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April 21, 2014

April 21, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Pre Calculus **

Find an expression equivalent to sec theta tan theta/ sin theta.
*Saturday, December 14, 2013 at 9:12pm*

**Pre Calculus **

Find an expression equivalent to sec theta tan theta/ sin theta.
*Saturday, December 14, 2013 at 8:53pm*

**Calculus**

A spherical oil tank with a radius of 10 feet is half full of oil that weighs 60 pounds per cubic foot. Its top is 6 feet underground. How much work is needed to pump the oil out of the tank through a hole in its top? How do I solve this? Does the fact that it its top is 6 ...
*Saturday, December 14, 2013 at 8:13pm*

**Calculus (Please check my work.)**

p(x) =-4x^2+48x-18 P'(x)= -8x +48 0 = -8x +48 8x =48 x = 6 5)C(x) = 20x+600 C(31)-C(30) = 20(31)+600-([20(30)+600]) 1220- 1200= 20 C(x) 20x +600 C'(x) = 20
*Saturday, December 14, 2013 at 1:14pm*

**Calculus (Please check my work.)**

Also, for 5, is the marginal cost at 31 $186?
*Saturday, December 14, 2013 at 12:11pm*

**Calculus (Please check my work.)**

Thanks so much for pointing those things out! But I do have a question on #2? How'd you get the 6 for the maximum profit? I got the equation -4x^2+48x-18, but I couldn't get the 6. (It's probably some really stupid mistake. I make a lot of those. hahaha)
*Saturday, December 14, 2013 at 11:48am*

**Calculus (Please check my work.)**

#1. ok #2. Max R is at (15,450) -- ok P = R-C = -4x^2+48x-18 max P is at (6,126) avg cost is C(x)/x = 2x+12+18/x min avg cost is at (3,24) -- ok #3. ok #4. ok #5. ok, but you did not figure the marginal cost at x=31 #6. ok #7. ok, but you could note that the slope of R' is...
*Saturday, December 14, 2013 at 11:29am*

**Calculus (Please check my work.)**

This is my last chance to make at least a C in my AP Calculus course, so please tell me if I got these right or wrong so that I can fix them before it's due. Also, there was one I wasn't sure how to do, so if you can help me with that, it would be much appreciated! 1)...
*Saturday, December 14, 2013 at 9:25am*

**Calculus**

How do you show that the minimum average cost occurs when the average cost equals the marginal cost?
*Saturday, December 14, 2013 at 8:40am*

**Calculus**

since x^2-6x+9 = (x-3)^2 the denominator -> 0 while the numerator does not. So, you are correct that the limit is undefined.
*Friday, December 13, 2013 at 12:12am*

**Math (pre calculus)**

if so then (3x-1) ln 2 = (2x+1) ln 5 3 x ln 2 - 2 x ln 5 = ln 5 + ln 2 x (ln 8 - ln 25) = ln 10 x ln (8/25) = ln 10 x = ln 10 / ln (8/25)
*Thursday, December 12, 2013 at 6:56pm*

**Math (pre calculus)**

do you mean 2^(3-x) = 5^(2x+1) ??????
*Thursday, December 12, 2013 at 6:53pm*

**Math (pre calculus)**

Solve 2^3-x = 5^2x+1 Give an exact solution.
*Thursday, December 12, 2013 at 6:48pm*

**Math (pre calculus)**

p = po e^(kt) 10,000 = 10,000 e^0 if t = 0 at 1982 1995 - 1982 = 13 35,000 = 10,000 e^(k*13) ln(3.5) = 13 k k = 1.253/13 = .09637 300,000 = 10,000 e^(.09637 t) ln 30 = .09637t t = 35.3 1982 + 35.3 = 2017 almost there :)
*Thursday, December 12, 2013 at 6:47pm*

**Math (pre calculus)**

The population of a certain town grows from 10,000 in 1982 to 35,000 in 1995. At this rate, in what year will the population reach 300,000? Tip* Population growth formula: P = P(subzero)multiplied by e^(k)(t)
*Thursday, December 12, 2013 at 6:37pm*

**Calculus**

Often if you do not trust your answer, try numbers for example say x = 3.1 -9.61 / (9.61 - 18.6 + 9) = - 9.61 / .01 = -961 next try x = 3.01 but you get the idea. I think your answer is correct. The result is -9/0 which is undefined or infinite.
*Thursday, December 12, 2013 at 5:44pm*

**Calculus**

Lim x to 3 (-x^2)/((x^2)- 6x + 9) substitution i get -9/0 factoring i get same answer would that mean the answer is infinity? if not how could i prove that it isn't infinity
*Thursday, December 12, 2013 at 5:32pm*

**calculus**

Not sure how your "sign chart" method works, I will solve it the traditional way. x^2 - 2x +6 < 0 Just looking at the graph of y = x^2 - 2x + 6 we can tell that there is no solution. http://www.wolframalpha.com/input/?i=Plot+y+%3D+x%5E2+-+2x+%2B+6
*Thursday, December 12, 2013 at 5:01pm*

**calculus**

x^2-2x+6 < 0 no solutions, which means that x^2+x > 2x for all x http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2B6%2Cy%3D2x+where+x%3D-3..3
*Thursday, December 12, 2013 at 4:55pm*

**calculus **

use a sign chart to solve inequality. Express answers in interval notation. x^2+6<2x
*Thursday, December 12, 2013 at 4:46pm*

**Calculus PLEASE HELP**

r = Revenue c = cost p = profit = r-c u = number of units sold dr/du = marginal revenue dc/du = marginal cost so dp/du = dr/du - dc/du = marginal profit at peak profit, changer in profit for another unit sold is zero so 0 = dr/du-dc/du or dr/du = dc/du
*Thursday, December 12, 2013 at 7:37am*

**Calculus PLEASE HELP**

Show that the maximum profit occurs when the marginal revenue equals the marginal cost. I'm pretty sure this is supposed to be some sort of proof, but I have no idea how to even start it...
*Thursday, December 12, 2013 at 5:29am*

**Calculus**

Since P(x) is a cubic with negative leading coefficient, it becomes more and more negative. However, I'm not sure what a negative population means. equilibrium means dP/dt = 0, so t = 300 or 430 Hard to imagine a quadratic rate of growth. Usually it some kind of negative ...
*Wednesday, December 11, 2013 at 11:26pm*

**Calculus**

Assume that the rate of population growth of ants is given by the equation: dP/dt=.1P(1 - P/730)-(129000/7300) a. If P(0)=150, what happens to P as t gets very large? b. If P(0)=800, what happens to P as t gets very large? c. If P(0)=800, what happens to dP/dt as t gets very ...
*Wednesday, December 11, 2013 at 10:35pm*

**Calculus**

Show that the maximum profit occurs when the marginal revenue equals the marginal cost. I'm pretty sure this is supposed to be some sort of proof, but I have no idea how to even start it...
*Wednesday, December 11, 2013 at 8:21pm*

**Calculus**

as you can see from the graph, they do not intersect, so we simple take the integral from 1 to 2 area = ∫ e^x - x dx from 1 to 2 = [e^x - x^2/2] from 1 to 2 = e^2 - 2 - (e^1 - 1/2) = e^2 - e - 3/2
*Wednesday, December 11, 2013 at 7:19pm*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to three decimal places.) Between y = e^x and y = x for x in [1, 2]
*Wednesday, December 11, 2013 at 6:58pm*

**Calculus**

Here is a sketch http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+−+4x+%2B+1+and+y+%3D+−x%5E2+%2B+4x+−+5 Easy to find their intersection points: (1, -2) and (3, -2) effective height between x = 1 and x = 3 = (-x^2 + 4x - 5) - (x^2 - 4x + 1...
*Wednesday, December 11, 2013 at 6:57pm*

**Calculus**

thank you so much
*Wednesday, December 11, 2013 at 6:56pm*

**Calculus**

the curves intersect at x=1,3 so the area is ∫[1,3] (−x^2 + 4x − 5 )-(x^2 − 4x + 1) dx = ∫[1,3] -2x^2 + 8x - 6 dx = -2/3 x^3 + 4x^2 - 6x [1,3] = 8/3
*Wednesday, December 11, 2013 at 6:44pm*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Enclosed by y = x^2 − 4x + 1 and y = −x^2 + 4x − 5
*Wednesday, December 11, 2013 at 6:35pm*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Enclosed by y = x^2 − 4x + 1 and y = −x^2 + 4x − 5
*Wednesday, December 11, 2013 at 6:34pm*

**pre calculus**

X^2 - 3X - 7 = 0 This Eq cannot be factored. Use the Quadratic Formula: X = (-B +- sqrt(B^2-4AC))/2A X = (3 +- Sqrt(9+28))/2 X = (3 +- 6.08)/2 = 4.54, and -1.54.
*Wednesday, December 11, 2013 at 3:49pm*

**Calculus**

xy' + 1 = y^2 xy' = y^2-1 y' = (y^2-1)/x dy/(y^2-1) = dx/x arctanh(y) = log(x) or, log (1-y)/(1+y) = 2log(x)+c and you can massage that into exponentials and wind up with y = 1-e^(2cx^2) / 1+e^(2cx^2)
*Tuesday, December 10, 2013 at 11:33pm*

**Calculus**

Solve the differential and initial value problem: x(dy/dx) + 1 = y^2 y(1)=0 I tried using bernoulli and it didn't quite work.
*Tuesday, December 10, 2013 at 8:39pm*

**pre calculus**

better use the quadratic formula for this one.
*Tuesday, December 10, 2013 at 8:13pm*

**pre calculus**

You simply have to ask a question.
*Tuesday, December 10, 2013 at 6:46pm*

**pre calculus**

You need to write out the question. I have no idea what "pre calculus" this could be, maybe zeroes, however, that is standard fare for an algebra course.
*Tuesday, December 10, 2013 at 6:44pm*

**pre calculus**

You need to write out the question. I have no idea what "pre calculus" this could be, maybe zeroes, however, that is standard fare for an algebra course.
*Tuesday, December 10, 2013 at 6:42pm*

**pre calculus**

Is there a question here?
*Tuesday, December 10, 2013 at 6:41pm*

**pre calculus**

2k^2+5k-9
*Tuesday, December 10, 2013 at 6:30pm*

**pre calculus**

2k^2+5k-9=0
*Tuesday, December 10, 2013 at 6:26pm*

**pre calculus**

x^2-3x-7=0
*Tuesday, December 10, 2013 at 6:03pm*

**pre calculus**

x^2-3/4x+1/8=0
*Tuesday, December 10, 2013 at 5:58pm*

**pre calculus**

Pre-Calculus? I wonder what the question is. In a basic algebra course, you would be looking for zeroes, and probably get those by factoring. 2x^2+11x-21=0 (2x-3)(x+7)=0 x= 1.5, or x=-7
*Tuesday, December 10, 2013 at 1:04pm*

**pre calculus**

2x^2+11x-21=0
*Tuesday, December 10, 2013 at 12:39pm*

**calculus**

If the tent ends have side s and the tent has length t, the area is a = 2(√3/4 s^2) + 3st since the volume of the tent is 2.2, √3/4 s^2 t = 2.2, and so t = 8.8/(√3 s^2) If the cost of the floor is 1, then the cost of the whole tent is c = 2(√3/4 s^2)(1....
*Tuesday, December 10, 2013 at 6:06am*

**calculus**

Imagine making a tent in the shape of a right prism whose cross-section is an equilateral triangle (the door is on one of the triangular ends). Assume we want the volume to be 2.2 m3, to sleep two or three people. The floor of the tent is cheaper material than the rest: assume...
*Tuesday, December 10, 2013 at 1:01am*

**calculus**

d ( 1 / x ^ n ) / dx = - n * x ^ ( - n - 1 ) d ( 1 / x ) / dx = d ( 1 / x ^ 1 ) / dx = - 1 * x ^ ( - 1 - 1 ) = - 1 * x ^ - 2 = - 1 / x ^ 2 d ( 3x ) / dx = 3 d [ 3 x + 2 ( 37,500,000 / x ) ] / dx = 3 + 2 * 37,500,000 * ( - 1 ) / x ^ 2 = 3 - 75,000,000 / x ^ 2
*Monday, December 9, 2013 at 10:40pm*

**calculus**

Got it thanks anyways... I for some reason wasn't putting 3 over x^2 to simplify. dumb mistake.
*Monday, December 9, 2013 at 10:36pm*

**calculus**

Differentiate 3x+2(37,500,000/x) I know the answer is suppose to be 3(x^2-25,000,000)/x^2 but I'm not sure how to get there.
*Monday, December 9, 2013 at 10:27pm*

**calculus**

dy/dx = 2x at (a,a^2) , dy/dx = 2a equation of tangent: y - a^2 = 2a(x-a) at 2,1) 1 - a^2 = 2a(2-a) 1 - a^2 = 4a - 2a^2 a^2 - 4a + 1 = 0 let's complete the square: a^2 - 4a + 4 = -1 + 4 (a-2)^2 = 3 a-2 = ± √3 a = 2 ± √3 or let point of contact be...
*Monday, December 9, 2013 at 5:34pm*

**Calculus**

:)
*Monday, December 9, 2013 at 5:30pm*

**Calculus**

2x = -2y dy/dx dy/dx = -x/y at (3,4) dy/dx = -3/4 differentiate again, 2 = -2y (d^2 y/dx^2) + (dy/dx)(-2dy/dx) 2 = -2y (d^2 y/dx^2) - 2(dy/dx)^2 2 = -2(4)(d^2 y/dx^2) - 2(9/16) (d^2 y/dx^2) = (2 + 9/8)/-8 = -25/64 or y^2 = 25-x^2 y = (25-x^2)^(1/2) dy/dx = (1/2)(25-x^2)^(-1/2...
*Monday, December 9, 2013 at 5:25pm*

**Calculus**

Never mind, I got it!
*Monday, December 9, 2013 at 5:17pm*

**Calculus **

Y=5x^4 -24x^3+24x^2+17 is concave down for which interval? A) x<0 B) x>0 C) x<-2 or x>-2/5 D) x<2/5 or x>2 E) 2/5 < x < 2 I know you take the second derivative...but I keep getting stuck with something that won't factor right?
*Monday, December 9, 2013 at 5:10pm*

**Calculus**

If x^2=25-y^2, what is the value of [(d^2)(y)]/(dx^2) at the point (3,4)?
*Monday, December 9, 2013 at 4:54pm*

**calculus**

the tangent line to the curve y=x^2 at the point (a,a^2) passes through the point (2,1) find all possible values of a. Thank you so much!!
*Monday, December 9, 2013 at 4:28pm*

**CALCULUS!!!!!**

done, check your earlier post
*Monday, December 9, 2013 at 8:51am*

**Differential Calculus**

place ship B 65 miles east of ship A and mark them that way. Let the time passed be t hrs Draw a line BP , so that P is between A and B showing the distance traveled in those t hours. Draw a line downwards AQ showing the distance traveled by ship A. Join PQ to get the right-...
*Monday, December 9, 2013 at 8:50am*

**CALCULUS!!!!!**

At 9am, ship B was 65 miles due east of another ship, A. Ship B was then sailing due west at 10 miles per hour, and A was sailing due south at 15 miles per hour. If they continue their respective courses, when will they be nearest one another?
*Monday, December 9, 2013 at 6:44am*

**Differential Calculus**

At 9am ship B is 65 miles due east of another ship A. Ship B is then sailing due west at 10mi/h and A is sailing due south at 15 mi/hr if they continue in their respective course when will they be nearest to one another? and how near?
*Monday, December 9, 2013 at 6:05am*

**calculus**

presumably, he wants to use all of the fencing, if he wants as large a pen as possible... 2x+y=200 a = xy = x(200-2x) = 200x-2x^2 da/dx = 200-4x da/dx=0 when x=50 the pen is 50x100 In this kind of problem, the fencing is always divded equally among the lengths and widths.
*Monday, December 9, 2013 at 12:04am*

**Pre calculus**

I promise you that no physical harm will come to you when you use either one of these harmless procedures.
*Sunday, December 8, 2013 at 11:07pm*

**Pre calculus**

Why is it dangerous to use the law of sines to find an angle but is not dangerous to use the law of cosines? Please explain this thoroughly.
*Sunday, December 8, 2013 at 10:00pm*

**calculus**

a farmer wants to make a rectangular pen from 200 feet of fencing. he plans to use a 100 foot wall along with some of the fencing to make one side of the pen. find the dimensions of the pen that will make the enclosed area as large as possible.
*Sunday, December 8, 2013 at 9:17pm*

**Calculus PLEASE HELP**

You are correct, both in your calculations and your understanding of the concepts. C is increasing when C' > 0 C' = 6x^2-42x+36 = 6(x^2-7x+6) = 6(x-1)(x-6) C'>0 for x<1 or x>6 The marginal cost is C', so C' is increasing when C">0. C&...
*Saturday, December 7, 2013 at 6:26pm*

**Calculus PLEASE HELP**

Given that C(x)=2x^3-21x^2+36x+1000 is a cost function, determine the intervals for which the cost is increasing. Determine any intervals for which the marginal cost is increasing. Marginal cost is when you derive the cost function, correct? So do you do the second derivative ...
*Saturday, December 7, 2013 at 6:23pm*

**Calculus**

Given that C(x)=2x^3-21x^2+36x+1000 is a cost function, determine the intervals for which the cost is increasing. Determine any intervals for which the marginal cost is increasing. Marginal cost is when you derive the cost function, correct? So do you do the second derivative ...
*Saturday, December 7, 2013 at 4:01pm*

**Calculus (check my work)**

no, R(11) is just that. No derivative. It's showing you how the marginal revenue can be approximated by noting the increase in revenue from one point to another. R(11) = 759 R(10) = 700 ∆R = 59, which is pretty close to R'(10)
*Saturday, December 7, 2013 at 3:52pm*

**Calculus**

max R when R'=0 R' = -2x+400 R'=0 when x=200 In a way, you are on the right track. R is a parabola, whose vertex is midway between the roots. That is, at x=200. Using R' always works, though, even when R is not a nice easy function like a quadratic. This is a ...
*Saturday, December 7, 2013 at 3:49pm*

**Calculus**

P(x) = R(x) - C(x) P(x) =-x^2 +400x -x^2 -40x -100 P(x) -2x^2 + 360x -100 P'(x) = -4x +360 solve for x 0 = -4x + 360 R'(x ) -2x + 400 0 = -2x + 400 2x = 400 x = 200 R(200) = -(200)^2 + 400(200) -40000 + 80000 = 40000
*Saturday, December 7, 2013 at 3:25pm*

**Calculus**

P(x) = R(x) - C(x) P(x) =-x^2 +400x -x^2 -40x -100 P(x) -2x^2 + 360x -100 P'(x) = -2x +360 solve for x 0 = -2x + 360
*Saturday, December 7, 2013 at 3:12pm*

**Calculus**

Using the equations R(x)=-x^2+400x and C(x)=x^2+40x+100, find the Maximum Revenue
*Saturday, December 7, 2013 at 2:43pm*

**Calculus (check my work)**

Yes! Thank you! I do have another question, though. The second part of the problem says to "compare R(11) -R(10) with that result." Does that mean that I derive it before plugging in 11 and 10 or...?
*Saturday, December 7, 2013 at 2:36pm*

**Calculus (check my work)**

R(x) = -x^2 + 80x R'(x) = -2x + 80 R(10) = -2(10) + 80 = -20 + 80 = 60 Correct
*Saturday, December 7, 2013 at 2:32pm*

**Calculus (check my work)**

Find the marginal revenue at x=10 using the equation R(x)=-x^2+80x Is the answer 60?
*Saturday, December 7, 2013 at 2:04pm*

**Calculus**

Find the maximum revenue using the following equations: R(x)=-x^2 +400x and C(x)=x^2+40x+100. What I've done so far is use R(x) and solve for x, which got me 0 and 400. Is that how you're supposed to start it? What happens next? (I apologize; I'm really bad at ...
*Saturday, December 7, 2013 at 1:38pm*

**Calculus**

Ohhhh. I see what I did wrong in that second one! Thanks for all your help!(:
*Saturday, December 7, 2013 at 12:43pm*

**Calculus**

P(x) = R(x)-C(x) = -2x^2+360x-100 P'(x) = -4x+360 P'=0 at x=90 P(90)=16100 good work AvgCost A(x) = C(x)/x = x+40+100/x A'(x) = 1-100/x^2 A'=0 at x=10 A(10) = 60
*Saturday, December 7, 2013 at 12:37pm*

**Calculus**

if you can find the critical numbers, that is where the derivative is zero. That is the point of min/max of the function. Check some of the related links below; you will find such problems worked out. Also, a net search for minimum revenue and such will produce many examples ...
*Saturday, December 7, 2013 at 12:33pm*

**Calculus**

Check my work? The equations given were R(x)=-x^2+400x and C(x)=x^2+40x+100. We were asked to find the maximum profit and minimum average cost. (I'm having trouble remembering how to find a minimum and maximum, so please help with that, if you can!) For Max Profit, I got $...
*Saturday, December 7, 2013 at 12:31pm*

**Calculus**

I don't remember how to did min/max. I can get as far as the critical numbers...
*Saturday, December 7, 2013 at 12:25pm*

**Calculus**

revenue is price * quantity profit is revenue - cost avg cost is totalcost/quantity as usual, max/min is found via the derivative
*Saturday, December 7, 2013 at 11:50am*

**Calculus**

How do you find maximum revenue, maximum profit, and minimum average cost? Our teacher gave us a project to teach ourselves the section, but I can't learn very well even with her teaching, so I don't know how I'm supposed to do this...
*Saturday, December 7, 2013 at 11:39am*

**Calculus**

f undefined for x=0 f' = 2-6/x^2 f'=0 when x=±√3 I'll let you sort them out.
*Saturday, December 7, 2013 at 5:29am*

**Calculus- please help**

i cant help
*Saturday, December 7, 2013 at 12:48am*

**Calculus**

Consider the function f(x)=2x+6x^(-1). For this function, there are three numbers A<B<C which are either critical or not in the domain of the function. A= B= C=
*Saturday, December 7, 2013 at 12:36am*

**calculus**

The function crosses the origin so I see it as A = ∫ x(x^2+16)^(1/2) dx from 0 to 3 = [ (1/3)(x^2+16)^(3/2) ] from 0 to 3 = (1/3)(25)^(3/2) - 0 = (1/3)(125) = 125/3
*Saturday, December 7, 2013 at 12:22am*

**calculus**

find the area of y=x sqrt(x^2+16), bound by the x-axis and the vertical line x=3 I got 22.5 is that correct?
*Saturday, December 7, 2013 at 12:07am*

**Pre-Calculus**

Solve the equation 2x + 5y - 3z = -1. Write the general solution as a matrix equation. I did some guessing and ended up getting this much: [x]= [-1/2 _ _ ] X [1] [y]= [ 0 1 0 ] X [s] [z]= [ 0 0 1 ] X [t] however I still can't figure out the last two values in the first row
*Friday, December 6, 2013 at 7:39pm*

**Pre-Calculus**

I still don't really understand how to do it... I did some guessing and ended up getting this much: [x] [-1/2 _ _] [1] [y]= [0 1 0] X [s] [z] [0 0 1] [t] however I still can't figure out the last two values in the first row.
*Friday, December 6, 2013 at 7:00pm*

**Pre-Calculus**

(a) 4log5 = log625 (b) x = ln y (c) 10^x = -3
*Friday, December 6, 2013 at 5:55am*

**Pre-Calculus**

(2 5 -3) (x y z)T = -1 The transpose is for readability, since it has to be a column vector.
*Friday, December 6, 2013 at 5:53am*

**Pre-Calculus**

okay i put it wrong I'm sorry the question said Convert to the Alternate form (Exponential INTO logarithm)
*Friday, December 6, 2013 at 1:34am*

**Pre-Calculus**

[x] [...] [1] [y]= [...] X [s] [z] [...] [t] to align it better I put dots in the blanks...and I think s and t are free variables that be be any real number
*Friday, December 6, 2013 at 1:31am*

**Pre-Calculus**

Solve the equation 2x + 5y - 3z = -1. Write the general solution as a matrix equation. [x] [ ] [1] [y]= [ ] X [s] [z] [ ] [t] I'm supposed to fill in that blank in the middle, but I don't understand how to solve this if I'm only given one equation. please explain
*Friday, December 6, 2013 at 1:28am*

**Pre-Calculus**

convert to the alternate form to exponential form. a) 5^4=625 b) e^x=y c) log.001=x I really need help on these problems i have no idea how to do it? pls help me i need to study for it on my exam.
*Friday, December 6, 2013 at 1:25am*

**Math - Calculus I**

Draw a side view of the situation (The cone will look like an isosceles triangle with a rectangle (the cylinder) sitting on its base and touching the sides) let the radius of the cylinder be r and let the height of the cylinder be h Look at the small right-angled triangle at ...
*Thursday, December 5, 2013 at 11:26pm*

**Math - Calculus I**

Optimization Problem: Find the dimensions of the right circular cylinder of greatest volume inscribed in a right circular cone of radius 10" and height 24"
*Thursday, December 5, 2013 at 9:36pm*

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