/sci/ - Science & Math

[Anonymous No.3117271]

What should I do? I'll keep this shit short:
>Going into sophomore year
>Took easy courses to stay on Dean's List
>Just this summer decided on going to med school/pharm school/dent school (not sure which yet)
>Need to complete pre-reqs for all (most of which are very similar)
>Some of the schools I'm looking at don't accept CC credits when completing pre-reqs
>My college is retarded and has intro to bio and intro to chem both on MWF so I'd have bio from 9:05-10:00 and chem from 10:10 to 11:05... (Is this a terrible idea to have both these classes so close to together on the same day [I'm thinking about future exams]?
>CC deadline closes in a few days... Would taking a pre-calc class be a good idea even though I did shit poor in high school algebra (didn't exactly try either)? Or should I take intro to algebra or college algebra?

Thanks to anyone who gives some suggestions...

[Anonymous No.3117307]

.

[Anonymous No.3117327]

Well...you're asking 4chan for advice.

I would pick one of the science courses and take liberal arts math.

[Anonymous No.3117361]

>>3117327
Yeah I was originally thinking about that but the math course I have to take before calc is only available on MWF and is at 8 am lol so at 8:55 I'd get out and then I'd have bio at 9:05... I guess it would be better in the sense that I wouldn't be taking two sciences?

[Anonymous No.3117367]

two difficult classes back to back is not a problem. suck it up bitch.

Do not take precalc if you can't do algebra.

[Anonymous No.3117383]

>>3117367
Jajaja okay

[Anonymous No.3117393]

Pre-med? Really

[Anonymous No.3117394]

>mfw I am mixing up MWF

So I would take that math course followed by biology. Biology is easy. That math course will probably be moderately difficult if you didn't pay attention to algebra in high school.

Chemistry will fuck you over no matter what level of math you're at. However, you stand a better chance if you know at least some algebra because there is a shit ton of algebraic manipulation in basic chem I.

Also word problems: word problems everywhere.

[Anonymous No.3117432]

Oh and an easy test.

1) Tell me the quadratic equation and why you use it (giving the conditions)
2) Tell me how to complete the square (giving the initial conditions)
3) Tell me what you know about imaginary numbers
4) Give me the equation of a line that passes through (-4,2) and (1,3)
5) Tell me the property of exponents (multiplication, addition) and logarithms (same thing as exponents)
6) Factor <span class="math">\displaystyle (x-2)^2[/spoiler]

Everything barring imaginary numbers is useful in calculus I and if you can't explain these all, you need to take basic algebra.

[Anonymous No.3117446]

>>3117394
Just realized that the math lab is at the same time as the biology lab so that math course is out of the picture lol. So I'm thinking bio+chem back to back might be my only option (along with a math course over the summer), unless I want to take physics instead of chem but that is probably significantly more difficult.

These are the CC math courses I can choose from:
https://my.smccme.edu/ICS/default.aspx?portlet=Course_Schedules
It goes on for a few pages..

[Anonymous No.3117493]

Bio and chem are doable. Bio is memorization and chem is quite a bit of application. Chem labs were harder as well.

[Anonymous No.3117517]

>>3117432
Lol shit

1) Hmm if I'm thinking of the right thing ax+bx+c = 0or something like that.. I don't remember what it is used for
2) I know this involves the quadratic equation and you turn the equation into something else lol
3)Uhh isn't that just a number that if you square it, it is negative
4) 1 over 5?
5)Multiplication? So like 2 squared x 2 squared = 16? Addition: 2 squared + 2 squared= 8? Logarithm is pretty much how many times does X need to be multiplied by itself to equal y?
6)x to the power of 2 + 4?

Yes I am this mathematically inept

[Anonymous No.3117522]

>poor in high school algebra
>med school
isn't calc II a pre req? And you're already in sophomore year?

[Anonymous No.3117550]

>>3117522
Med school is the least likely of the 3... As of now there is about a 15% chance that I actually take the MCAT

[Anonymous No.3117668]

>>3117517

1) Those are the initial conditions. The actual quadratic formula is used for finding x values in a quadratic equation that is not factorable.

<span class="math">\displaystyle x = \frac{-b \pm \sqrt{(b^2 - 4(a)(c)}}{2a}[/spoiler]

2) Not really. This is used to turn a certain factor in an equation into a perfect square. This comes up quite often in calculus I/II. It's another method for dealing with equations that were not otherwise factorable.

Ex. <span class="math">\displaystyle x^2 + 6x - 7 = 0[/spoiler]

Take half of 6 and then square it (9).

<span class="math">\displaystyle (x^2 + 6x +9) - 9 - 7 = 0[/spoiler]

The whole idea of this is that you add 9 and subtract 9 which is essentially adding zero.

<span class="math">\displaystyle (x + 3)^2 - 16 = 0[/spoiler]

3) Not quite. Imaginary numbers pop up when you get negatives under a radical. <span class="math">\displaystyle \sqrt{-1} = i[/spoiler] and <span class="math">\displaystyle i^2 = -1[/spoiler].

[Anonymous No.3117673]

4) 1 over 5 what? The equation of a standard line is <span class="math">\displaystyle y = mx + b[/spoiler] so you have two points and are able to calculate the slope (m) for it:

<span class="math">\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}[/spoiler]
<span class="math">\displaystyle m = \frac{3 - 2}{1 - (-4)} = \frac{1}{5}[/spoiler]

Great you have m. Now...

<span class="math">\displaystyle y = \frac{1}{5}x + b[/spoiler]

Find b. Pick one of the points and then solve.

<span class="math">\displaystyle 3 = \frac{1}{5}*1 + b[/spoiler]

<span class="math">\displaystyle b = 3 - \frac{1}{5}[/spoiler]

<span class="math">\displaystyle b = \frac{14}{5}[/spoiler]

<span class="math">\displaystyle y = \frac{1}{5}x + \frac{14}{5}[/spoiler]

5) Sort of. What I was hoping to see:

<span class="math">\displaystyle e^x*e^x = e^{x+x} = e^{2x}[/spoiler]

<span class="math">\displaystyle (e^x)^x = e^{x^2}[/spoiler]

<span class="math">\displaystyle \frac{1}{2}ln(x) - ln(x^2) = 0[/spoiler]

<span class="math">\displaystyle ln\frac{x}^{\frac{1}{2}}{x^2} = 0[/spoiler]

6) Factor: <span class="math">\displaystyle x^2 - 4x + 4[/spoiler]
<span class="math">\displaystyle \mathbb{You\;need\;to\;take\;college\;algebra}[/spoiler]

[Anonymous No.3117678]

4) 1 over 5 what? The equation of a standard line is <span class="math">\displaystyle y = mx + b[/spoiler] so you have two points and are able to calculate the slope (m) for it:

<span class="math">\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}[/spoiler]
<span class="math">\displaystyle m = \frac{3 - 2}{1 - (-4)} = \frac{1}{5}[/spoiler]

Great you have m. Now...

<span class="math">\displaystyle y = \frac{1}{5}x + b[/spoiler]

Find b. Pick one of the points and then solve.

<span class="math">\displaystyle 3 = \frac{1}{5}*1 + b[/spoiler]

<span class="math">\displaystyle b = 3 - \frac{1}{5}[/spoiler]

<span class="math">\displaystyle b = \frac{14}{5}[/spoiler]

<span class="math">\displaystyle y = \frac{1}{5}x + \frac{14}{5}[/spoiler]

5) Sort of. What I was hoping to see:

<span class="math">\displaystyle e^x*e^x = e^{x+x} = e^{2x}[/spoiler]

<span class="math">\displaystyle (e^x)^x = e^{x^2}[/spoiler]

<span class="math">\displaystyle \frac{1}{2}ln(x) - ln(x^2) = 0[/spoiler]

<span class="math">\displaystyle ln\frac{x}{x^2}^{\frac{1}{2}} = 0[/spoiler]


6) Factor: <span class="math">\displaystyle x^2 - 4x + 4[/spoiler]
<span class="math">\displaystyle \mathbb{You\;need\;to\;take\;college\;algebra}[/spoiler]