Here are the class notes with the revisions that we made in class. We will talk about FX16-19 during the next class.
Friday, December 7, 2007
Class Notes 12/07/07
Here are the class notes with the revisions that we made in class. We will talk about FX16-19 during the next class.
Thursday, December 6, 2007
YemolaR 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.
Wednesday, December 5, 2007
Jhendarto: BB-70
Explanation:
What I did to get the answer for BB-70 is...for example, A, the answer is 1.03y because
-I simplified y+0.03y...... is..... y (1+0.03)....... because the y in (0.03 y) is equal to 1
-Then I would get Y (1+0.03) and I simplify it again to get Y(1.03)
-Multiply Y to 1.03 and you get 1.03y
This is how you get the answeres for BB-70. This method works for all three problems.
What I did to get the answer for BB-70 is...for example, A, the answer is 1.03y because
-I simplified y+0.03y...... is..... y (1+0.03)....... because the y in (0.03 y) is equal to 1
-Then I would get Y (1+0.03) and I simplify it again to get Y(1.03)
-Multiply Y to 1.03 and you get 1.03y
This is how you get the answeres for BB-70. This method works for all three problems.
BB-103
Arithmetic
t(n)=100-50n
n t(n)
0 100
1 50
2 0
3 -50
We know our initial term, 100, and we know the difference between the initial number and the number right after that, which is 100-50, which equals 50. And since this sequence is decreasing, we know that our generator is also negative. Since we are only increasing once between the independent variables, we know that our generator is -50.
f(x)=50+25x
x f(x)
0 50
1 75
2 100
3 125
We know that our initial term, which is 50, and we know that when x equals 2, f(x) is equivalent to 100. Using the elimination process in which we substitute each set of numbers in the sequence to the equation, f(x)=dx, knowing that d is our generator, x is the independent variable and f(x) is the dependent variable. So we have 50=0d and 100=2d. Thus when we subtract each, 100-50 & 2-0, we get 50 and 2. We divide 2 into 50 to get the number by which we add, or d, and we get 25 to be that number. Thus we can now finish our equation knowing the initial number and the generator. f(x)=50+25x
Geometric
t(n)=100/(2^n)
n t(n)
0 100
1 50
2 25
3 12.5
We know that when we branch from 100, the initial term, to 50, and we know that the sequence is geometric, meaning that we multiply or divide by a generator to the nth power, that we are going from a higher number, 100, to a lower number, 50, we are using division. Thus to figure out the rest of the terms in the sequence we must then divide 100 and 50 to find our generator, which is 2. Since our generator is 2 and our initial number is 100, we can make the equation stated above the table, and fill out the rest of the chart.
f(x)=50*(2^1/2)^n
x f(x)
0 50
1 50*(2^1/2)
2 100
3 100(2^1/2)
Since we know that f(x)=a+d^x, and we know that a is the initial number, the d is the generator of the sequence, and the x is the independent variable, i.e. the number of times you multiply the generator, d, due to the fact that is is a geometric sequence, we can plug in a set of numbers to help us figure out the generator of the sequence, which we do not know in this case. Since the initial number is always when x is equivalent to 0, we know that in this sequence our initial number is 50. Now instead of plugging that in, we'll use our second set of numbers, 2 and 100, which will be much easier to work with. Since 100 is equivalent to f(x) and 2 is equivalent to x, we can plug those numbers in along with our initial number, so that we get a good 100=50*d^2.
t(n)=100-50n
n t(n)
0 100
1 50
2 0
3 -50
We know our initial term, 100, and we know the difference between the initial number and the number right after that, which is 100-50, which equals 50. And since this sequence is decreasing, we know that our generator is also negative. Since we are only increasing once between the independent variables, we know that our generator is -50.
f(x)=50+25x
x f(x)
0 50
1 75
2 100
3 125
We know that our initial term, which is 50, and we know that when x equals 2, f(x) is equivalent to 100. Using the elimination process in which we substitute each set of numbers in the sequence to the equation, f(x)=dx, knowing that d is our generator, x is the independent variable and f(x) is the dependent variable. So we have 50=0d and 100=2d. Thus when we subtract each, 100-50 & 2-0, we get 50 and 2. We divide 2 into 50 to get the number by which we add, or d, and we get 25 to be that number. Thus we can now finish our equation knowing the initial number and the generator. f(x)=50+25x
Geometric
t(n)=100/(2^n)
n t(n)
0 100
1 50
2 25
3 12.5
We know that when we branch from 100, the initial term, to 50, and we know that the sequence is geometric, meaning that we multiply or divide by a generator to the nth power, that we are going from a higher number, 100, to a lower number, 50, we are using division. Thus to figure out the rest of the terms in the sequence we must then divide 100 and 50 to find our generator, which is 2. Since our generator is 2 and our initial number is 100, we can make the equation stated above the table, and fill out the rest of the chart.
f(x)=50*(2^1/2)^n
x f(x)
0 50
1 50*(2^1/2)
2 100
3 100(2^1/2)
Since we know that f(x)=a+d^x, and we know that a is the initial number, the d is the generator of the sequence, and the x is the independent variable, i.e. the number of times you multiply the generator, d, due to the fact that is is a geometric sequence, we can plug in a set of numbers to help us figure out the generator of the sequence, which we do not know in this case. Since the initial number is always when x is equivalent to 0, we know that in this sequence our initial number is 50. Now instead of plugging that in, we'll use our second set of numbers, 2 and 100, which will be much easier to work with. Since 100 is equivalent to f(x) and 2 is equivalent to x, we can plug those numbers in along with our initial number, so that we get a good 100=50*d^2.
Tuesday, December 4, 2007
BB 107 ~ Ashirah,Narayan,Selarra and Jacob
For these sets of problems in BB 107, you will need to KNOW the Laws of Exponents to solve them.
a).png)
* The first step would be to perform the operation 2^4, or 2x2x2x2 and get the number 16. Next, multiply the exponents 2x4 (one of the laws of exponents) and get 8 for the 8th power. For the last part the y^4 remains the same.
b).png)
*First we solved - 6x ^2 and got 36x^2 for the denominator, but the numerator stays the same. In this equation, x^2 cancels each other out on the top and bottom of the fraction. From this, you only have y^3 left on top. Then you know that -3x can go into 36x twelve times (12 is negative because a positive divided by a negative is negative).
c).png)
* The first part's answer would also be 16x^8y^4 like in the previous problem and the denominator stays the same. Because of the law of exponents, you would subtract 1 (on the bottom) from the number 8 exponent and get 7 but the 16 remains. 4 - 5 is -1 so you would bring the "y" to the bottom, giving you this answer.
d) 5(5xy)^2 (x^3y)
5(25x^2y^2) (x^3y)
answer: 125x^5y^3
* You have to break the first line's variables down to their own exponents to get the next line but the 2nd parentheses stays the same. Next you multiply 25 by 5x to get 125x on the last line. Then you would add the exponents 3 + 2 (Laws of exponents) to get 5 and the y^3 remains.
a)
.png)
* The first step would be to perform the operation 2^4, or 2x2x2x2 and get the number 16. Next, multiply the exponents 2x4 (one of the laws of exponents) and get 8 for the 8th power. For the last part the y^4 remains the same.
b)
.png)
*First we solved - 6x ^2 and got 36x^2 for the denominator, but the numerator stays the same. In this equation, x^2 cancels each other out on the top and bottom of the fraction. From this, you only have y^3 left on top. Then you know that -3x can go into 36x twelve times (12 is negative because a positive divided by a negative is negative).
c)
.png)
* The first part's answer would also be 16x^8y^4 like in the previous problem and the denominator stays the same. Because of the law of exponents, you would subtract 1 (on the bottom) from the number 8 exponent and get 7 but the 16 remains. 4 - 5 is -1 so you would bring the "y" to the bottom, giving you this answer.
d) 5(5xy)^2 (x^3y)
5(25x^2y^2) (x^3y)
answer: 125x^5y^3
* You have to break the first line's variables down to their own exponents to get the next line but the 2nd parentheses stays the same. Next you multiply 25 by 5x to get 125x on the last line. Then you would add the exponents 3 + 2 (Laws of exponents) to get 5 and the y^3 remains.
Problem BB-66 Algebra 2
If you have an equation where y isn't equal to something, like your equation x+y=5, then you would have to solve for y. Then, to find out the intersection of two linear equation or lines, you have to set them equal to each other. By doing this, you will find out the x-coordinate. To find out the y-coordinate, plug in your answer for x and solve for y. When you have the x-coordinate and y-coordinate, then you have the intersection of the two graphs.
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