EVILNess needs algebra help hopefully for the last time

[EVILNess]

It's time for finals, and I am doing these practice tests like crazy for my final on the 28th. I have gotten 40+ of the possible questions right multiple times, but a few are hit or miss to say the least. If a question shows up on the actual final it will look identical except have different numbers.

Long story short I need someone to work some algebra for me so I can compare what I am doing to what is correct. ;)

http://i1223.photobucket.com/albums/dd508/treble0096/2.jpg

http://i1223.photobucket.com/albums/dd508/treble0096/1.jpg

Any help is much appreciated and I promise I will never ask again. Mainly because I will be done and will never have to do math in an academic setting ever again!

[TDK]

Well ln(e^9) is just 9.

ln (natural log) is the same as log base e of the value. So log base e of e^9 is 9 because e^9 = e^9

ln and e pretty much always cancel out. So like, if you have something like

5 = e^x

Solving for x, you would take ln of both sides.

ln5 = ln(e^x) = x

x = ln5

and vice versa.

5 = lnx solving for x, take e to the power of both sides

e^5 = e^(lnx) = x

x = e ^ 5

[Eltargrim]

Question 2:

|y-7.8| <= 1.5
y - 7.8 = +- 1.5
y = +-1.5 + 7.8
y = [6.3, 9.3]

[Toastburner B]

I did that wrong. The formula I was using was for dilutions, not mixing of two solutions. Sorry, my bad.

I can help with question one, I think.

I had to do similar problems with my studies. My teacher taught us a very simple formula for it:

Concentration1 X Volume 1 = Concentration 2 X Volume 2 (C1XV1 = C2XV2)

So, in the case of this problem you know the concentration and volume you have for one, but only the concentration for the other, so

EDIT: I plugged in the wrong numbers. See my next post for the right numbers.

9 X V1 = 60 X 7

Simplified down to

9(V1) = 420

Just solve for V1:

V1 = 420/9

V1 = 42.67 mL

Just round it out to what decimal place you need.

[Krylo]

The answer to all of them is five.

Except the last one.

That is C.

[EVILNess]

Thank you everyone. Much appreciated!

Except Krylo. Fuck that guy.

[Eltargrim]

Question 1:

Let x = volume of 9% solution
Let y = volume of 7% solution = 60 mL + x

All units are hereby neglected for typing convenience

2*60 + 9*x = 7*y
120 + 9x = 7(x + 60)
120 + 9x = 420 + 7x
-300 = -2x
300 = 2x
x = 150

The pharmacist should add 150 mL of the 9% solution

Question 4:

S_n = (n/2) * (a_1 + a_n)
S_18 = (18/2) * (a_1 + a_18)
252 = 9*(a_1 + 31)
252/9 - 31 = a_1
a_1 = -3

a_18 = a_1 + (18-1)*d
31 = -3 + 17d
34 = 17d
d = 2

[Professor Smarmiarty]

Last question solved by elimination:
20x-4+3y+9=120
20x+3y=115

3x+3-2x+2y=18
x+2y=15

20x+3y=115
x+2y=15

40x+6y=230
3x+6y=45
37x=185
X=5

5+2y=15
2y=10
Y=5
So answer A with X=5 y=5

But the answers make no sene to me. Either I did something really wrong or they hoping other people do something really wrong.

[TDK]

I did that wrong. The formula I was using was for dilutions, not mixing of two solutions. Sorry, my bad.

If you do the problem in terms of the 2% solution, it would be a dilution problem. :I

[Professor Smarmiarty]

Also the real answer to the dilution problem- hit an undergrad till they do it for you.

[Eltargrim]

If you do the problem in terms of the 2% solution, it would be a dilution problem. :I

Not of the kind that his formula would work for :P

And Smarty, someone needs to teach the undergrads how to do it. Once.

[Pip Boy]

If the slope of the function f(x) can be determined on an interval by the formula [f(x+h)-f(x)]/[(x+h)-(x)], then the slope at an instantaneous point can be determined by the limit of the slope formula as "h" approaches 0. Therefore, if f(x)=x^2, then lim as h -> 0 [(x+h)^2-x^2]/(x+h-x) = lim h -> 0 (x^2+2xh+h^2-x^2)/h= lim as h -> 0 h(2x+h)/h= 2x+h. Now use direct substitution to insert 0 for "h" and see that the slope of the function x^2 is always 2 times its x coordinate value (2x). Stick that in your calculator and calculate it. Then burn it. Because its witchcraft.