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Discussion Starter · #1 ·
looking for advice, reaction, etc

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I really don't know what to do about my math situation.

The short version of it is, I am horrible at it.

It took me three years to get through two years of it in high school, and five attempts to pass the Regents. Even then, I only managed a 57, which singlehandedly prevented me from getting a highly-thought-of Regents diploma.

Then in college, I thought I only needed one math class ... and my advisor never told me different. But then I am hit with the bombshell that I need that one, plus two pre-reqs, because I didn't get a 65 or better on my Math Regents.

Well, I am in a low-level algebra class now ... and is likely going to either drop it or fail it. My first tutor quit on me, and then I tried to get one through school, and they told me I didn't meet TRIO requirements. To add to that, I'm just clueless in most of the lectures...

I don't know if this hinders me or not, but the way he teaches his class isn't really helpful for someone like me. I have math for three 50 minute periods a week. Each class, he'll teach for 25-30 minutes (though it's been as short as 15 min once) and the rest of the time is spent on group problems (usually 3-4) ... that are supposed to heighten your understanding while working with others.... well, not only do i not work well in groups (no shock...) ... I never get anything out of it. Most people are in a hurry to leave, so the one person who knows what he/she is doing does the work and the rest of us put our name on the paper .... sometimes i wonder if i would have fared better if i had a professor who used the entire class time to lecture/do examples.....

now if i fail this, i wont get my 2 yr til spring 07 .... when i should be getting my bachelors..... (besides math, in which my 9th grade math teacher said i was "hopeless" in, i am a 3.0 student.... but i just CANT do the math)
 

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Its hard to explain math online, but there are tons of smart people that would do their best to help you here. What parts are giving you the most trouble?
 

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If you're talking about regents, I assume you're from new york.
I was horrible at math as well. I managed to pass Math I regents, but failed II and III. But luckily for me, passing the regents was not a requirement for graduating. I miss the cutoff by one year =).

I also got lucky in my two required speech credits. I took two classes where the teacher did not require us to do any presentations.
Intro to theatre (we didnt actually have to DO theatre) and some shakespeare class where all we did was read.

But as far as the regents go, you can always get a GED. I don't think GEDs are worse than diplomas since a few of my friends who got GEDS ended up going to good schools in SUNYs.
 

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Discussion Starter · #4 ·
GED

Not sure why I would want to go for a GED when I'm already in college, but...

Here's 4 that are giving me big problems -

"Write an equation of the line that passes through the point and had the specified slope. Sketch the line."

(0, 10), m = negative 1/4
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"Use the point-slope form to write an equation of a line that passes through the oiunt and has the specified slope. Sketch the line."

(0, 2), m = 3/5
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"Write the point-slope form of the equation of the line."

(3, 1), m = 3/2
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"Write an equation of the line that passes through the points. When possible, write the equation in slope-intercept form."

(6, -1), (3, 3)
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"Write an equation of the line passing through the points"

(3, 5), (1, 6)
 

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Do you know the equation y=mx+b? and do you know how to get m (the slope) when given (x, y), (x, y)?

Here's what I've learnt since reentering the world of math: Math is doable as long as you do the homework (do EVERY question at the end of the section/chapter). DO NOT rely on your teacher to teach you math. I find that math is less confusing when I study the material on my own at my own pace (just make sure you understand the material before class). Textbooks are pretty good with giving you step-by-step instructions/examples of a particular type of problem. Study guide helps too.
 

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The first on you just need to put it in y = mx + b form. They gave you the m. B is what y equals when x is zero, The coordinate has a x as zero, so b would 10. The equation would be y = -1/4(x) + 10.

Another way would be to substitute x, y, and m into y = mx + b and solve for b.



Im gonna have to look up point slope form to remember it better, but its going to be pretty much the same.

The equation is y = m(x - X) + Y, where X and Y are the point that you know. Its going to pretty much look like y = mx + b when you're done substituting.
 

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Discussion Starter · #8 ·
re

He puts his lectures online, and I often look at them, but they never do me any good. I'm that dense when it comes to this crap.
 

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Re: GED

rdf8585 said:
"Write an equation of the line that passes through the point and had the specified slope. Sketch the line."

(0, 10), m = negative 1/4
m = rise/run
y = mx + b
10 = -(1/4)0 +b
10 = b

y = -(1/4)x + 10 -> if x = 0, y = 10; then plug in different X values to get the Y value.

"Use the point-slope form to write an equation of a line that passes through the oiunt and has the specified slope. Sketch the line."
(0, 2), m = 3/5
m = (y2-y1)/(x2-x1)
y = mx + b
(3/5) = (y-2)/(x) -> cross-multiply
3x = 5(y-2)
3x = 5y - 10
3x + 10 = 5y
(3/5)x + 2 = y -> y = mx + b; y = (3/5)x +2

"Write the point-slope form of the equation of the line."

(3, 1), m = 3/2
y = mx + b -> 3/2 is m, use this as a check
where m = (y2-y1)/(x2-x1)

let x1 = 3, y1 = 1 (starting point = x1, y1)
m = rise/run
m = (y2 - y1)/(x2 - x1)
(3/2) = (y - 1)/(x - 3) cross-multiply
3(x - 3) = 2(y - 1)
3x - 9 = 2y - 2
3x - 7 = 2y
(3/2)x - (7/2) = y -> y = mx + b -> y = (3/2)x + (7/2)

Check with the direct method: y = mx + b -> solve for b
1 = (3/2)3 + b
1 = (9/2) + b
(2/2) = (9/2) + b
-(7/2) = b
therefore y = (3/2)x - (7/2)

"Write an equation of the line that passes through the points. When possible, write the equation in slope-intercept form."

(6, -1), (3, 3)
First thing you are given is the two points -> solve for slope m
m = (y2-y1)/(x2-x1); let x1 = 6, y1 = -1, x2 = 3, y2 = 3 from the points given.
m = (3 - (-1))/(3 - 6)
m = (3 + 1)/(3 - 6)
m = (4/-3)
m = -(4/3) take either point, use y = mx + b to solve for b.
y = mx + b <- (3, 3) or (6, -1) -> y = mx + b
3 = -(4/3)3 + b -1 = -(4/3)6 + b
3 = -4 + b -(3/3) = (-24/3) + b
7 = b -1 = -8 + b ----> 7 = b!

Put the pieces together, leaving x and y as variables ("for any X and Y")
y = mx + b
y = -(4/3) + 7

"Write an equation of the line passing through the points"

(3, 5), (1, 6)
Same deal -> you are given two points, solve for the slope between them.
m = (y2-y1)/(x2-x1) -> let x1 = 3, y1 = 5, x2 = 1, y2 = 6
m = (6 - 5)/(1 - 3)
m = -(1/2) -> now plug either point into y= mx + b to solve for b.
y = mx + b <- (3, 5) (1, 6) -> y = mx + b
5 = -(1/2)3 + b 6 = -(1/2)1 + b
5 = -(3/2) + b 6 = -(1/2) + b
(10/2) = -(3/2) + b (12/2) = -(1/2) + b
(13/2) = b (13/2) = b

Plug back into y = mx +b for any x and y
y = -(1/2)x + (13/2)

Whew! Boy, that took a while to type out! I hope this makes sense. Fractions...well, I did the best I could with them! :lol
 

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Zephyr said:
And as far as sketching goes, it's supposed to be easiest from slope-intercept form
I think it's easiest if you just find the x and y-intercepts. Then you have two points, hence a line.
Which one do you use y=mx+b to graph? I think that one's easier.
 

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geek said:
Zephyr said:
And as far as sketching goes, it's supposed to be easiest from slope-intercept form
I think it's easiest if you just find the x and y-intercepts. Then you have two points, hence a line.
Which one do you use y=mx+b to graph? I think that one's easier.
Using y=mx+b, the x intercept would be -b/m and the y intercept would just be b, so there you go. Your two points are (-b/m,0) and (0,b).
 

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I find a good way to look at math is to see how it applies to real life situations.

Most math books will give real life examples to put some equations into perspective.

The slope is nothing more than a rate. For example, if you walk 10 miles in 2 hours, the rate will be 10/2 = 5 miles per hour. Hence y2-y1 is the total distance traveled (10) and x2-x1 is the total time traveled. Dividing y2-y1/x2-x1 is the same as saying distance/time. Which is a rate, which is the slope.

Since you travel at a certain rate, multiplying that rate by the time should give you the total distance you travel. For example 5 miles/hr (m) x 2 hr (x) = 10 miles. B in y=mx+b is simply the y intercept (The position you are in when x (time) equals 0)

So if you are traveling 5 miles an hour and start off at a distance of 3 miles, the equation would be y=5x + 3
 
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