#$&*
course PHY 231
(7/22 )
Precalculus Initial Questions003. PC1 questions
*********************************************
Question: `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For the first line it moves from (3,5) to (7, 17) so the slope is 12/4 =3; for the second line the slop is 12/3=4, which is higher slop than the 1st line and hence steeper
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
`
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): no
------------------------------------------------
Self-critique Rating: 3
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
In testing the given data, If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero and If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. We now know that the only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.
The reason this happens is PEMDAS (order of operations)
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
`
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): no
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
A classic example. To do this you must set all the equations set to zero; so we now have 3x-6 = 0 or x+4=0 or x^2-4=0. For the 1st one 3x-6 = 0 is arranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and hence entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. The same logic is applied to the next two. So for x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. Also so x^2-4 = 0 is arranged to x^2 = 4 which has solutions x = + - square root of (4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.
x = 2, or -4, or -2.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
`
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): no
------------------------------------------------
Self-critique Rating: 3
*********************************************
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'.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Though to the naked eye the second trapozid looks bigger with math we can show that is 4 units wide (obtained from 7-4) while the second runs from has a width of 40 units (obtained from 50-10).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. Hence from this data you are able to prove that the second trapezoid has a larger area.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): wording could be better
------------------------------------------------
Self-critique Rating: 2
*********************************************
Question: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:
As we move from left to right the graph increases as its slope increases.
As we move from left to right the graph decreases as its slope increases.
As we move from left to right the graph increases as its slope decreases.
As we move from left to right the graph decreases as its slope decreases.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Solution:
For x = 1, 2, 3, 4:
The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases proportionally at an increasing rate.
The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing but less and less each time. This shows that it is a negative slope but not directly proportionally related
For y = square root of (x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): No
------------------------------------------------
Self-critique Rating: 3
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
At 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.
So after 300 months we will have multiplied by 1.1 a total of 300 times. This would give us 20 * 1.1^300=5.234x10^13
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): No
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
If x = .1, for example, 1 / x = 1 / .1 = 10 the patter we get is .1, .2, .3, .4, ... .9, 10.
So if x = .01, 1/x = 100; If x = .001 then 1/x = 1000, etc..
The values of 1/x don't just increase, they increase without bound. (there is no limit) x approaching 0 through the values .1, .01, .001, ..., and reciprocals 10, 100, 1000, 10000 etc. there is no limit. So when graphed it looks like it gets closer and closer to y each time but DOES NOT touch the y axis
The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): No
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question: `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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For t=5, v = 3 t + 9 = (3*5) + 9 = 24.
E = 800 * 24^2 = 460800.
confidence rating #$&*: 3 (basic)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): No
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2.
confidence rating #$&*: 3 (basic)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): No
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question:
Suppose you invest $1000 and, at the end of any given year, 10% is added to the amount. How much would you have after 1, 2
and 3 years?
What is an expression for the amount you would have after 40 years (give an expression that could easily be evaluated using a calculator, but don't bother to actually evaluate it)?
What is an expression for the amount you would have after t years?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
1000*.1=1100 first year
For the rest you use compounding interest formula
2 year= 1210
3 years= 1331and on and on
confidence rating #$&*:2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
------------------------------------------------
Self-critique Rating:3
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
@&
Your solutions are very good. Though you deleted information, I'm not going to ask you to rebusmit this document. In future submissions, I'm sure you'll submit the documents as specified.
*@