The only problem I lost points on was the derivation of the law of cosines. I got the equations x^2+h^2=c^2 and (a-x)^2+h^2=b^2, but I couldn't connect them.
Here is how to connect the two equations:
First, factor out the (a-x)^2 to make the equation a^2-2ax+x^2+h^2=b^2
Then, replace the x^2+h^2 with c^2 using substitution to make the equation a^2-2ax+c^2=b^2.
Using the picture, it is possible to determine that cosB=x/c, or x=c(cosB).
Therefore, we can substitute this for x, making the final equation a^2+c^2-2ac(cosB)=b^2, or the law of cosines.
I am good at geometry so I liked this unit. I only had to memorize the formulas and apply them, which is what I am good at in math.
Wednesday, December 19, 2012
Tuesday, December 4, 2012
Trigonometry and Unit Circle Test Corrections
1) In this problem, I made the mistake of saying that the radius of the circle was 100 rather than 10. I forgot that the equation of a circle is x^2+y^2=r^2. Knowing this, the radius would be the square root of 100, or 10.
7) In this problem, I lost all of my points for not paying attention to negatives. The point was in the second quadrant, therefore everything except for the sine should be negative. I made them all positive.
9) e. I did not simplify -2(squareroot2)/2 to -(squareroot2) by cancelling the 2's.
g. 5pi/6 is in the second quadrant, therefore the sine should be positive, but I wrote -1/2.
After memorizing the unit circle, this unit was not too bad. It is easy to make silly mistakes, however.
7) In this problem, I lost all of my points for not paying attention to negatives. The point was in the second quadrant, therefore everything except for the sine should be negative. I made them all positive.
9) e. I did not simplify -2(squareroot2)/2 to -(squareroot2) by cancelling the 2's.
g. 5pi/6 is in the second quadrant, therefore the sine should be positive, but I wrote -1/2.
After memorizing the unit circle, this unit was not too bad. It is easy to make silly mistakes, however.
Friday, November 9, 2012
Test: Angles and Velocity Corrections
2) Find the measure of angle CAB in radians. Explain how you know.
I got this question wrong because I didn't see the arc length so I didn't know how to do the problem.
Now that I see the arc length, here is my answer:
The arc length formula for radians is the central angle*radius, therefore we can derive the formula that the central angle is the arc length/radius.
Therefore, since the arc length is 15 cm and the radius is 5 cm, the measure of the central angle is 3 radians.
7) For this problem I forgot to put in the angle measure of each reference angle, so here are the angle measures.
a) The reference angle for 500 degrees is 40 degrees, since the 500 degree angle goes around the cirlce once and then goes around an additional 140 degrees, making its terminal side in the 2nd quadrant. The acute angle formed with the x-axis is therefore 40 degrees.
b) The reference angle for -5pi/3 is pi/3 because the angle -5pi/3 lies in the first quadrant, pi/3 radians away from the x-axis.
I missed one critical day during this section so I didn't ever fully learn the formulas on my own, which made this unit difficult for me.
I got this question wrong because I didn't see the arc length so I didn't know how to do the problem.
Now that I see the arc length, here is my answer:
The arc length formula for radians is the central angle*radius, therefore we can derive the formula that the central angle is the arc length/radius.
Therefore, since the arc length is 15 cm and the radius is 5 cm, the measure of the central angle is 3 radians.
7) For this problem I forgot to put in the angle measure of each reference angle, so here are the angle measures.
a) The reference angle for 500 degrees is 40 degrees, since the 500 degree angle goes around the cirlce once and then goes around an additional 140 degrees, making its terminal side in the 2nd quadrant. The acute angle formed with the x-axis is therefore 40 degrees.
b) The reference angle for -5pi/3 is pi/3 because the angle -5pi/3 lies in the first quadrant, pi/3 radians away from the x-axis.
I missed one critical day during this section so I didn't ever fully learn the formulas on my own, which made this unit difficult for me.
Saturday, November 3, 2012
Functions and Critical Values Test Corrections
6) Given the function y=20x^7-10x^3/5x^3
a) State the end behavior of the function and explain how you can determine the end behavior without graphing the function.
What I had:
end behavior: as x approaches infinity f(x) approaches infinity
as x approaches negative infinity f(x) approaches negative infinity
If the greatest exponent is odd and the leading coefficient is positive, the above will be the end behavior.
Why this is wrong:
I did not simplify the function.
Simplified, the function is 4x^4-2.
The right answer:
end behavior: as x approaches infinity f(x) approaches infinity
as x approaches negative infinity f(x) approaches infinity
If the greatest exponent is even and the leading coefficient is positive, the above will be the end behavior.
b) State the intervals where the function is increasing and decreasing.
What I had: increasing for all real numbers
Why this is wrong: I put the equation into the calculator without parentheses around the numerator and denominator.
The right answer: decreasing {x/0>x} increasing {x/0<x}
This unit was fine as it was pretty much a review from last year, but I didn't ever memorize the end behavior rules.
a) State the end behavior of the function and explain how you can determine the end behavior without graphing the function.
What I had:
end behavior: as x approaches infinity f(x) approaches infinity
as x approaches negative infinity f(x) approaches negative infinity
If the greatest exponent is odd and the leading coefficient is positive, the above will be the end behavior.
Why this is wrong:
I did not simplify the function.
Simplified, the function is 4x^4-2.
The right answer:
end behavior: as x approaches infinity f(x) approaches infinity
as x approaches negative infinity f(x) approaches infinity
If the greatest exponent is even and the leading coefficient is positive, the above will be the end behavior.
b) State the intervals where the function is increasing and decreasing.
What I had: increasing for all real numbers
Why this is wrong: I put the equation into the calculator without parentheses around the numerator and denominator.
The right answer: decreasing {x/0>x} increasing {x/0<x}
This unit was fine as it was pretty much a review from last year, but I didn't ever memorize the end behavior rules.
Monday, September 24, 2012
Functions and Graphs Test Corrections
Test Corrections: Functions and Graphs
9a)
This graph was a parabola going through the points (2,1) (vertex), (-1, 3) and (5,3). If this graph had no stretch or shrink it would have had the two additional points at (1, 2) and (3,2), so I knew on the test that this had a stretch/shrink. At this point, I only eyeballed and estimated 1/2 was about right, but I was wrong. What I should have done was to algebraically calculate the stretch/shrink. I could do this by considering that on the parent graph, a point three away from the vertex would have a y-value of 3^2, or 9. This point, however, on this graph, was only 2 y-values up from the origin. Therefore it had a shrink of 2/9.
9f)
This graph was a translated graph of the parent graph 1/x. This graphs "vertices," for my lack of a better term, should go through the points (1,1) and (-1,-1) on the parent graph. On the graph on the test, however, the "vertices" go through no distinct point. This is how I knew there was a stretch/shrink. I therefore correctly put 2/x-1 as part of my answer. Where I went wrong was by adding a -1 to the end. I saw the graph as shifted down 1 because the portion that would normally be in the 3rd quadrant appeared to me at the time as having shifted down one. This was just my perception being deceived however, as I had already accounted for the shift in the graph in the x-1.
I found this unit to be pretty easy, but sometimes I got the discontinuities wrong.
9a)
This graph was a parabola going through the points (2,1) (vertex), (-1, 3) and (5,3). If this graph had no stretch or shrink it would have had the two additional points at (1, 2) and (3,2), so I knew on the test that this had a stretch/shrink. At this point, I only eyeballed and estimated 1/2 was about right, but I was wrong. What I should have done was to algebraically calculate the stretch/shrink. I could do this by considering that on the parent graph, a point three away from the vertex would have a y-value of 3^2, or 9. This point, however, on this graph, was only 2 y-values up from the origin. Therefore it had a shrink of 2/9.
9f)
This graph was a translated graph of the parent graph 1/x. This graphs "vertices," for my lack of a better term, should go through the points (1,1) and (-1,-1) on the parent graph. On the graph on the test, however, the "vertices" go through no distinct point. This is how I knew there was a stretch/shrink. I therefore correctly put 2/x-1 as part of my answer. Where I went wrong was by adding a -1 to the end. I saw the graph as shifted down 1 because the portion that would normally be in the 3rd quadrant appeared to me at the time as having shifted down one. This was just my perception being deceived however, as I had already accounted for the shift in the graph in the x-1.
I found this unit to be pretty easy, but sometimes I got the discontinuities wrong.
Saturday, September 8, 2012
Summer Assignment Test
I got half a point off of problem number 3. Here is the problem redone:
2/2x-5+3
2/2x-2
1/x-1
I also got number 6 wrong. Here is the problem redone:
A. 13x-20y+40=0
13x+40=20y
13/20x+2=y wrong slope and right y-intercept
***answer I originally put because I only looked at the y-intercept***
B. 3x+2y-1=0
3x-1=-2y
3/2x+1/2=y wrong slope and wrong y-intercept
C. 4x-3y+6=0
4x+6=3y
4/3x+3=y right slope and right y-intercept
***correct answer***
D. 6x-3y=0
6x=3y
2x=y wrong slope and wrong y-intercept
These are the only two problems that I got points off on. I found the test to be fairly easy because most of it was review from Algebra I and II.
2/2x-5+3
2/2x-2
1/x-1
I also got number 6 wrong. Here is the problem redone:
A. 13x-20y+40=0
13x+40=20y
13/20x+2=y wrong slope and right y-intercept
***answer I originally put because I only looked at the y-intercept***
B. 3x+2y-1=0
3x-1=-2y
3/2x+1/2=y wrong slope and wrong y-intercept
C. 4x-3y+6=0
4x+6=3y
4/3x+3=y right slope and right y-intercept
***correct answer***
D. 6x-3y=0
6x=3y
2x=y wrong slope and wrong y-intercept
These are the only two problems that I got points off on. I found the test to be fairly easy because most of it was review from Algebra I and II.
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