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Algebra II Honors - Help

Synthesis

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I'm currently receiving a D average for my Alg. 2 honors class, which is killing my GPA (4.3 down to a 3.5 something). Does anyone have/know of any comprehensive websites or literature that could help me? A D in any class is...just unacceptable. We just got done with logarithms and I don't have any idea as to what their function is or how to solve them, if someone could erm, help me...I'd be very grateful. Futhermore, I'm also taking a Geometry (the title of the thread is deceptive) Honors course - part of the International Bacculaureate (IB) requirement - and am currently covering trig. functions; I could use some help with determining when/where to use tangent/cosine/sine, if anyone would explain their signifigance and mechanics please?

(I'm a general failure at mathematics, so my questions may be slightly dense at times :()
 

Latro

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The point of a logarithm is to provide an inverse operation to exponentiation. In this way it is analogous to division or subtraction. Formally:
a^b = c
is the same as
log (base a) c = b

That is, the log (base a) of a number gives you the number to which you must raise a to get back the number.

The properties of exponents, combined with this definition, give us a few rules:
a^(b+c)=a^b*a^c
we can take the log (base a) of both sides. (The reason is because log (base a) has an inverse, namely a^, which means that, having done that step, we can do another step to get back to where we were initially, every time, and so we can do the log operation without breaking equality.) On the left side we get b+c (from above) which means that on the right side we must also get b+c. This suggests that log(xy) = log(x)+log(y), and indeed this holds.

The main other rule is for powers, which also comes from the properties of exponents:
(a^b)^c = a^(bc)
again, take the log (base a) of both sides. This time we know the answer on the RHS, however, and it is bc. This means that when we raise something to the power of c, we should expect its logarithm to be scaled by c, and indeed this is what happens. In other words, log(x^c) = c*log(x).

The only other main rule that isn't just a special case of these two is the change of base formula, which I don't think I can quite cover here.

The most basic purpose of logs is to give us a way to solve equations in which a variable is in the exponent. For example:
2^x = 7
take the log(base 2) of both sides:
x = log(base 2) 7
is the solution.


That's really all there is to logs for purposes of Algebra II, I think. As for trig, there's a lot more structure, and so I couldn't probably summarize in that way. If you can understand the geometric meanings of sine, cosine, tangent, etc., then the only identities you need to solve most problems in actual trig are pretty much:
sin^2(x)+cos^2(x) = 1
sin(2x) = 2*sin(x)*cos(x)
cos(2x) = cos^2(x)-sin^2(x)

Sometimes you might have to use others (like sin(a+b) where a != b for example) but most problems boil down to these or something that can be fairly easily derived from these. An example of an easy derivation comes in the first one if we divide both sides by cos^2(x):
sin^2(x)/cos^2(x) + 1 = 1/cos^2(x)
tan^2(x)+1=sec^2(x)
 

BigApplePi

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Synthesis, Latro is correct.

Just remember a logarithm is an exponent. So since it is an exponent it behaves like one.

Sine and cosine -- learn their definitions. Tangent too.

After that it's a matter of being exposed to the formulas Latro was talking about. Usually the formulas are first proven. That helps you understand them. Then they are applied over and over.

Makes sense? Questions?
 

Vrecknidj

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Get a tutor, spend a LOT of time learning the stuff. Then, do a ton of problems. Really. Do hundreds and hundreds of problems.

There's no other way.

Dave
 

Latro

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Oh, and for trig (not working with triangles (which I still call geometry) but working with the trig functions without triangles present), you do need to know the basic algebraic relationships between the trig functions. tan = sin/cos. csc=1/sin. etc. You need them mainly because one of the usual things to do in a trig proof (which is one of the standard trig problems), especially if you don't immediately see the answer, is to get everything on both sides of the given equation in terms of just sin(x) and cos(x) and then make the sides look the same. But if you don't know, say, that tan=sin/cos, you won't be able to get back to sin and cos-land to simplify matters.
 

BigApplePi

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TheHmmmm

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The way I've always remembered the functions is SOH-CAH-TOA

Sin=O/H
Cos=A/H
T=O/A
 

Jah

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Synthesis

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Apologies on the late response on my part, I've been very busy with school/homework (IB student) lately. Thank you everyone for answering my queries and providing links/examples and the breakdowns of logs et cetera. I'm thinking that I get it now, just need to review the material on this page, the links, and previous notes/work; then I should be considerably better off. Thanks all :)
 

BigApplePi

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Apologies on the late response on my part, I've been very busy with school/homework (IB student) lately. Thank you everyone for answering my queries and providing links/examples and the breakdowns of logs et cetera. I'm thinking that I get it now, just need to review the material on this page, the links, and previous notes/work; then I should be considerably better off. Thanks all :)

If you need anything more or better, drop back in.:)
 

TheHmmmm

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Just for clarificative purposes, with a right triangle whose hypotenuse is 12 and opposite side (the shortest side?) is 8, one would find the third side (base) length by: Cos = 1/h[12]? (or would it be 8/12, given the contextual information?)

in this case, you would use a^2+b^2=c^2.

12^2=8^2+x^2

144=64+x^2

80=x^2

x=the square root of 80 (sorry, too lazy to find out how to input a radical sign).

Sin, Cos, and Tangent are used when you have one angle (other than the 90 degree one) and one side in a right triangle.
 

Synthesis

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x=8.94427-->8.9. Damn, I knew I was doing it right the first time. Thank you TheHmm. Ive another trigonometry question (the teacher is...not so good, Ive found from my experience and that of my classmates):
Is the hypotenuse always the tangent, the shortest side always the sine, and the base side always the cosine? Or are they relative to the side/angle you are attempting to find? ~confusion~ Thank you for your aid, by the by. :)
 

TheHmmmm

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x=8.94427-->8.9. Damn, I knew I was doing it right the first time. Thank you TheHmm. Ive another trigonometry question (the teacher is...not so good, Ive found from my experience and that of my classmates):
Is the hypotenuse always the tangent, the shortest side always the sine, and the base side always the cosine? Or are they relative to the side/angle you are attempting to find? ~confusion~ Thank you for your aid, by the by. :)

Each side doesn't have a corresponding sin, cos, or tan. These are just ratios relating the location of the sides of a triangle to a specific angle.

Sin(angle)=(side opposite the angle)/(hypotenuse)
Cos(angle)=(side adjacent the angle)/(hypotenuse)
Tan(angle)=(side opposite the angle)/(side adjacent to the angle)

To answer your question in short, yes they are relative to the location of the angle.

Sorry if that's hard to understand; I really need to draw it out for you.

As far as identifying the sides, the only constant characteristic is that the hypotenuse is ALWAYS the longest side of a right triangle.
 

Synthesis

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Ah, that makes sense. I can visualise it in my mind, thank you once more; this helps. Aye, I figured that the hypotenuse would always be the longest side...else there would be odd lengths sticking out of the triangle's joints.
 

TheHmmmm

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Ah, that makes sense. I can visualise it in my mind, thank you once more; this helps. Aye, I figured that the hypotenuse would always be the longest side...else there would be odd lengths sticking out of the triangle's joints.

Indeed. That clear everything up then?
 

Synthesis

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For the moment, aye. I *may* be needing aid with Algebra II once it rolls along in my scheduale (how is this spelled? I can never remember for the life of me), a D in any class is just unacceptable, aside from it killing my GPA (4. something to 3.5). I will likely post as is necessary in this thread, but I see no reason for you to check again this day. My thanks for your aid in Geometry :)
 

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I found a unit circle to be invaluable in visualizing trig. I have no idea if it will help you, but take a look. This is a very geometric way of looking at things.

A unit circle has a radius of 1, and of course you know that similar triangles will have similar ratios.

If you look at my little attached drawing (see thumbnail), the angle that is marked has a sin of the opposite over the hypotenuse (and the hypotenuse is the radius and equal to 1). That means the sin is the height of that line segment. As the angle gets smaller toward zero, the sin gets shorter and goes to zero. As the angle goes to 90 degrees traveling counterclockwise, the sin goes to a max of 1, the radius. Then the angle gets to the other side, negative x, and the sin goes back toward zero and continues toward negative one.

The cos does the same. When the angle is zero, cos is 1, and it shrinks to zero as the angle goes to 90 degrees, and so on.

Now for the tan. There is a similar triangle there, with the same angle we had before. If you look, sin/cos = tan/1, or sin/cos = tan. Also, when the angle is zero, tan = 0, and goes up to + infinity when the angle is 1.

When it is a 45 degree angle, sin = cos, and there is an isosceles triangle with tan = 1.

I also found it handy to learn how to draw sin and cos waves, with a few memorized points:

sin 0 = 0
sin 30 = .5
sin 45 = .707 or (√2)/2
sin 60 = .866 or (√3)/2
sin 90 = 1

and the corresponding values for cos, plus where they go negative.

I know, memorizing isn't as good as deriving something, but a few handy facts really do make things faster.

As for logarithms, look at some log charts. For example, Google Finance will show charts of any given stock, and you can set them to have a log scale. You can just see how the scale changes with the value. The cool thing about logs is that if you double (or triple, or half) your value, the distance on the graph is the same regardless of where you started. So if a stock goes up by 10%, it looks the same whether it goes from $30 to $33 or $9000 to $9900.

For a visual example, go to http://www.google.com/finance?q=goog, a fairly volatile stock. Set the chart to show 5 years, and click Settings. Change between linear and logarithmic to see the difference. A move from $200 to $400 will either look the same as a move from $400 to $600 (linear, same dollar difference) or the same as a move from $300 to $600 (logarithmic, same percentage move).

Play around with the Richter scale, and try to figure out how much bigger a 6.5 is than a 6.0. Decibels are logarithmic, too. The more you see logs, the less they feel confusing. None of this will teach you techniques for using logarithms, but they will help in visualizing what they do. Exposure really matters, so I second Vrecknidj about doing lots of examples.
 

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Synthesis

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@Trebuchet,
I will be coming back to your post for some time - my Geometry class is approaching circles with Trig. functions soon, so it will be of immeasurable value to me. As far as logarithms go, I'll follow your advice for a time and observe any changes. Thank you for the information...would you happen to know of sigma notation? (I.e., 2 + 4 + 6 + . . . + 1000) I am currently snagged on this as far as my homework goes, and I know not how to do it.
 

TheHmmmm

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@Trebuchet,
I will be coming back to your post for some time - my Geometry class is approaching circles with Trig. functions soon, so it will be of immeasurable value to me. As far as logarithms go, I'll follow your advice for a time and observe any changes. Thank you for the information...would you happen to know of sigma notation? (I.e., 2 + 4 + 6 + . . . + 1000) I am currently snagged on this as far as my homework goes, and I know not how to do it.

Sigma notation is used when finding the sum of a series. Are you trying to put that series into sigma notation? Or are you actually finding the sum of the series?
 

Synthesis

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I have to put it into series notation. We just encroached upon this subject last week and I have absolutely no idea what/how to do it. Eh, I am a little uncertain as to how to sum them, Im assuming you apply the variable (i.e, n=2) to a given equation until you reach a certain number?
 

TheHmmmm

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First, let's start with series notation.

On top of the sigma symbol is the number of terms in the series. After fiddling with some formulas, I figured out how to find this with your example series (which is arithmetic, btw).

I use this formula (only works for arithmetic series'):
(Last term-First term)/(NT-1)=D

NT= the number of terms in the series.
D=the difference between each term

Thus, it comes out to this:

(1000-2)/(NT-1)=2
998/2=NT-1
449=NT-1
NT=500

Put this on top of the symbol.

To the immediate right of the symbol, you place the formula for the terms of the series. To find this, find the common difference between each term and multiply it by n (2n). Underneath the sigma symbol is the index number of the series or what you plug into the expression first (assumed to be 1 in this case).

That's all there is to it.
 

Synthesis

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Ah, that makes a lot of sense. I will be writing all of that down in my notes. I knew it was arithmetic, but wasn't sure how to proceed...in any area. Thank you, once more. You've helped me quite a bit today with my homework -sheepish grin- :o:worship:
 

BigApplePi

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@Trebuchet,
I will be coming back to your post for some time - my Geometry class is approaching circles with Trig. functions soon, so it will be of immeasurable value to me. As far as logarithms go, I'll follow your advice for a time and observe any changes. Thank you for the information...would you happen to know of sigma notation? (I.e., 2 + 4 + 6 + . . . + 1000) I am currently snagged on this as far as my homework goes, and I know not how to do it.
Sigma notation is neat:
http://www.math.montana.edu/frankw/ccp/general/sigma/learn.htm

In your case if "n" is the variable you want to sum n from 2 to 1000 by twos. You write it like this:
Sigma 2m from m=1 to m = 500 where n = 2m. You stick the m=1 below the "Sigma sign" and the 500 at the top. Let's see if Hmmmm agrees.

Added: I forgot to say when using Sigma and saying m=1 to m=500, one always increases m by one, like so: m = 1, 2, 3, 4, ....., 499, 500.
 
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TheHmmmm

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Sigma notation is neat:
http://www.math.montana.edu/frankw/ccp/general/sigma/learn.htm

In your case if "n" is the variable you want to sum n from 2 to 1000. You write it like this:
Sigma 2m from m=1 to m = 500 where n = 2m. You stick the m=1 below the "Sigma sign" and the 500 at the top. Let's see if Hmmmm agrees.

Wow, I can't believe I screwed up that bad. My bad synthesis, I got a few things backwards.

Yes, n=1 goes on the bottom and yes, 2n is what goes to the right of the sigma notation. I'm going to edit my post for correctness. Thanks Big.

Synthesis, please go re-check my post for my corrections.
 

Synthesis

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Rechecked the post. Just to make sure I understand it properly: on top one puts the number of terms in a series (derived via the arithmetic forumla/methodology you showed), on the bottom goes the index number, on the immediate right goes the common difference - which is the difference between every term in the series, and is then multiplied by n...which is what? ~confusion~ Thank you BAP for catching what you did/help, I will check the site in the morn.
 

TheHmmmm

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Rechecked the post. Just to make sure I understand it properly: on top one puts the number of terms in a series (derived via the arithmetic forumla/methodology you showed), on the bottom goes the index number, on the immediate right goes the common difference - which is the difference between every term in the series, and is then multiplied by n...which is what? ~confusion~ Thank you BAP for catching what you did/help, I will check the site in the morn.

The common difference is what you add/subtract to get to the next term.

Technically, the formula looks like this:

Formula=D(n)+X
D=common difference
X=(the first term)-(the common difference)

So for the example series you gave (2+4+6...+1000), you would get:

Formula= 2n+(2-2)
or just 2n
 

Synthesis

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If any are kind enough to aid me once more with understanding my Algebra homework...

Objective: Write the first three terms of the infinte geometric series satisfying the given condition.

I. t1 = 8, S = 12

II. r = -1/3, S = 30

t1= first term in the series
S = sum of the series
R = rate between each term in the series

If anyone could drop a hint or two as to how I'd go about solving this, I'd appreciate it. At first I thought I could just find a second term in the first series, but I do not have a rate to go by; and in the second series listed I have no first term to go by in order to apply the given rate. I'm assuming one would have to turn these into a sigma notation eventually to derive a process of finding the next term? ~confusion~ :(
 

TheHmmmm

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If any are kind enough to aid me once more with understanding my Algebra homework...

Objective: Write the first three terms of the infinte geometric series satisfying the given condition.

I. t1 = 8, S = 12

II. r = -1/3, S = 30

t1= first term in the series
S = sum of the series
R = rate between each term in the series

If anyone could drop a hint or two as to how I'd go about solving this, I'd appreciate it. At first I thought I could just find a second term in the first series, but I do not have a rate to go by; and in the second series listed I have no first term to go by in order to apply the given rate. I'm assuming one would have to turn these into a sigma notation eventually to derive a process of finding the next term? ~confusion~ :(

There's a formula for this

S= T1/(1-r)
Where -1<r<1

If r does not satisfy those boundaries, then the sum does not exist.
 

Synthesis

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Hm. I see; to find the rate one would subtract the 1 from S, then multiply by r, then solve for t1 (isolation of the variable(s))? My understanding of this material is that the forumlae are highly adaptable and can be changed to fit the user's needs. Thank you once more for your help TheHmmm
 

TheHmmmm

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Hm. I see; to find the rate one would subtract the 1 from S, then multiply by r, then solve for t1 (isolation of the variable(s))? My understanding of this material is that the forumlae are highly adaptable and can be changed to fit the user's needs. Thank you once more for your help TheHmmm

This principle is true for all of algebra and that's why I like it. You can twist variables around to fit situational needs.

To see the work for problem 1:

12=8/(1-r)
1-r=8/12
r=(1/3)

8, 8/3, 8/9

Problem 2:

30=x/(1-(-1/3))
1-(-1/3)=x/30
(4/3)(30)=x
x=40
40, (-40/3), (40/9)
 

Synthesis

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Oh. Haha, that's much more simple than I thought...seems I was overanalyzing and making it a bit more complex than it was. *shakes head bemusedly* D'oh moment, apparently. Thanks Hmmm :)
 
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