#$&*
course Phy 201
6/3 9
Question: `q001. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
8 * 4 = 32
If you earn $8 dollars an hour and work for 4 hours, you would need to multiply 8 times 4 and get the solution of $32 for 4 hours work.
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q002.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
168 / 12 = 14
If you work 12 hours and earn $168 then you would need to divide 168 by 12 to find out the hourly rate you were making, which is $14 per hour.
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q002.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: (type in your solution starting in the next line)
168 / 12 = 14
If you work 12 hours and earn $168 then you would need to divide 168 by 12 to find out the hourly rate you were making, which is $14 per hour.
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q003.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
72 / 8 = 9
If you earn $8 an hour and you want to know how many hours you will have to work to earn $72, you need to divide 72 by 8 giving you the solution of 9 hours.
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q004
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(8 + 3) * 5 8 + 3 * 5
11 * 5 = 55 8 + 15 = 23
In the first problem you must calculate what is in the parenthesis before continuing with the multiplication. So 8 + 3= 11 and then you multiply that by 5 resulting in the solution of 55. In the second problem you must calculate the multiplication before continuing with the addition. So 3 * 5 = 15 and then 15 + 8 gives the solution of 23.
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q005.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(2^4) * 3 2^(4*3)
16 * 3 = 48 2^12 = 4096
In the first problem you must work what is in parenthesis first and 2^4 = 16 and then you multiply 16 by 3 giving the solution of 48. In the second solution you must do what is in parenthesis first and 4 * 3 = 12 and then 2^12 = 4096.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique (if necessary):OK
Question: `q006
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
3 * 5 - 4 * 3 ^ 2 3 * 5 -( 4 * 3)^2
3 * 5 - 4 * 9 3 * 5 -12^2
15 - 36 = -21 15 - 144 = -129
In the first problem you must work the exponents first 3^2 = 9 and then you must work the multiplication 3 * 5 = 15 and 4 * 9 = 36, and then you work the subtraction 15 - 36 = -21. In the second problem you must work what is in the parenthesis first 4 * 3 = 12, then you must work the exponents 12^2 = 144 and then you work the multiplication 3 * 5 = 15, then you work the subtraction 15 - 144 = -129.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique Rating:OK
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Question: `q007.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
X Y Y = 2x + 3
-2 -1 Y = 2(-2) + 3 = -4 + 3 = -1
-1 1 Y = 2(-1) + 3 = -2 + 3 = 1
0 3 Y = 2(0) + 3 = 0 + 3 = 3
1 5 Y = 2 (1) + 3 = 2 + 3 = 5
2 7 Y =2(2) + 3 = 4 + 3 = 7
Before we are able to graph we must be complete table by calculating the y coordinates. We do this by putting the x coordinates into the problem. Once we complete the problem by placing all the x coordinates into the problem to get the y coordinates. Once we have both the x and y coordinates, we can begin to graph. The graph increases in a straight line and looks most like the linear graph.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique Rating: OK
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Question: `q008
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
X Y Y = x ^2 + 3
-2 7 Y = -2^2 + 3= 4 + 3 = 7
-1 4 Y = -1^2 + 3 = 1 + 3 = 4
0 3 Y = 0^2 + 3 = 0 + 3 = 3
1 4 Y = 1^2 + 3 = 1 + 3 = 4
2 7 Y = 2^2 + 3 = 4 + 3 = 7
We must first replace the x with the values of x to be able to complete the table with the y values. Once we graph this we see that both sides of the x axis are the same and the graph looks most like the quadratic or parabolic.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique Rating: OK
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Question: `q009
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
X Y Y = 2^x + 3
1 5 Y = 2^1 + 3 = 2 + 3 = 5
2 7 Y = 2^2 + 3 = 4 + 3 = 7
3 11 Y = 2^3 + 3 = 8 + 3 = 11
4 19 Y = 2^4 + 3 = 16 + 3 = 19
We must first place the x values in the problem to get the y values. Once we get both values we are able to graph. The points in the graph increase and form a straight line and looks most like the exponential.
confidence rating #$&*:
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Self-critique (if necessary):
#### Is there any way that this graph could be considered linear as well or is it just exponential because it has the slight curve or just exponential because the formula contains exponents???
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Self-critique Rating:OK
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Question: `q010.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
It is equal to the original number.
confidence rating #$&*:
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Self-critique Rating: OK
Question: `q011.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
It will be less than the original number
confidence rating #$&*:
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Self-critique Rating: OK
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Question: `q012
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The result will be greater than the original number.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique Rating: OK
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Question: `q013
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
####Is there any way that this graph could be considered linear as well or is it just exponential because it has the slight curve or just exponential because the formula contains exponents???
"
Self-critique (if necessary):
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Self-critique rating:
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Question: `q013
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
####Is there any way that this graph could be considered linear as well or is it just exponential because it has the slight curve or just exponential because the formula contains exponents???
"
@&
The curvature becomes very pronounced as you move to the right.
Any function of the form y = b^x is exponential. The key is that the variable is in the exponent.
For example the function y = x^2 would not be exponential. The variable is the base, not the exponent.
On the other hand y = 2.3^x is exponential, with the variable in the exponent.
The complete definition of an exponential function is a little more involved, but this is the basic idea.
*@
Self-critique (if necessary):
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Self-critique rating:
#*&!
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Question: `q013
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
####Is there any way that this graph could be considered linear as well or is it just exponential because it has the slight curve or just exponential because the formula contains exponents???
"
@&
The curvature becomes very pronounced as you move to the right.
Any function of the form y = b^x is exponential. The key is that the variable is in the exponent.
For example the function y = x^2 would not be exponential. The variable is the base, not the exponent.
On the other hand y = 2.3^x is exponential, with the variable in the exponent.
The complete definition of an exponential function is a little more involved, but this is the basic idea.
*@
Self-critique (if necessary):
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Self-critique rating:
#*&!#*&!
&#This looks good. See my notes. Let me know if you have any questions. &#