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.
Tuesday, December 18, 2007
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.
continuous and Discontiuous funciton bb 44 D
Fx 43 and 44 mainly focused on 44 d:
Using fx 43 we fill in the table. When you graph the table this is what it looks like(look below)
(this is a continuous function)
44D asks if the table graphs into a discrete or continuous function. You can tell this is a continuous function because the points connect and is a polynomial function and all polynomial functions are continues. In problem c the exponent becomes more and more negative, x goes from -1 to negative 4 or gets smaller and smaller. The value of y gets smaller and smaller and closer to 0. When the input changes the output also changes in a like manner. That is another way you can tell it is a continuous function.
Using fx 43 we fill in the table. When you graph the table this is what it looks like(look below)
(this is a continuous function)44D asks if the table graphs into a discrete or continuous function. You can tell this is a continuous function because the points connect and is a polynomial function and all polynomial functions are continues. In problem c the exponent becomes more and more negative, x goes from -1 to negative 4 or gets smaller and smaller. The value of y gets smaller and smaller and closer to 0. When the input changes the output also changes in a like manner. That is another way you can tell it is a continuous function.
EF-122- Square root functions
Alison Campbell
Square root functions
You cannot take the square root of a negative number as it will be an imaginary number.
When dealing with square root equations you have to think of the problem as two separate equations, other wise it will not be an equation, due to the vertical line rule. The vertical line rule tells us that we cannot have a line graphed that a vertical line would go through twice. Therefore you have to treat it as two equations.
When it is y= √(x-4) the x- intercept is 4, when it is y=√(x+4) the x intercept is -4. This is because to find the x-intercept of a line you solve the equation for y=0. See Greg’s post on intercepts for more info on finding x and y intercepts.
When you divide a square root, the slope of the line will flatten out towards the x-axis.
The graph looks like half a parabola, turned 90 degrees
Regular problem:
y= √(4x2) Solve for x
y2= +/-4x2 To get ride of the square root you square both sides and add +/-
y=+/- 2x Simplify and get ride of the squares
x=y/2 Get ride of the 2x by dividing both sides by 2
EF-122:
y=√x+4 Solve for x intercept
0=√x+4 set y to 0
0=√-4+4 find the number that will make the right side 0
x= -4
y=√0+4 Solve for y intercept
y=√4 set x to 0
y= +/-2 Simplify and get ride of the square root
Square root functions
You cannot take the square root of a negative number as it will be an imaginary number.
When dealing with square root equations you have to think of the problem as two separate equations, other wise it will not be an equation, due to the vertical line rule. The vertical line rule tells us that we cannot have a line graphed that a vertical line would go through twice. Therefore you have to treat it as two equations.
When it is y= √(x-4) the x- intercept is 4, when it is y=√(x+4) the x intercept is -4. This is because to find the x-intercept of a line you solve the equation for y=0. See Greg’s post on intercepts for more info on finding x and y intercepts.
When you divide a square root, the slope of the line will flatten out towards the x-axis.
The graph looks like half a parabola, turned 90 degrees
Regular problem:
y= √(4x2) Solve for x
y2= +/-4x2 To get ride of the square root you square both sides and add +/-
y=+/- 2x Simplify and get ride of the squares
x=y/2 Get ride of the 2x by dividing both sides by 2
EF-122:
y=√x+4 Solve for x intercept
0=√x+4 set y to 0
0=√-4+4 find the number that will make the right side 0
x= -4
y=√0+4 Solve for y intercept
y=√4 set x to 0
y= +/-2 Simplify and get ride of the square root
FX-1 arithmitic sequences
Fx-1
Patrick Woodburn
A)
To get the awnser for a you add the percentage of the four years together witch would be 32%.
B)
To get the awnser for b you can ether add $8 to the previous amount in the table.
C)
To get the awnser for c you first find the initial value(b) witch is 100
Then you find how much the amount goose up by(m) witch is 8
Then x is equal to the amount of time
Now you know y=mx+b so you plug in the values to get y=8x+100
Then you graph it buy putting point on the y axis equal to the b after that every time you add a move over 1 on the x axis and up m or in this case up 8
Patrick Woodburn
A)
To get the awnser for a you add the percentage of the four years together witch would be 32%.
B)
To get the awnser for b you can ether add $8 to the previous amount in the table.
C)
To get the awnser for c you first find the initial value(b) witch is 100
Then you find how much the amount goose up by(m) witch is 8
Then x is equal to the amount of time
Now you know y=mx+b so you plug in the values to get y=8x+100
Then you graph it buy putting point on the y axis equal to the b after that every time you add a move over 1 on the x axis and up m or in this case up 8
FX-36 Elimination Method
Melanie Thomas
5x-4y=7
2y+6x=22
To use the elimination method to solve this system of linear equations, first you have to eliminate y, to find x.
To do this, you have to add its additive inverse.
First, multiply both sides of the second equation by 2 in order to get 4y so that the -4y from the first equation cancels out when added to the 4y from the second equation.
2(2y+6x)=22(2)
4y+12x=44
Once you've done this, in order to solve the system, add 4y+12x to the left side of the first equation and then add 44 to the right side of that same equation. Now the equations are in standard form but the coefficients and variables need to be lined up. The system of equations should now look like this:
5x-4y=7
12x+4y=44
-----------------
17x+0y=51
Once you've done this you should get 17x=51 because 5x+12x=17x, the -4y+4y cancel out and you're left with 0y, and 7+44=51. After this, you solve for x.
To solve for x, you divide both sides of the equation by 17:
17x/17=51/17
x=3
On the left side, you're left with an isolated x variable and on the right side you're left with 3 because 51/17=3.
Now you have to solve for y, and to do this, the first step would be to substitute 3 for x in either of the original equations. In the first equation:
5x-4y=7, it would be
5(3)-4y=7 then you'd simplify and it would be:
15-4y=7
After this, you subtract 15 from both sides to isolate the 4y, which is the second step of solving for y.
15-4y=7
-15 -15
-----------------
-4y= -8
You're left with -4y=-8 and to isolate y, you divide both sides by -4.
-4y/-4 =-8/-4
y=2
You're left with 2.
Now use the numbers that were found for the x and y variables and the solution of the system is (3, 2).
5x-4y=7
2y+6x=22
To use the elimination method to solve this system of linear equations, first you have to eliminate y, to find x.
To do this, you have to add its additive inverse.
First, multiply both sides of the second equation by 2 in order to get 4y so that the -4y from the first equation cancels out when added to the 4y from the second equation.
2(2y+6x)=22(2)
4y+12x=44
Once you've done this, in order to solve the system, add 4y+12x to the left side of the first equation and then add 44 to the right side of that same equation. Now the equations are in standard form but the coefficients and variables need to be lined up. The system of equations should now look like this:
5x-4y=7
12x+4y=44
-----------------
17x+0y=51
Once you've done this you should get 17x=51 because 5x+12x=17x, the -4y+4y cancel out and you're left with 0y, and 7+44=51. After this, you solve for x.
To solve for x, you divide both sides of the equation by 17:
17x/17=51/17
x=3
On the left side, you're left with an isolated x variable and on the right side you're left with 3 because 51/17=3.
Now you have to solve for y, and to do this, the first step would be to substitute 3 for x in either of the original equations. In the first equation:
5x-4y=7, it would be
5(3)-4y=7 then you'd simplify and it would be:
15-4y=7
After this, you subtract 15 from both sides to isolate the 4y, which is the second step of solving for y.
15-4y=7
-15 -15
-----------------
-4y= -8
You're left with -4y=-8 and to isolate y, you divide both sides by -4.
-4y/-4 =-8/-4
y=2
You're left with 2.
Now use the numbers that were found for the x and y variables and the solution of the system is (3, 2).
Thursday, December 13, 2007
The Fixed BB-71
Robert Yemola
BB-71
A- 1, 4, 7, 10, 13
The first thing that you want to do is put this in a chart so it is easier to understand.
N 0 1 2 3 4
T(N) 1 4 7 10 13
The first thing that you want to do is look at the numbers and see if there is a pattern to get from one number to the next. In this case each time you have to +3 each time.
Then to make an equation you to find out if there is a starting point, in this case there is and the starting point is 1. Then you have to plug in the information, T(N)=1+3N you can plug in one to make sure it is correct T(N)=1+3(3) and you get 10 so we know that this equation is correct. Then finally since it adds by the same number each time the equation is Arithmetic.
B- 0, 5, 12, 21, 32
Once again put the numbers into a chart so it is easier to understand.
N 0 1 2 3 4
T(N) 0 5 12 21 32
The first thing you want to do is see if you can find a pattern, you do see a pattern and it is that every time it adds by an odd number. But the thing about that is that, that is neither Arithmetic or Geometric so the Answer to this one is Neither.
C- 2, 4, 8, 16, 32
Put all of the numbers into a table
N 0 1 2 3 4
T(N) 2 4 8 16 32
The first thing you do is look to see if you can find any patterns and you can each time you are multiplying by 2. To make the equation what you would have to do is you would first have to find the starting point which is 2. Then you would put all of the info in and the equation would be T(N)= 2*2^N. Then finally we know that this equation is Geometric because the Equation has an Exponent in it.
D- 5, 12, 19, 26,
Put the numbers into a chart so that they are easier to understand
N 0 1 2 3
T(N) 5 12 19 26
The first thing that you would want to do is try to find a pattern and in this case we can see that every time you have to add 7 to get to the next number. Then to get the Equation you need to find the starting point and in this case the starting point is 5. So the equation for this set of numbers would be T(N)=5+7N. Finally we know that this set of numbers is Arithmetic because each time you have to add the same number.
E- x, x+1, x+2, x+3
Put the numbers in a chart so they are easier to understand
N 0 1 2 3
T(N) X X+1 X+2 X+3
First you want to see if you can find a pattern and you see that the X stays the same. The N is what is added each time. To make this equation you need to find the starting point and this starting point is X. So this equation would be pretty simple T(N)=X+N. Finally we know that this is an Arithmetic equation because each time you add the same thing in this case you add N every time.
F- 3, 12, 48, 192
Put the numbers in a chart so that they are easier to understand
N 0 1 2 3
T(N) 3 12 48 192
The first thing that you want to look at is if there is a pattern to get from one number to the next number. In this case we can see that each time you have to multiply the previous number by 4. Then to get the equation, you need to find the starting point and in this case the starting point is 3. So the equation would be T(N)= 3*4^N. Finally this equation is Geometric because the equation has an exponent in it.
BB-71
A- 1, 4, 7, 10, 13
The first thing that you want to do is put this in a chart so it is easier to understand.
N 0 1 2 3 4
T(N) 1 4 7 10 13
The first thing that you want to do is look at the numbers and see if there is a pattern to get from one number to the next. In this case each time you have to +3 each time.
Then to make an equation you to find out if there is a starting point, in this case there is and the starting point is 1. Then you have to plug in the information, T(N)=1+3N you can plug in one to make sure it is correct T(N)=1+3(3) and you get 10 so we know that this equation is correct. Then finally since it adds by the same number each time the equation is Arithmetic.
B- 0, 5, 12, 21, 32
Once again put the numbers into a chart so it is easier to understand.
N 0 1 2 3 4
T(N) 0 5 12 21 32
The first thing you want to do is see if you can find a pattern, you do see a pattern and it is that every time it adds by an odd number. But the thing about that is that, that is neither Arithmetic or Geometric so the Answer to this one is Neither.
C- 2, 4, 8, 16, 32
Put all of the numbers into a table
N 0 1 2 3 4
T(N) 2 4 8 16 32
The first thing you do is look to see if you can find any patterns and you can each time you are multiplying by 2. To make the equation what you would have to do is you would first have to find the starting point which is 2. Then you would put all of the info in and the equation would be T(N)= 2*2^N. Then finally we know that this equation is Geometric because the Equation has an Exponent in it.
D- 5, 12, 19, 26,
Put the numbers into a chart so that they are easier to understand
N 0 1 2 3
T(N) 5 12 19 26
The first thing that you would want to do is try to find a pattern and in this case we can see that every time you have to add 7 to get to the next number. Then to get the Equation you need to find the starting point and in this case the starting point is 5. So the equation for this set of numbers would be T(N)=5+7N. Finally we know that this set of numbers is Arithmetic because each time you have to add the same number.
E- x, x+1, x+2, x+3
Put the numbers in a chart so they are easier to understand
N 0 1 2 3
T(N) X X+1 X+2 X+3
First you want to see if you can find a pattern and you see that the X stays the same. The N is what is added each time. To make this equation you need to find the starting point and this starting point is X. So this equation would be pretty simple T(N)=X+N. Finally we know that this is an Arithmetic equation because each time you add the same thing in this case you add N every time.
F- 3, 12, 48, 192
Put the numbers in a chart so that they are easier to understand
N 0 1 2 3
T(N) 3 12 48 192
The first thing that you want to look at is if there is a pattern to get from one number to the next number. In this case we can see that each time you have to multiply the previous number by 4. Then to get the equation, you need to find the starting point and in this case the starting point is 3. So the equation would be T(N)= 3*4^N. Finally this equation is Geometric because the equation has an exponent in it.
Tuesday, December 11, 2007
Triangles
The two triangles are similar. They both share a common angle and the smaller one is inside the larger one. For instance two triangles of the same angle degrees intwine. The small triangle has sides of 3 and 5. The large triangle has sides of 8 and an unknown number we should call "z". Our job is to find out what "z" is.
8/3 = z/5
8/3 = 2.6
2.6 * 5 = 13.3
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