, 2008
Michael Doyle
Simplifying Expressions
BB-35
A) (2m3)(4m2)
B) 6y5/3y2
C) -4y2/6y7
D) (-2x2)3
A) When simplifying expression A, 2m3 and 4m2 have to be multiplied. In order to do this, the first thing to do is to check for like terms. m and m are like terms so they can be multiplied together resulting in 2m*4m=8m. Though, the original terms in this expression both have exponents. The law for multiplying with exponents is that whatever power each term is to, the exponential value is added together. So in this expression it would be m3+m2=m5. So when simplified, this expression is 8m5. Parenthesis is not needed anymore after multiplication.
B) When simplifying expression B, 6y5 is being divided by 3y2. Dividing terms with exponents is similar to multiplying, just the other way around. Rather than adding the power the each term is to, they are subtracted. Normal division is held for the rest of the terms. So, 6y/3y=2y. And since the exponents are being subtracted and not added, so it is y5-y2=y3. So when simplified, this expression is 2y3.
C) Expression C is very much similar to expression B, just with a negative term. The same rules apply as they always do, -4y/6y=-0.667y and for the exponents –y2-y7=-y-5. So the full expression simplified is -0.667-5.
D) For expression D, a term with and exponent is being put to the 3rd power. When this occurs, the exponential factors are multiplied. So the term -2x remains the same, but the 2 in -2x2 is multiplied by 3. Which then is seen as 2*3=6. Thus, the simplified expression is -2x6. Parenthesis is not needed anymore after multiplication.
BB-35
A) (2m3)(4m2)
B) 6y5/3y2
C) -4y2/6y7
D) (-2x2)3
A) When simplifying expression A, 2m3 and 4m2 have to be multiplied. In order to do this, the first thing to do is to check for like terms. m and m are like terms so they can be multiplied together resulting in 2m*4m=8m. Though, the original terms in this expression both have exponents. The law for multiplying with exponents is that whatever power each term is to, the exponential value is added together. So in this expression it would be m3+m2=m5. So when simplified, this expression is 8m5. Parenthesis is not needed anymore after multiplication.
B) When simplifying expression B, 6y5 is being divided by 3y2. Dividing terms with exponents is similar to multiplying, just the other way around. Rather than adding the power the each term is to, they are subtracted. Normal division is held for the rest of the terms. So, 6y/3y=2y. And since the exponents are being subtracted and not added, so it is y5-y2=y3. So when simplified, this expression is 2y3.
C) Expression C is very much similar to expression B, just with a negative term. The same rules apply as they always do, -4y/6y=-0.667y and for the exponents –y2-y7=-y-5. So the full expression simplified is -0.667-5.
D) For expression D, a term with and exponent is being put to the 3rd power. When this occurs, the exponential factors are multiplied. So the term -2x remains the same, but the 2 in -2x2 is multiplied by 3. Which then is seen as 2*3=6. Thus, the simplified expression is -2x6. Parenthesis is not needed anymore after multiplication.
Saturday, January 19, 2008
Friday, December 21, 2007
Thursday, December 20, 2007
Exponential Function by. Maxx Kim
pg. 88
FX- 60 a) f(x)= 3x^2
This is not a exponential function because there is a ^2 instead of ^x. All exponential functions look the similar on a graph. Below is what a) looks like. As you can see its a parabola.
b) h(x)= (3)^x
This is a exponential function because the exponent is x. As long as it has the x as a exponent it is a exponential function. My proof that it is a exponential function is because of the graph.
Every exponential Function will be graphed like this.
Another example, c)
y=5(4)^x
again this is another exponential function because it has the x as a exponent. This graphed looks like this.
again the graph looks the same at b)
One more example, d)
g(x)=(2.46) ^x
Again this is a exponential function because it has the x as an exponent. This is what the graph looks like. Again you will see that it is similar.
FX- 60 a) f(x)= 3x^2
This is not a exponential function because there is a ^2 instead of ^x. All exponential functions look the similar on a graph. Below is what a) looks like. As you can see its a parabola.
b) h(x)= (3)^x
This is a exponential function because the exponent is x. As long as it has the x as a exponent it is a exponential function. My proof that it is a exponential function is because of the graph.
Every exponential Function will be graphed like this.
Another example, c)
y=5(4)^x
again this is another exponential function because it has the x as a exponent. This graphed looks like this.
again the graph looks the same at b)
One more example, d)
g(x)=(2.46) ^x
Again this is a exponential function because it has the x as an exponent. This is what the graph looks like. Again you will see that it is similar.
BB-116
Multiplying Fractions with Variables
Example: BB-116a
1. First, you need to multiply straight across
2. Then you need to factor out the numerator and denominator.
3. Next, you make zero pairs with the X’s (cross out an X from the top and bottom)
4. Then divide by largest factor, which is 5
5. Simplify some more and divide by largest factor (which is 3)
Subtracting Fractions with Variables
Example: BB-116d
1. You need to get the same denominator,so you need to multiply the left side by 2x.
2. Yet your not done so you need to multiply the right side by y
3. Next, you cannot combine like terms because there are none so you must put everything on the same line.
Adding Fractions with Variables
Example: BB-116b
1. You need to get the same denominator, so you multiply the right side by t
2. Since you cannot combine like terms you put everything on the same line
Example: BB-116a
1. First, you need to multiply straight across
2. Then you need to factor out the numerator and denominator.
3. Next, you make zero pairs with the X’s (cross out an X from the top and bottom)
4. Then divide by largest factor, which is 5
5. Simplify some more and divide by largest factor (which is 3)
Subtracting Fractions with Variables
Example: BB-116d
1. You need to get the same denominator,so you need to multiply the left side by 2x.
2. Yet your not done so you need to multiply the right side by y
3. Next, you cannot combine like terms because there are none so you must put everything on the same line.
Adding Fractions with Variables
Example: BB-116b
1. You need to get the same denominator, so you multiply the right side by t
2. Since you cannot combine like terms you put everything on the same line
BB-69
Mithun Das
12/6/07
BB-69.
Write the range and domain for each of the following graphs.12/6/07
BB-69.
• To find the range you need to look the y values on the graphs. In each graph you need to check what y values the graph goes through.
1. In the first graph the x-values that represent the function are real numbers.
So, the domain consists of real numbers. The domain is (-∞.∞)
All y–values greater than or equal to –1 represent the function.
So, the range is y –1. The range is [-1,∞)
2. In the second graph the x values that function are real numbers.
So, the domain consists of real numbers. The domain is (-∞.∞)
The range is y–values less than or equal to 1 represent the function.
So, the range is y ≥1. The is (-∞, 1]
3. In the third graph the x values that represent the function are real numbers.
So, the domain consists of real numbers. The domain is (-∞,∞)
The y–values less than or equal to 0 represent the function.
So, the range is y 0. The range is (-∞,0]
Factoring EMcAninch
Factoring Equations
FX-14
A)
x^2-49
x^2-49=(x)^2-(7)^2
x^2-(7)^2=(x+7)(x-7)
so
x^2-49=(x+7)(x-7)
B)
4x^2-1
4x^2-1=(2x)^2-(1)^2
(2x)^2-(1)^2=(2x+1)(2x-1)
so
4x^2-1 = (2x+1)(2x-1)
C)
x^2y^2-81z^2
x^2y^2-81z^2=(xy)^2-(9z)^2
(xy)^2-(9z)^2=(xy+9z)(xy-9z)
so
x^2y^2-81z^2=(xy-9z)(xy-9z)
D)
2x^3-8x
2x(x^2-4)
2x(x^2-4)=2x((x)^2-(2)^2)
2x((x)^2-(2)^2)=2x(x+2)(x-2)
so
2x^3-8x=2x(x+2)(x-2)
FX-14
A)
x^2-49
x^2-49=(x)^2-(7)^2
x^2-(7)^2=(x+7)(x-7)
so
x^2-49=(x+7)(x-7)
B)
4x^2-1
4x^2-1=(2x)^2-(1)^2
(2x)^2-(1)^2=(2x+1)(2x-1)
so
4x^2-1 = (2x+1)(2x-1)
C)
x^2y^2-81z^2
x^2y^2-81z^2=(xy)^2-(9z)^2
(xy)^2-(9z)^2=(xy+9z)(xy-9z)
so
x^2y^2-81z^2=(xy-9z)(xy-9z)
D)
2x^3-8x
2x(x^2-4)
2x(x^2-4)=2x((x)^2-(2)^2)
2x((x)^2-(2)^2)=2x(x+2)(x-2)
so
2x^3-8x=2x(x+2)(x-2)
BB- 30
Andrey Chekis
Silver
BB-30
2x^2 goes up faster and is more steeper then then 1/2x^2, which goes up slower. X^2 is average.
Silver
BB-30
2x^2 goes up faster and is more steeper then then 1/2x^2, which goes up slower. X^2 is average.
Monday, December 17, 2007
Domain, Range, independent and dependent variables
Instructions: To get the domain of a graph you have to look at the line. If it is a line that keeps going forever you look at where it hits the x, once you see that you can put [whatever the x is, oo) (-oo, whatever the x is]. Now if the line stops at a certain point, you get the x of the two tips of the line and you put the higher point first such as [65,-43] Now these numbers are also the independent variables.
To get the range of a graph it is a lot like doing the domain but with the y instead of x. All you really have to do is see where the y starts and stops, it could be infinitive or it could stop, but that is what you look at. When you find that out you have also found the dependent variable.
Examples:
105A
In this problem you can see that there is no line so all you have to find is the x and the y of the points and put them in chronological order and then you are done. So for the domain, 0, 1, 2, and since it goes from 0 - 2 you can write the domain like this (0,2) and this is also the independent variables. The range in this graph would be [1, -2] because the y goes from 1 to -2, which then is also the dependent variable.
105B
Domain / independent variable- [-1, 1]
Range / Dependent Variable- [-1, 2]
105C
Some people may think that this looks confusing but it really isn’t, because the two lines are on the same x value when the opened circle ends and the closed begins mean that you can just combine the two make the independent variable and the domain are [-2, 2]. The range/ dependent variables does the same thing so it would be [2,-2].
105D
In this problem you just look at where the farthest x is to the left and you have your first number, which is -2. The second number in the domain would be 1 because that’s the farthest to the right the x goes. So the final domain would be [-2,1]. The range would be [2, -2] because the highest point and the lowest point which all connects is 2 and -2 which makes it the range.0
To get the range of a graph it is a lot like doing the domain but with the y instead of x. All you really have to do is see where the y starts and stops, it could be infinitive or it could stop, but that is what you look at. When you find that out you have also found the dependent variable.
Examples:
105A
In this problem you can see that there is no line so all you have to find is the x and the y of the points and put them in chronological order and then you are done. So for the domain, 0, 1, 2, and since it goes from 0 - 2 you can write the domain like this (0,2) and this is also the independent variables. The range in this graph would be [1, -2] because the y goes from 1 to -2, which then is also the dependent variable.
105B
Domain / independent variable- [-1, 1]
Range / Dependent Variable- [-1, 2]
105C
Some people may think that this looks confusing but it really isn’t, because the two lines are on the same x value when the opened circle ends and the closed begins mean that you can just combine the two make the independent variable and the domain are [-2, 2]. The range/ dependent variables does the same thing so it would be [2,-2].
105D
In this problem you just look at where the farthest x is to the left and you have your first number, which is -2. The second number in the domain would be 1 because that’s the farthest to the right the x goes. So the final domain would be [-2,1]. The range would be [2, -2] because the highest point and the lowest point which all connects is 2 and -2 which makes it the range.0
Equilateral Triangles
A2+B2 = C2
BB-41-B- Find the height of an equilateral triangle with sides of 6.
An equilateral triangle is a triangle with all three sides of equal length a, corresponding to what could also be known as a "regular" triangle. An equilateral triangle is therefore a special case of an isosceles triangle having not just two, but all three sides equal. An equilateral triangle also has three equal 60 degrees angles.
The three angles of the triangle add up together equal 180o. As in the picture above each angle is 60o. And there are 3 sides so 3*60= 180o. The sides on equilateral triangles are all the same side as shown in the picture above. However if it were an isosceles triangle two sides would be the same and one would be different. However on a right triangle all the sides would be different with an angle of 90o.
90o
To answer this problem FX-41b finding the height of an equilateral triangle with sides of 6.
In order to find the length of the height of the triangle you need to use the Pythagorean Theorem, which is A2+B2 =C2.When we use the Pythagorean Theorem we use half of the triangle so the bottom is cut into half. So instead of the length of 6 on the bottom it is 3. This is a given number. And we want to find the height so we use the hypotenuse as another given number, which is 6. So now we plug these to numbers in the equation A2+B2 = C2. 6 is the hypotenuse so we substitute it for c2 and the 3 can be substituted for the A or B. In this case we used 3 to substitute the A and the height “would be the b and we substitute the b for the h. So the equation would be 32+h2=62. Then you square the 3 and the 6 so the next equation would be 9+h2=36. After that you subtract the 9 from both sides of the equation which would be h2=27. Next you solve for h which would be h=3√3≈5.2
BB-41-B- Find the height of an equilateral triangle with sides of 6.
An equilateral triangle is a triangle with all three sides of equal length a, corresponding to what could also be known as a "regular" triangle. An equilateral triangle is therefore a special case of an isosceles triangle having not just two, but all three sides equal. An equilateral triangle also has three equal 60 degrees angles.
The three angles of the triangle add up together equal 180o. As in the picture above each angle is 60o. And there are 3 sides so 3*60= 180o. The sides on equilateral triangles are all the same side as shown in the picture above. However if it were an isosceles triangle two sides would be the same and one would be different. However on a right triangle all the sides would be different with an angle of 90o.
90o
To answer this problem FX-41b finding the height of an equilateral triangle with sides of 6.
In order to find the length of the height of the triangle you need to use the Pythagorean Theorem, which is A2+B2 =C2.When we use the Pythagorean Theorem we use half of the triangle so the bottom is cut into half. So instead of the length of 6 on the bottom it is 3. This is a given number. And we want to find the height so we use the hypotenuse as another given number, which is 6. So now we plug these to numbers in the equation A2+B2 = C2. 6 is the hypotenuse so we substitute it for c2 and the 3 can be substituted for the A or B. In this case we used 3 to substitute the A and the height “would be the b and we substitute the b for the h. So the equation would be 32+h2=62. Then you square the 3 and the 6 so the next equation would be 9+h2=36. After that you subtract the 9 from both sides of the equation which would be h2=27. Next you solve for h which would be h=3√3≈5.2
32+h2=62
9 +h2=36
-9 -9
h2=27
h=3√3≈5.2
So the height of the triangle is ≈5.2.
9 +h2=36
-9 -9
h2=27
h=3√3≈5.2
So the height of the triangle is ≈5.2.
Probability
Probability-the extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
BB-108
Toss three coins in the air. Make a list of all the possible outcomes, and find the probability that:
A) all of them land on heads up
B) two of them land with tails up
• List of all possible outcomes H= heads T=tails
• HHH; HHT; HTH; THH; HTT; TTH; THT; TTT
• A) There are only 8 possible outcomes of rolling the three coins. The probability of all the coins landing on heads is only 1 outcome out of 8 outcomes. Therefore the probability is 1/8.
• B) There are only 8 outcomes of rolling three coins. Only 3 of the outcomes have 2 coins being rolled a tails. Therefore the probability of rolling 3 coins and 2 of them landing on heads is 3/8.
BB-108
Toss three coins in the air. Make a list of all the possible outcomes, and find the probability that:
A) all of them land on heads up
B) two of them land with tails up
• List of all possible outcomes H= heads T=tails
• HHH; HHT; HTH; THH; HTT; TTH; THT; TTT
• A) There are only 8 possible outcomes of rolling the three coins. The probability of all the coins landing on heads is only 1 outcome out of 8 outcomes. Therefore the probability is 1/8.
• B) There are only 8 outcomes of rolling three coins. Only 3 of the outcomes have 2 coins being rolled a tails. Therefore the probability of rolling 3 coins and 2 of them landing on heads is 3/8.
Linear Equations
BB-66 : Page 56
.png)
and
3. Then you add 'x' to both sides of the equation leaving you with: 5=(1/3)x+1
4. Then you must do additive inverse so you subtract 1 from both sides leaving you with: 4=(1/3)x+x
5. Then you simplify that equation and end up with: 4=(4/3)x
6. To get 'x' by itself, multiply 3/4 to both sides of the equation.
7. You should then end up with x=3
8. To get the 'y' coordinate you must substitute 'x' into one of the two original equations.
9. y=5-3 : 5-3=2 : y=2
10. So you can tell that the lines do cross at the coordinates (3,2)
.png)
and
Graphing the Equations on the Same Graph
- First thing you have to do is simplify the first equation. After that you should end up with y=5-x .
- You then have to plot it on a graph. After you've completed that then you put the second equation on the same graph as the first equation.
- You then continue extending both lines until they intersect each other. They intersect at the point (3, 2). (Refer to Picture 1 at the bottom)
Solving the Problems Algebraically
1. Once you see where they intersect you must then solve for 'y' and 'x' algebraically to ensure that you've got the right coordinates. You have to substitute for 'Y' from the first equation into the second equation.
2. Your equation should look like this:.png)
.png)
3. Then you add 'x' to both sides of the equation leaving you with: 5=(1/3)x+1
4. Then you must do additive inverse so you subtract 1 from both sides leaving you with: 4=(1/3)x+x
5. Then you simplify that equation and end up with: 4=(4/3)x
6. To get 'x' by itself, multiply 3/4 to both sides of the equation.
7. You should then end up with x=3
8. To get the 'y' coordinate you must substitute 'x' into one of the two original equations.
9. y=5-3 : 5-3=2 : y=2
10. So you can tell that the lines do cross at the coordinates (3,2)
Picture 1
Sunday, December 16, 2007
Substitution Method
FX-50
y=-x-2 and 5x-3y=22
5x-3(-x-2)=22
5x+3x+6=22
8x+6=22
8x=16
x=2
y=-(2)-2
y=-2-2
y=-4
x=2 and y=-4
You have a system of equations, y=-x-2 and 5x+3y=22. One of them looks easier to solve than the other, so first substitute the easier equation into the harder one and simplify. Once you do that, you should have the equation 5x+3x+6=22. Now, go ahead and solve for x. You found that x=2, so substitute 2 for x in the first equation and you can solve for y. You should get y=-4. So x=2 and y=-4. Now you have solved the system of equations.
y=-x-2 and 5x-3y=22
5x-3(-x-2)=22
5x+3x+6=22
8x+6=22
8x=16
x=2
y=-(2)-2
y=-2-2
y=-4
x=2 and y=-4
You have a system of equations, y=-x-2 and 5x+3y=22. One of them looks easier to solve than the other, so first substitute the easier equation into the harder one and simplify. Once you do that, you should have the equation 5x+3x+6=22. Now, go ahead and solve for x. You found that x=2, so substitute 2 for x in the first equation and you can solve for y. You should get y=-4. So x=2 and y=-4. Now you have solved the system of equations.
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