Looking for some math help

[Aylee]

Yes I know this is wrong to do but taking 18 hours + working full time I just dont have the time to sit down and do this on my own. Combined with the fact that we only have 300 points in the class ( 2 test and a final) I just need the easy points.

Taking Pre calculus this semester and i 100% suck at it. Looking to hire someone to help me get through this class. Basically All I need is someone to give the questions to, you answer them with steps. There are 3 sets of questions ~14 sets of questions per set. I am willing to pay if you want (paypal please). Questions are stuff like: Derivatives, evaluate limits, cost/profit functions, graphing derivatives, integrals, marginal cost. Please PM me your price if your interested.

[BerenTebogo]

Things you won't do if you do this: Pass the class.

Seriously, if you think you're gonna get through exams on limits, derivative, and integrals without taking the time to sit down and properly go through the exercises and learn all the little tricks you've either taken the class before (so why not do the easy homework?) or you're delusional.

Also, lol posting this here. If I could I'd screen cap this and send it to your professor.

[Xajii]

Also, lol posting this here. If I could I'd screen cap this and send it to your professor.

elphaba pls

[BerenTebogo]

elphaba pls

I look damn good in a plaid skirt.

[Aylee]

Things you won't do if you do this: Pass the class.

Seriously, if you think you're gonna get through exams on limits, derivative, and integrals without taking the time to sit down and properly go through the exercises and learn all the little tricks you've either taken the class before (so why not do the easy homework?) or you're delusional.

Also, lol posting this here. If I could I'd screen cap this and send it to your professor.

We don't have easy homework. Basically how he lays out the class is 3 test. If you don't answer a question on the test 100% perfect you get 0 points. Then he gives you the test back and lets you rework it for 6 points/question (10 if you did it perfect first time). I don't want to take the risk of working them out and probably doing them wrong and losing out on those points (and some I just have 0 clue on). Figured it was a lot easier to just pay someone to work the ones i missed then post here constantly asking for advice on where I went wrong on like 20 questions.

[Xajii]

Figured it was a lot easier to just pay someone to work the ones i missed then post here constantly asking for advice on where I went wrong on like 20 questions.

Because these are clearly your only two options.

[Aylee]

Because these are clearly your only two options.

At this point they pretty much are. No one else around me is good with this type of math to help me. The teacher help he just gives you extremely vague hints. tutoring is closed for spring break. idk what else to do, so if you have any suggestions im up for trying them.

[BerenTebogo]

http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx Paul's Online Math Notes

http://www.youtube.com/watch?v=riXcZ...8E3C9C7791BD2D Khan Academy Pre-Calc playlist.

http://kickass.to/calculus-early-tra...-t3494019.html

http://www.amazon.com/Mead-Filler-Pa.../dp/B00005C55O

http://www.amazon.com/Dixon-Ticonder...G9H82APTEGYKAE

[Salodin]

What you need help with dude, you got my number and I'm taking calc I now. I'll help ya out bro.

[Aylee]

with saldoins help I managed to work through most of them... Hopefully they are right. The one question neither of us could figure out is:

Let C(x) = 200 +.3x +0.001x^2 be the cost of producting x widgets. Let the revenue be given by R(X) = 2x. What is the profit function? What is the marginal profit? Approximately how much more profit would you receive for selling 20th widget?

Profit function is P(x) = R(X) - c(x) I know that
Marginal Profit is where im clueless
For 20th widget Would it be:

2x(200+.3x+0.001x^2)
c'(x) = 2(.3x +.002x)
c'(19) = 2(.3(19) + .002(19))
= 2(5.7 + 0.38)
=11.476?

[Xerlic]

Iirc, Marginal Profit is just the derivative of the Profit function.

Once you have that, you solve it for x=19 to find the profit for selling the 20th widget.

I'm not sure if the formula that you came up with is correct, but I can tell you that the derivative of 2x(200+.3x+0.001x^2) is not 2(.3x +.002x). You have to use the product rule.

[Jefe]

No product rule needed because they are using the derivative as a function (read: limit). But yes, the marginal profit can be found by the derivative of the profit and plugging w/e is the fixed number.

[Aylee]

Profit Function is P(x) = R(x) - C(x)

So would Marginal Profit be:

p(x) = 2x - (200 + .3x + .001x^2) '2x is the Revenue Function - (200+ .3x + .001x^2) is the cost function'
P'(x) = 2 - (.3 + .002) 'Examples from class just use the power rule to take the derivative, do I need to use the power rule instead?'
p'(x) = 1.698

Then Profit for 20'th would be:

c'(19) = 2 - (.3 + .002(19))
c'(19) = 2 - .638)
C'(19) = 1.362


Does this satisfy the question? Or am i still making a mistake?

[Jefe]

The derivative is correct, so plug in whatever value of x you want to find. Yes, you have to use the power rule when you see x raised to the n power: x^n => e.g. y = x^2, so dy/dx = x^2 = 2x.

Remember that the derivative is just a shortcut of this:

Spoiler: show

[f(x+h)-f(x)]/h as h ->0.
Then: [(x+h)^2-(x)^2]/h = [(x^2+2xh+h^2)-(x^2)/h] = h(2x+h)/h = 2x+0 = 2x.

Did your class skip the fundamentals or something? >_> Anyway, good luck!

[Xerlic]

Then Profit for 20'th would be:

c'(19) = 2 - (.3 + .002(19))
c'(19) = 2 - .638)
C'(19) = 1.362

You might want to double check your arithmetic.

[Aylee]

You might want to double check your arithmetic.

Yeah idk where that 6 came from

[Cadsuane]

There's two possibilities here: either you are a person for whom the simplest most menial task is impossible, or you need to man the fuck up and put a non-zero degree of effort into your baby shit easy math course. Either way, don't be a dickmilk and cheat.

[Acevalefor]

More math help? This one should be simple but something is wrong for me. It is a story problem in basic algebra:
You are choosing between two plans at a discount warehouse. Plan A costs $300 annually and you pay 70% of the manufacturer's list price. Plan B costs $40 annually and you pay 90% of the list price. How much merchandise would you have to purchase per year to pay the same amount on both plans? How much does each plan cost?

I have: Let y = annual cost. Let x = Manufacturer's List Price

Plan A Cost: y=.7x+300
Plan B Cost: y=.9x+40

Need to spend about $1400/year for the plans to cost the same amount.

The whole problem is marked wrong. We are not supposed to graph it, just solve the system and find the intersect. I did that...I thought. What did I do wrong? I will be asking the instructor tonight at class, but I wanted feedback from the geniuses here first.

[Blubbartron]

You failed to math.

.9x + 40 = .7x + 300
.2x = 260
x = 1300

[quannum]

wolfram alpha can check your math for you. you can type in "y=.7x+300; y=.9x+40" to the search box and it will solve it

https://www.wolframalpha.com/input/?...++y%3D.9x%2B40

it also knows the how to draw the bat symbol, http://www.wolframalpha.com/input/?i=batman+equation