course MTH 158
Hello
*********************************************
Your solution:
The slope will be greater in (7<170 (10,29) because there is the same rise but a one unit less
*********************************************
Given Solution:
`aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.
Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3.
Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
ok
Self-critique Rating:
ok
Question: **** `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
*********************************************
Your solution:
(x-2)*(2x+5)
(2-2)*(2*-2.5+5)
(0)*(-5+5)
0*0 = 0
this is the only set of numbers because only 2-2 will give you 0 and only –2.5 will give you –5 when multiplied and then added to 5 to give you 0
*********************************************
Given Solution:
`aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero.
If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero.
The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.
Note that (x-2)(2x+5) can be expanded using the Distributive Law to get
x(2x+5) - 2(2x+5). Then again using the distributive law we get
2x^2 + 5x - 4x - 10 which simplifies to
2x^2 + x - 10.
However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
ok
Self-critique Rating:
ok
Question: **** `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
*********************************************
Your solution:
X=2
(3*2-6)*(2+4)*(2^2-4)
(6-6)*(6)*(4-4)
0*6*0
when x=2 the expression will equal zero
X=-2
(3*-2-6)*(-2+4)*(-2^2-4)
(-6-6)*(2)*(4-4)
(-12)*(2)*(0)
when x=-2 the expression will equal zero
x=-4
(3*-4-6)*(-4+4)*(-4^2-4)
(-18)*(0)*(-16-6)
when x=-4 the expression is zero
Any number that will give you a set that equals zero will work because once a number is multiplied by zero it will all be zero
*********************************************
Given Solution:
`aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0.
3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.
x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.
x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.
We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
ok
Self-critique Rating:
ok
Question: **** `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
*********************************************
Your solution:
With the segments given the (10,2) and (50,4) points will give you a trapezoid with a greater area because the distance from the points is larger.
*********************************************
Given Solution: this solution given shows me how I could have analyzed more and gave more details. This was a very good example
`aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area.
To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.
This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
* `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:
As we move from left to right the graph increases as its slope increases.
As we move from left to right the graph decreases as its slope increases.
As we move from left to right the graph increases as its slope decreases.
As we move from left to right the graph decreases as its slope decreases.
*********************************************
Your solution:
I am unsure about this problem
*********************************************
Given Solution:
`aFor x = 1, 2, 3, 4:
The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate.
The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero.
Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases.
We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate.
For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
This makes the problem make a little more sense but I am still a little unclear
&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).
&#
Self-critique Rating:
Question: **** `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
*********************************************
Your solution:
10% of 20 is 2
month 1 – 2 more 22 frogs
month 2- 24.2 frogs
month 3- 26.6 frogs
*********************************************
Given Solution:
`aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs.
The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic).
A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
This critique was better than the one I provided it actually showed the math steps
Self-critique Rating:
Question: **** `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
*********************************************
Your solution:
1/1 = 1
1/.1=10
1/.01=100
1/.001= 1000
the values go farther away as we continue to approach 0
*********************************************
Given Solution:
`aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1.
So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc..
Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere.
The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become.
The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.
This is what it means to say that the y axis is a vertical asymptote for the graph .
* `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
•
*********************************************
Your solution:
V=3*5+9
V=24
E=800*24^2
E=800*576
E=460800
*********************************************
Given Solution:
`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
•
* `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
*********************************************
Your solution:
E=800 v^2= 800v=3t+9
*********************************************
Given Solution:
`aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here.
For further reference, though, note that this expression could also be expanded by applying the Distributive Law:.
Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get
E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I did not do the problem in the correct order but I see that I needed the paranthesis and to raise it to the 2nd power at the end
Self-critique Rating:
2
Overall you did pretty well on these problems.
It appears that the graphing problem gave you a little trouble. We'll get to that topic in your course. However be sure to see my note on that problem.