Math

[Magus]

Not exactly sure where to ask this, but I need some help in math. (I haven't done math in so long)

Ratio's... Proportions. I don't even think I was taught this back in middle or high school so it's a little new to me.

Question. In the following proportion, if [i]y[/i] is larger than 8, what statement can be made about the size of [i]x[/i]?

y/4 = x/3

I'm just now getting the hang of ratios/proportions and then this book want to go ahead and kill me with questions like that.

[Ikillion]

[FONT="Trebuchet MS"][SIZE="1"]I'm pretty sure that x > 6 there.[/SIZE][/FONT]

[Nerdsy]

[quote name='Magus']Not exactly sure where to ask this, but I need some help in math. (I haven't done math in so long)

Ratio's... Proportions. I don't even think I was taught this back in middle or high school so it's a little new to me.

Question. In the following proportion, if [i]y[/i] is larger than 8, what statement can be made about the size of [i]x[/i]?

y/4 = x/3

I'm just now getting the hang of ratios/proportions and then this book want to go ahead and kill me with questions like that.[/QUOTE]

[color=deeppink]In that equation, if Y is large than 8, then x is larger then 6.

If you plug 8 into y, you'll find that you get 8/4, whcih reduces to two. Therefore, you must set x/3 equal to 2 and find the x.

x/3=2
(3)(x/3)=(3)(2)
x=6.

Since y cannot equal 8, x can't actually equal 6, but using the information we can find that if y is any higher than 8, x must also be higher than 6.

Ratios are basically division problems.[/color]

[Magus]

[quote name='Nerdsy'][color=deeppink]In that equation, if Y is large than 8, then x is larger then 6.

If you plug 8 into y, you'll find that you get 8/4, whcih reduces to two. Therefore, you must set x/3 equal to 2 and find the x.

x/3=2
(3)(x/3)=(3)(2)
x=6.

Since y cannot equal 8, x can't actually equal 6, but using the information we can find that if y is any higher than 8, x must also be higher than 6.

Ratios are basically devision problems.[/color][/QUOTE]
Wow, that's some good stuff right there. So I'm pretty much using the same stuff I would use for solving equations? Thanks a lot.

[Drizzt Do'urden]

hmmm, though my post can't help with your question at all. I do have a question for those that answered it, if y's limits are stated as GREATER than 8, how can you set Y?


Wouldn't X just = 3/4 Y?

[Allamorph]

[QUOTE=Drizzt Do'urden][I]f y's limits are stated as GREATER than 8, how can you set Y?


Wouldn't X just = 3/4 Y?[/QUOTE]
[FONT=Arial]It would, but that's not the last step for what you're asked to examine.

[B]y[/B]'s limits are set by the statement [B]y > 8[/B], since that's what's you're asked to assume with the 'if' statement.. This means that [B]y[/B] could be any number from 8 to infinity. The only definite concept you can bring out of this is that the lower boundary of [B]y[/B] is 8. [B]y[/B] can be 8.01, 8.001, 8.0001, etc., as long as it does not actually make it to 8.

But that's ridiculous to think about. So we [U]temporarily[/U] come at the problem another way; namely, 'if [B]y[/B] = 8, what is [B]x[/B]?'. Solving this equation will give us 's lower boundary, which is 6 . . . but since [B]y[/B] is the set of all numbers greater than 8, [B]x[/B] has to be the set of all real numbers greater than 6.

And you still don't know either of the suckers. Ain't that fun? :animesmil[/FONT]

[Vicky]

[SIZE=1]Oh my god O_O...

You lost me at 'proportion'.[/size]

[Roxie Faye]

[quote name='Allamorph'][FONT=Arial]And you still don't know either of the suckers. Ain't that fun? :animesmil[/FONT][/QUOTE][color=#9933cc]lol. You are hilarious.

I'm also mildly proud of myself that I can still answer the question although I haven't taken math in a while... *_*

If I could minor in math at college I would, just for the hell of it.[/color]

[Magus]

[quote name='Vicky'][SIZE=1]Oh my god O_O...

You lost me at 'proportion'.[/size][/QUOTE]
Same thoughts exactly when I was given the assignment (ratio actually). Apparently, the teacher teach us just enough to cover other stuff... (But leave off other types of problems that we don't cover at all..) They kill me with this "you should already know it" stuff...

We're covering shapes now.... That length x width x height... crap... (PIE x whatever) I think I'm gonna die.

[Nerdsy]

[quote name='Magus']We're covering shapes now.... That length x width x height... crap... (PIE x whatever) I think I'm gonna die.[/QUOTE]


[color=deeppink]Simple volume problems aren't incredibly difficult once you get the hang of 'em. It's mostly memorization of formulae.

What level math class are you in? If you're finding the volume of shapes made by rotating graph around a line, that might be a bit trickier, heh.[/color]

[Magus]

Problem Solving... I think I take college math 1 afterwards.. I'm at ITT just incase anyone wanted to know.

And that's pretty much been my issue. Remembering the formulas.