10 additional problem

#$&*

course Mth173

Question 3-If you are earning 8 dollars / hour, how long will it take you to earn $72? The answer may well be obvious, but explain as best you can how you reasoned out your result

My solution-I have a 8 dollar salary which I got paid 72.I would just divide the 72 dollar over my 8 dollar salary which gives me 9 days

Given Solution: Many students simply know, at the level of common sense, that if we divide $72 by $8 / hour we get 9 hours, so 9 hours are required.

Confidence Rating - 3

Self -Critique -Rating-OK

Self-Critique -OK

Question 4- Calculate (8 + 3) * 5 and 8 + 3 * 5, indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results.

My Solution-In school I always remember PEMDAS ,First we add the parenthesis 8+3 which would give us 11 and you*by 5 and it should give you 55

First you multiply 3*5 and it should give you 15 then add 8 and you should get 23

.............................................

Given Solution: (8 + 3) * 5 and 8 + 3 * 5

To evaluate (8 + 3) * 5, you will first do the calculation in parentheses. 8 + 3 = 11, so

(8 + 3) * 5 = 11 * 5 = 55.

To evaluate 8 + 3 * 5 you have to decide which operation to do first, 8 + 3 or 3 * 5. You should be familiar with the order of operations, which tells you that multiplication precedes addition. The first calculation to do is therefore 3 * 5, which is equal to 15. Thus

8 + 3 * 5 = 8 + 15 = 23

Confidence Assessment -3

Self-Critique Rating -OK

Self -Critique-OK

Question 5- Calculate (2^4) * 3 and 2^(4 * 3), indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results.

We will used the PEMDAS step first we will multiply 2 4 times 2*2*2*2 in it would equal 16 then you multiply by 3 and get 48

First we multiply 3*4 and we get 12 we times 2 12 times 2*2*2*2*2*2*2*2*2*2*2*2 and get 4096

.............................................

Given Solution:

To evaluate (2^4) * 3 we first evaluate the grouped expression 2^4, which is the fourth power of 2, equal to 2 * 2 * 2 * 2 = 16. So we have

(2^4) * 3 = 16 * 3 = 48.

To evaluate 2^(4 * 3) we first do the operation inside the parentheses, obtaining 4 * 3 = 12. We therefore get

2^(4 * 3) = 2^12 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 4096.

Self-Critique Rating -OK

Self -Critique -OK

Confidence Assessment-3

Question 6- Calculate 3 * 5 - 4 * 3 ^ 2 and 3 * 5 - (4 * 3)^2 according to the standard order of operations, indicating the order of your steps. Explain, as best you can, the reasons for the difference in your results.

My Solution-

First we look at the exponent 3^2=9 3*5=15 9*-4=-36

-36+15=-21

First we multiply 4*3 which is in the parenthesis which equal12 then you time it by 12 2 times=144 then you 5*3 which give you 15 144-15=129

.............................................

Given Solution:

To calculate 3 * 5 - 4 * 3 ^ 2, the first operation is the exponentiation operation ^.

The two numbers involved in the exponentiation are 3 and 2; the 4 is 'attached' to the 3 by multiplication, and this multiplication can't be done until the exponentiation has been performed.

The exponentiation operation is therefore 3^2 = 9, and the expression becomes 3 * 5 - 4 * 9.

Evaluating this expression, the multiplications 3 * 5 and 4 * 9 must be performed before the subtraction. 3 * 5 = 15 and 4 * 9 = 36 so we now have

3 * 5 - 4 * 3 ^ 2 = 3 * 5 - 4 * 9 = 15 - 36 = -21.

To calculate 3 * 5 - (4 * 3)^2 we first do the operation in parentheses, obtaining 4 * 3 = 12. Then we apply the exponentiation to get 12 ^2 = 144. Finally we multiply 3 * 5 to get 15. Putting this all together we get

3 * 5 - (4 * 3)^2 =

3 * 5 - 12^2 =

3 * 5 - 144 =

15 - 144 = -129.

Self-Critique Rating-OK

Self-Critique-OK

Confidence Assessment-3

Question 7- Let y = 2 x + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it).

Evaluate y for x = -2. What is your result? In your solution explain the steps you took to get this result.

Evaluate y for x values -1, 0, 1 and 2. Write out a copy of the table below. In your solution give the y values you obtained in your table.

My Solution-The graph is a linear function we solve this problem by plugging in -2,-1,0,1,2 in for the x value which would give you the Y

For -2 the y value is -1

For -1 the y value is 1

For 0 the y value is 3

For 1 the y value is 5

For 2 the y value is 7

.............................................

Given Solution: Two slightly different explanations are give below, one by a student and one by the instructor. Neither format is inherently better than the other. GOOD SOLUTION BY STUDENT:

First we need to complete the table. I have added a column to the right of the table to show the calculation of “y” when we us the “x” values as given.

x y Calculation: If y = 2x + 3

-2 -1 If x = -2, then y = 2(-2)+3 = -4+3 = -1

-1 1 If x= -1, then y = 2(-1)+3 = -2+3 = 1

0 3 If x= 0, then y = 2(0)+3 = 0+3 = 3

1 5 If x= 1, then y = 2(1)+3 = 2+3 = 5

2 7 If x= 2, then y = 2(2)+3 = 4+3 = 7

Once an answer has been determined, the “y” value can be filled in. Now we have both the “x” and “y” values and we can begin our graph. The charted values continue on a straight line representing a linear function as shown above.

INSTRUCTOR'S SOLUTION:

We easily evaluate the expression:

When x = -2, we get y = 2 x + 3 = 2 * (-2) + 3 = -4 + 3 = -1.

When x = -1, we get y = 2 x + 3 = 2 * (-1) + 3 = -2 + 3 = 1.

When x = 0, we get y = 2 x + 3 = 2 * (0) + 3 = 0 + 3 = 3.

When x = 1, we get y = 2 x + 3 = 2 * (1) + 3 = 2 + 3 = 5.

When x = 2, we get y = 2 x + 3 = 2 * (2) + 3 = 4 + 3 = 7.

Filling in the table we have

x y

-2 -1

-1 1

0 3

1 5

2 7

When we graph these points we find that they lie along a straight line.

Only one of the depicted graphs consists of a straight line, and we conclude that the appropriate graph is the one labeled 'linear'.

Confidence Assessment-3

Self-Critique-OK

Self-Critique Rating-OK

Question 8- `q008. Let y = x^2 + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it).Evaluate y for x = -2. What is your result? In your solution explain the steps you took to get this result.Evaluate y for x values -1, 0, 1 and 2. Write out a copy of the table below. In your solution give the y values you obtained in your table.

My Solution-The graph is a quadratic function .For the x value we plug them in to get the y value.

For the x value -2 the y is -1

For the x value -1 the y is 2

For the x value 0 the y is 3

For the x value 1 the y is 4

For the x value 2 the y is 7

Confidence Assessment-3

Self-Critique Rating-OK

Self-Critique-OK

Given Solution- Evaluating y = x^2 + 3 at the five points:

If x = -2 then we obtain y = x^2 + 3 = (-2)^2 + 3 = 4 + 3 = 7.If x = -1 then we obtain y = x^2 + 3 = (-1)^2 + 3 = ` + 3 = 4.If x = 0 then we obtain y = x^2 + 3 = (0)^2 + 3 = 0 + 3 = 3.If x = 1 then we obtain y = x^2 + 3 = (1)^2 + 3 = 1 + 3 = 4.If x = 2 then we obtain y = x^2 + 3 = (2)^2 + 3 = 4 + 3 = 7.The table becomes

x y

-2 7

-1 4

0 3

1 4

2 7

We note that there is a symmetry to the y values. The lowest y value is 3, and whether we move up or down the y column from the value 3, we find the same numbers (i.e., if we move 1 space up from the value 3 the y value is 4, and if we move one space down we again encounter 4; if we move two spaces in either direction from the value 3, we find the value 7).

A graph of y vs. x has its lowest point at (0, 3).

If we move from this point, 1 unit to the right our graph rises 1 unit, to (1, 4), and if we move 1 unit to the left of our 'low point' the graph rises 1 unit, to (-1, 4).

If we move 2 units to the right or the left from our 'low point', the graph rises 4 units, to (2, 7) on the right, and to (-2, 7) on the left.

Thus as we move from our 'low point' the graph rises up, becoming increasingly steep, and the behavior is the same whether we move to the left or right of our 'low point'. This reflects the symmetry we observed in the table. So our graph will have a right-left symmetry.

Two of the depicted graphs curve upward away from the 'low point'. One is the graph labeled 'quadratic or parabolic'. The other is the graph labeled 'partial graph of degree 3 polynomial'.

If we look closely at these graphs, we find that only the first has the right-left symmetry, so the appropriate graph is the 'quadratic or parabolic' graph.

Question 9- Let y = 2 ^ x + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it).Evaluate y for x = 1. What is your result? In your solution explain the steps you took to get this result.Evaluate y for x values 2, 3 and 4. Write out a copy of the table below. In your solution give the y values you obtained in your table.

x y

1

2

3

4

My solution-The graph is a Even negative power.For the graph 2^x+3 , we just plug 1,2,3,4, into the x value and we should get our y value

1 y value 5

2 y value 7

3 y value 11

4 y value 19

Confidence Assessment-3

Self-Critique Rating-OK

Self-Critique-OK

.............................................

Given Solution:

Recall that the exponentiation in the expression 2^x + 1 must be done before, not after the addition.

When x = 1 we obtain y = 2^1 + 3 = 2 + 3 = 5.

When x = 2 we obtain y = 2^2 + 3 = 4 + 3 = 7.

When x = 3 we obtain y = 2^3 + 3 = 8 + 3 = 11.

When x = 4 we obtain y = 2^4 + 3 = 16 + 3 = 19.

x y

1 5

2 7

3 11

4 19

Looking at the numbers in the y column we see that they increase as we go down the column, and that the increases get progressively larger. In fact if we look carefully we see that each increase is double the one before it, with increases of 2, then 4, then 8.

When we graph these points we find that the graph rises as we go from left to right, and that it rises faster and faster. From our observations on the table we know that the graph in fact that the rise of the graph doubles with each step we take to the right.

The only graph that increases from left to right, getting steeper and steeper with each step, is the graph labeled 'exponential'.

Question 10- `q010. If you divide a certain positive number by 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number?

.............................................

Given Solution: If you divide any number by 1, the result is the same as the original number. Doesn't matter what the original number is, if you divide it by 1, you don't change it.

My Solution- when you divide ant whole you are going to get the same number back

Self-Critique Rating-OK

Self-Critique-OK

Confidence Assessment-2, I think it is saying like if you get a number like 2 and you divide it by 1 you still going to get 2

Question 11- If you divide a certain positive number by a number greater than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number?

Given Solution- If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by another number is similar. The bigger the number you divide by, the less you get.

Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a number greater than 1, what you get has to be smaller than the original number. Again it doesn't matter what the original number is, as long as it's positive.Students will often reason from examples. For instance, the following reasoning might be offered:OK, let's say the original number is 36. Let's divide 36 be a few numbers and see what happens:36/2 = 18. Now 3 is bigger than 2, and36 / 3 = 12. The quotient got smaller. Now 4 is bigger than 3, and36 / 4 = 9. The quotient got smaller again. Let's skip 5 because it doesn't divide evenly into 36.36 / 6 = 4. Again we divided by a larger number and the quotient was smaller

My Solution-I guess when you divide a number like 110/10=11

Confidence Assessment-2 somewhat confuse hopefully my solution is right

Self-critique Rating-OK

Self-Critique-OK

Question 12- If you divide a certain positive number by a positive number less than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number?

.............................................

Given Solution: If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by some other number is similar. The bigger the number you divide by, the less you get. The smaller the number you divide by, the more you get. Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a positive number less than 1, what you get has to be larger than the original number. Again it doesn't matter what the original number is, as long as it's positive.

My Solution-when you divide a positive your answer should also stay positive, if it is negative the number is negative

Confidence Assessment-Just a little confuse with this particular question I think I understand it

Self-Critique Rating-OK

Self-Critique-OK

"

@& Your responses need to be inserted into a complete copy of the part of the document you were instructed to copy. Nothing should be deleted from that copy. Your responses should be inserted in the specified manner.

I was able to review enough of this document to tell that you're doing OK, so I won't ask you to resubmit. However if you want the posted document to be complete, and/or if you want me to give it a more thorough review, you are welcome to insert your answers into a complete copy and resubmit. In any case, in future submissions be sure to include everything.*@