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April 20, 2014

April 20, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Calculus**

An observer stands at a point P, one unit away from a track. Two runners start at the point S in the figure and run along the track. One runner runs 2 times as fast as the other. Find the maximum value of the observer's angle of sight è between the runners. Thank ...
*Saturday, November 23, 2013 at 10:31pm*

**Calculus**

just as a check, the distance from point P(h,k) to the line ax+by+c=0 is |ah+bk+c|/√(a^2+b^2) In this case, that is |24+5+1|/√(36+25) = 30/√61 Hmmm. What went wrong above? Oops. Forgot to multiply -6 by 6
*Saturday, November 23, 2013 at 7:17pm*

**Calculus**

Let the point be P(x,y) , point (-4,1) be A Since the shortest distance must be where AP is perpendicular to the given line. slope of given line = 6/5 so slope of AP is - 5/6 equation of AP: y-1 = (-5/6)(x+4) times 6 6y - 6 = -5x- 20 5x + 6y = -14 , #2 6x - 5y = -1 , #1, the ...
*Saturday, November 23, 2013 at 3:53pm*

**Calculus**

Find the point on the line -6 x + 5 y - 1 =0 which is closest to the point ( -4, 1 )
*Saturday, November 23, 2013 at 3:22pm*

**Calculus Area between curves**

Actually your answer is correct if it's indefinite integral. :) Now that you got the answer, we can now evaluate it at the given bounds: -(1/3)(8-2x)^(3/2) , at x = -7 to 0 = -(1/3)(8-2(0))^(3/2) - [-(1/3)(8-2(-7))^(3/2)] = -(1/3)(8)^(3/2) - [-(1/3)(22)^(3/2)] = (22/3)*(22...
*Friday, November 22, 2013 at 8:23pm*

**Calculus Area between curves**

Evaluate the definite integral: sqrt(8-2x) lower limit=-7 upper limit=0 I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!
*Friday, November 22, 2013 at 8:05pm*

**Calculus**

Yes I figured it out thanks
*Thursday, November 21, 2013 at 11:16pm*

**Calculus Help!!!**

notice that one parabola opens up, the other down. their intersection ..... x^2 - c^2 = c^2 - x^2 2x^2 = 2c^2 x^2 = c^2 x = ± c ---- y = 0 in both they intersect at (-c,0) and (c,0) using symmetry area = 2∫(2c^2 - 2x^2) dx from 0 to c = 2(2c^2 x - (2/3)x^3) | from...
*Thursday, November 21, 2013 at 11:16pm*

**Calculus-Area between curves**

Thanks I figured it out my answer was 9, and it was correct
*Thursday, November 21, 2013 at 11:15pm*

**Kelly - Amy - Chrissy Calculus**

Why are you switching names ???? Follow the method I showed you in the other two questions of this type. Did you look at the solutions? Let me know how far you get.
*Thursday, November 21, 2013 at 11:08pm*

**Chris - Amy -Calculus Area between curves**

Why are you switching names ?? find the intersection; from the first: x = 3 - 3y into the 2nd: y^2 - (3-3y) = 1 y^2 + 3y - 4 = 0 (y+4)(y-1) = 0 y = -4 or y = 1 in x = 3-3y .... if y = 1, x = 0 ---->(0,1) if y = -4 , x = 15 , ---> (15, -4) Your sketch should look like ...
*Thursday, November 21, 2013 at 10:56pm*

**Calculus-Area between curves**

first of all reduce the equation to simplest form: y = 2√x y = 5 y + 2x = 4 Your sketch should look like a "triangle" with one side slightly curved see: http://www.wolframalpha.com/input/?i=plot+y%3D2*sqrt%28x%29+%2C+y%3D5+%2C+y%2B2x%3D4 we need ...
*Thursday, November 21, 2013 at 10:41pm*

**Calculus Help!!!**

Find c > 0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 13.
*Thursday, November 21, 2013 at 10:32pm*

**Calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. x+y^2=42, x+y=0
*Thursday, November 21, 2013 at 10:29pm*

**Calculus Area between curves**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2-x=1
*Thursday, November 21, 2013 at 10:22pm*

**Calculus-Area between curves **

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't ...
*Thursday, November 21, 2013 at 10:16pm*

**Calculus - Integrals**

dE/dt = 6,000(t+1)^-3/2 E(t) = -12000(t+1)^-1/2 + c 3000 = -12000+c c = 15000 E(t) = 15000 - 12000/√(t+1) E(15) = 15000 - 12000/4 = 12000
*Thursday, November 21, 2013 at 7:54pm*

**Calculus - Integrals **

The projected rate of increase in enrollment at a new college is estimated by dE/dt = 6,000(t+1)^-3/2 where E(t) is the projected enrollment in t years. If the enrollment is 3,000 now (t=0), find the projected enrollment 15 years from now.
*Thursday, November 21, 2013 at 5:51pm*

**Fundamental Theorm of Calculus**

That would be ∫[3,6] 3x^2 - 2x dx --------------------------- 6-3
*Wednesday, November 20, 2013 at 8:02pm*

**Fundamental Theorm of Calculus**

that would be ∫[1,5] 7/x^2 dx ------------------------ 5-1
*Wednesday, November 20, 2013 at 8:00pm*

**Fundamental Theorm of Calculus**

Find the average value of f(x)=7x^{-2} on the interval [1,5]
*Wednesday, November 20, 2013 at 7:59pm*

**Fundamental Theorm of Calculus**

Find the average value of f(x)=3x^2-2 x on the interval [3,6]
*Wednesday, November 20, 2013 at 7:36pm*

**Fundamental Theorm of Calculus**

since the curve has x-intercepts at ±√7/2, we have ∫[-√7/2,√7/2] 7-4x^2 dx = 7x - 4/3 x^3 [-√7/2,√7/2] = 14√7 / 3
*Wednesday, November 20, 2013 at 5:27pm*

**Fundamental Theorm of Calculus **

Use a definite integral to find area of the region under the curve y=7-4x^2 and above the x-axis. Thanks in advance!
*Wednesday, November 20, 2013 at 5:13pm*

**Calculus Fundamental Theorem**

Hello,? By calculus time you should be well aware that |x-13| ≠ x+13 |x-13| makes a big difference. For x<13, |x-13| = -(x-13) For x>=13, |x-13| = x-13 So, that means that we have to break the interval into two parts: ∫[0,22] x-13 dx = ∫[0,13] -(x-13) ...
*Wednesday, November 20, 2013 at 4:59pm*

**Calculus Fundamental Theorem**

Thank you!
*Wednesday, November 20, 2013 at 4:54pm*

**Calculus Fundamental Theorem**

∫[-√2,√2] (t+8)(t^2+3) dt = ∫[-√2,√2] t^3 + 8t^2 + 3t + 24 dt = 1/4 t^4 + 8/3 t^3 + 3/2 t^2 + 24t [-√2,√2] = 176/3 √2
*Wednesday, November 20, 2013 at 4:53pm*

**Calculus Fundamental Theorem**

Original problem was: absolute value x-13, because of absolute value I made it x+13
*Wednesday, November 20, 2013 at 4:52pm*

**Calculus Fundamental Theorem**

I tried that way, my answer was wrong
*Wednesday, November 20, 2013 at 4:51pm*

**Calculus Fundamental Theorem**

∫[0,22] x+13 dx = 1/2 x^2 + 13x [0,22] = (1/2 * 22^2 + 13*22)-(1/2 * 0^2 + 13*0) = 528
*Wednesday, November 20, 2013 at 4:50pm*

**Calculus Fundamental Theorem**

Evaluate the definite integral. function: (t+8)(t^2+3) with respect to variable t lower limit: -sqrt(2) upper limit: sqrt(2)
*Wednesday, November 20, 2013 at 4:49pm*

**Calculus Fundamental Theorem**

Evaluate the definite integral. function: x+13 with respect to variable x lower limit:0 upper limit:22
*Wednesday, November 20, 2013 at 4:47pm*

**Pre-Calculus**

wow that was a silly mistake I didn't see the x next to 500 and thought I could just subtract it from 1000 thank you!
*Wednesday, November 20, 2013 at 1:03pm*

**Pre-Calculus**

well, you should see right off that since 51x^2 is always positive, you can never have 51x^2 = -500. So, let's see what happened. 50x^2 + 1000 = 500x - x^2 51x^2 - 500x + 1000 = 0 Now just use the quadratic formula to find the roots.
*Wednesday, November 20, 2013 at 6:35am*

**Math (Calculus)**

without a diagram, I assume point A is an anchor in between the towers. If its distance from T1 is x, then the cable length is d = √(60^2+x^2) + √(20^2+(140-x)^2) to find the minimum, we want dd/dx = 0 That happens when x=105
*Wednesday, November 20, 2013 at 6:30am*

**Math (Calculus)**

Tower 1 is 60 ft high and tower 2 is 20 ft high. The towers are 140 ft apart. A guy wire is to run from point A to the top of each tower. [See a picture of this situation on page 274 of the textbook.] How many feet from tower 1 should point A be so that the total length of guy...
*Wednesday, November 20, 2013 at 2:30am*

**Pre-Calculus**

A piece of equipment has cost function C(x)=50x^2 + 1000 and its revenue function is R(x)= 500x - x^2, where x is in thousands of items. What is the least number of items that must be sold in order to break even? 50x^2 + 1000 = 500x - x^2 50x^2 + x^2 + 500= 0 51x^2 + 500 = 0 ...
*Wednesday, November 20, 2013 at 1:38am*

**Pre-Calculus**

Wow, I didn't realize that finding (x-2) was already finding the equation of the line. Thanks!
*Tuesday, November 19, 2013 at 11:00pm*

**Pre-Calculus**

did you notice that f(x) = (x-9)(x-2)/(x-9) ?? = x-2, x ≠ 9 So for all values of x except x=9 the given function reduces to f(x) = x-2 which is a straight line. when x = 9, the original gives you 0/0 which is indeterminate, while the reduced function gives you (9,7) So ...
*Tuesday, November 19, 2013 at 10:34pm*

**Pre-Calculus**

The graph of f(x)= (x^2 - 11x + 18)/(x-9) consists of a line and a hole. Find the equation of the line and the coordinates of the hole. I really have no idea how to do this. I tried factoring it and ended up with (x-2), but I don't know how that helps me. please help
*Tuesday, November 19, 2013 at 10:05pm*

**Calculus**

f" = 5x+2sinx f' = 5/2 x^2 - 2cosx + c f'(0)=3, so 0-2+c=3 c=5 so, f' = 5/2 x^2 - 2cosx + 5 f = 5/6 x^3 - 2sinx + 5x + c f(0)=3, so 0-0+0+c=3 c=3 so, f=5/6 x^3 - 2sinx + 5x + 3 I assume you can now find f(3)
*Tuesday, November 19, 2013 at 9:29pm*

**Calculus**

f" = -4sin2x f' = 2cos2x + c 2+c = 4 c=2 so, f' = 2cos2x + 2 f = sin2x + 2x + c 0+0+c = -6 c = -6 so, f = sin2x + 2x - 6
*Tuesday, November 19, 2013 at 9:26pm*

**Calculus**

Given f''(x)= -4sin(2x) and f'(0)=4 and f(0) =-6. Find f(pi/2)
*Tuesday, November 19, 2013 at 9:17pm*

**Calculus**

Consider the function f(x) whose second derivative is f''(x)=5x+2sin(x). If f(0)=3 and f'(0)=3, what is f(3)?
*Tuesday, November 19, 2013 at 7:21pm*

**12th Calculus**

help me!
*Tuesday, November 19, 2013 at 2:15pm*

**Pre-Calculus**

oh yeah thanks I forgot that the reciprocal had to be negative
*Tuesday, November 19, 2013 at 1:33pm*

**Pre-Calculus**

since the line is perpendicular, their slopes are negative reciprocals of each other, that means, they must be OPPOSITE in sign so first slope = -3/2 2nd slope = +2/3 y-2 = (2/3)(x-4) 3y-6 = 2x-8 3y = 2x - 2 y = (2/3)x - 2/3
*Tuesday, November 19, 2013 at 10:43am*

**Pre-Calculus**

What is the equation of the line passing through (4, 2) and perpendicular to the line passing through the points (9,7) and (11,4)? Here's what I did: (7-4)/(9-11) = -3/2 y-2= (-2/3)(x-4) y-2=(-2x/3) + (8/3) y = (-2x/3) + 14/3 but the answer key says that it's y=(2x/3...
*Tuesday, November 19, 2013 at 8:22am*

**Calculus**

If the printed area has width x and height y, xy = 386 The total area is thus a = (x+8)(y+4) = (x+8)(386/x + 4) = 4x + 418 + 3088/x max area is where da/dx=0 da/dx = 4 - 3088/x^2 so, da/dx = 0 when x = 2√193 Now just evaluate y.
*Tuesday, November 19, 2013 at 5:25am*

**Calculus**

if the rectangular window has width 2x and height h, then the area of the whole shape is a = 2xh + pi/2 x^2 You also know that x+2h+pi*x = 25, so solve for h and plug into the formula for area. The set da/dx = 0.
*Tuesday, November 19, 2013 at 12:33am*

**Calculus**

since F(1) = 0, we have -7 + 6/5 + c = 0 which you can plug in to evaluate F(2)
*Tuesday, November 19, 2013 at 12:29am*

**Calculus**

We need the whole area to be twice the area under the line y=b. y(1/2) = 1 y(√b/2) = b So, integrating along x, we need ∫[0,1/2] 1-4x^2 dx = 2∫[0,√b/2] b-4x^2 dx Now just evaluate the integrals and solve the polynomial for b.
*Tuesday, November 19, 2013 at 12:24am*

**brief calculus**

You're right.
*Monday, November 18, 2013 at 11:30pm*

**brief calculus**

x/(x − 6)^2 I get ln(x-6)- 6/(x-6)+ C but it isn't correct, what should the answer be?
*Monday, November 18, 2013 at 11:22pm*

**Calculus**

Write it as f(x) = 7x^-2 - 6x^-6 F(x)= -7x^-1 + (6/5)x^-5 + c F(2) = -7/2 + (6/5)/2^5 + c = ......
*Monday, November 18, 2013 at 10:19pm*

**Calculus**

let the side of the square be x let the side of the triangle be y 4x + 3y = 40 x = (40-3y)/4 height of triangle from the 30-60-90 triangle ratio = √3/2 y area of triangle = (1/2)(y)(√3/2 y) = (√3/4)y^2 area of square = [(40-3y)/4]^2 = (1600 - 240y + 9y^2)/16...
*Monday, November 18, 2013 at 9:20pm*

**Calculus**

Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 1 into two regions with equal area. (Round your answer to two decimal places.)
*Monday, November 18, 2013 at 8:45pm*

**Calculus**

Consider the function f(x)=(7/x^2)-(6/x^6). Let F(x) be the antiderivative of f(x) with F(1)=0. Then F(2) equals _____.
*Monday, November 18, 2013 at 8:16pm*

**Calculus**

A piece of wire 40 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much of the wire should go to the square to minimize the total area enclosed by both figures?
*Monday, November 18, 2013 at 8:08pm*

**Calculus**

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 25 ft, find the dimensions of the window so that the greatest possible amount of light is ...
*Monday, November 18, 2013 at 8:07pm*

**Calculus**

The top and bottom margins of a poster are 2 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 386 square centimeters, find the dimensions of the poster with the smallest area.
*Monday, November 18, 2013 at 8:06pm*

**Calculus**

10x−6y√(x^2+1)dy/dx=0 6y√(x^2+1) dy/dx = 10x 6y dy = 10x/√(x^2+1) dx 3y^2 = 10√(x^2+1) + c Since y(0) = 4, 3*16 = 10+c c = 38 now you have it.
*Monday, November 18, 2013 at 2:04pm*

**Calculus**

Solve the separable differential equation 10x−6ysqrt(x^2+1)dy/dx=0. with the initial condition y(0)=4.
*Monday, November 18, 2013 at 1:55pm*

**Calculus**

I did plot the curve on Wolfram and got full curve across 3pi. Curve length is also shown full at 3pi but period is shown as 6pi.I can't understand this.
*Monday, November 18, 2013 at 6:47am*

**Calculus**

since the curve is a parabola, there is no inflection point. Since #1 #2 do not provide any useful data, the other items are hard to figure.
*Monday, November 18, 2013 at 6:07am*

**Calculus**

No idea. Did you visit wolframalpha.com and look at the graph? If you enter arc length sin^3(x/3), x = 0 .. 3pi it will give you the arc length, but that's only for 1/2 period.
*Monday, November 18, 2013 at 6:04am*

**Calculus**

For calculating complete length of curve r=sin^3(theta/3) I got right answer on integrating the arclength integrand from 0 to 3pi. How could that be?
*Monday, November 18, 2013 at 12:48am*

**Calculus**

2a is correct the period of sin^3(x) is the same as for sin(x). SO, sin^3(x/3) has period 2pi/(1/3) = 6pi sure. sin(x) does just that. sin(pi)=0, but it takes 2pi to complete the period. in polar coordinates, sin(theta/2) goes through the origin twice in its period of 4pi. ...
*Monday, November 18, 2013 at 12:24am*

**Calculus**

1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after theta=0, the function ...
*Monday, November 18, 2013 at 12:18am*

**Pre-Calculus**

thanks...though I asked for (-3,-1).
*Sunday, November 17, 2013 at 10:06pm*

**calculus**

f is a function from R to R such that f(a) is rational when a is irrational and f(a) is irrational when a is rational . prove that f cannot be continuous
*Sunday, November 17, 2013 at 7:00pm*

**calculus**

Find a function f from R to R such that f is continuous at only one point?
*Sunday, November 17, 2013 at 6:52pm*

**Calculus**

Suppose that a population develops according to the logistic equation dP/dt = 0.06P−0.0001P^2 where t is measured in weeks. 1) The carrying capacity is . 2) The growth rate k is . Use your calculator to sketch a direction field for this equation. Sketch the solutions ...
*Sunday, November 17, 2013 at 5:27pm*

**Calculus **

ohaganbooks is offering a wide range of online books, including current best-sellers. a colleague has determined that the demand for the latest best selling book is given by q=(-p^2)+33p+9 (18<p<28) copies sold per week when the price is p dollars. can you help me ...
*Sunday, November 17, 2013 at 4:59pm*

**calculus**

time to go a distance d is d/v so the cost is (d/v)(kv^3) = dkv^2 Looks like k=0 makes cost=0. I suspect a typo.
*Sunday, November 17, 2013 at 4:42pm*

**calculus**

(a) time = distance/speed, so a trip of 1000km takes 1000/v hours. So, the cost is c(v) = (1000/v)(160 + 1/100v^3) (b) Now just find minimum cost where dc/dv = 0 dc/dv = 20(v^3-8000)/v^2 so minimum cost when v=20 (c) Since min occurs at v=20, if top speed is 16, then minimum ...
*Sunday, November 17, 2013 at 4:39pm*

**Calculus - Optimization**

if the expensive side is x and the other dimension is y, then the cost c is c = 4(x+2y) + 12x But, we know the area is xy=800, so y = 800/x and the cost is now c = 4(x+1600/x) + 12x minimum cost when dc/dx=0, so we need dc/dx = -16(400-x^2)/x^2 dc/dx=0 when x=20, so the fence ...
*Sunday, November 17, 2013 at 4:24pm*

**Calculus - Optimization **

A fence is to be built to enclose a rectangular area of 800 square feet. The fence along 3 sides is to be made of material $4 per foot. The material for the fourth side costs $12 per foot. Find the dimensions of the rectangle that will allow for the most economical fence to be...
*Sunday, November 17, 2013 at 3:38pm*

**calculus**

The cost of running a ship at a constant speed of v km/h is 160 + 1/100*v^3 dollars per hour. a)Find the cost of a journey of 1000km at a speed of v km/h. b)Find the most economical speed for the journey, and the minimum cost. c)If the ship were to have maximum speed of 16 km/...
*Sunday, November 17, 2013 at 8:37am*

**calculus**

A traveller employs a man to drive him from Sydney to Melbourne. Running costs of the car, which are also paid by the traveller, are k*v^3 dollars per hour, v is the speed and k is a constant. Find the uniform speed that will minimize the total cost of the journey.
*Sunday, November 17, 2013 at 8:33am*

**Pre-Calculus**

in algebra I you learned that (a+b)^2 ≠ a^2 + b^2 That's still true, even if you're taking pre-cal! :-) 2(x^2+y^2)^2=25(x^2−y^2) 2 * 2(x^2+y^2) (2x+2yy') = 25(2x-2yy') y' = -x(4x^2+4y^2-25) -------------------------- y(4x^2+4y^2+25) So, at (-3,1...
*Sunday, November 17, 2013 at 7:40am*

**Pre-Calculus**

Find the slope of the tangent line to the curve 2(x^2+y^2)^2=25(x^2−y^2) at the point (−3,−1)? Here's what I did: 2(x^4 + y^4) = 25(x^2-y^2) 2x^4 + 2y^4 = 25x^2 - 25y^2 8x^3 + 8y^3(dy/dx) = 50x - 50y(dy/dx) d/dx(8y^3 + 50y) = 50x - 8x^3 d/dx = (50x-8x^3...
*Sunday, November 17, 2013 at 2:31am*

**Calculus**

height: 240 max height at t=11 the curve is just a parabola. Don't forget your algebra I just because you're taking calculus! :-)
*Saturday, November 16, 2013 at 10:46pm*

**Calculus**

I throw a ball off the roof. It travels s = 240+22t-t^2. S is the balls distance after I release it. How tall is the building? How high above the ground did the ball get?
*Saturday, November 16, 2013 at 10:20pm*

**Pre-Calculus**

thanks
*Saturday, November 16, 2013 at 9:47pm*

**Calculus**

(4n-7)/(4n+9) = [(4n+9) -16]/(4n+9) = 1 - 16/(4n+9) The limit thus exists and is equal to 1. Clearly, for every epsilon > 0, there exists an N such that A_n will be within epsilon of the limiting value for all n > N, taking N = 4/epsilon will do.
*Saturday, November 16, 2013 at 7:43pm*

**Calculus**

|a_n| = |sin(2/n)| Since |sin(x)| <= |x|, we have: |a_n| <= 2/n The limit if |a_n| for n to infinity is thus 0, which then implies that the limit of a_n is zero. You can easily make this rigorous, for every epsilon > 0, you can using the above inequality find an N ...
*Saturday, November 16, 2013 at 7:36pm*

**Calculus - Optimization**

Good catch, Steve. Thank you.
*Saturday, November 16, 2013 at 6:29pm*

**Calculus**

Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. A (sub n)=((4n-7)/(4n+9))
*Saturday, November 16, 2013 at 5:42pm*

**Calculus**

Determine whether the sequence is divergent or convergent. a(sub n)=(−1)^n(sin(2/n))
*Saturday, November 16, 2013 at 5:39pm*

**Calculus - Optimization**

but an 8x8x8 box has length+girth = 8+32 = 40 inches, so it will not work. We need to optimize s^2(24-4s) since a square has 4 sides. v = 24s^2 - 4s^3 v' = 48s - 12s^2 v'=0 when s=4 So, a 4x4x8 box has max volume. Do (B) similarly
*Saturday, November 16, 2013 at 5:25pm*

**Calculus - Optimization**

Let the size of the square (cross-section) be s. Then we need to maximize V=s²(24-2s) with respect to s. First find the derivative and equate to zero: dV/ds = 48s-6s²=0 means s=0 or s=8 s=0 corresponds to a minimum volume and s=8 corresponds to a maximum volume. So ...
*Saturday, November 16, 2013 at 5:18pm*

**Calculus - Optimization **

A parcel delivery service a package only of the length plus girth (distance around) does not exceed 24 inches. A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has a maximum volume. What is the maximum ...
*Saturday, November 16, 2013 at 4:24pm*

**Physics**

So f = (9.8)(153) and d=.794? And then do I solve for x? How would I solve for x? ( never taken calculus)
*Friday, November 15, 2013 at 8:07pm*

**Physics**

So f = (9.8)(153) and d=.794? And then do I solve for x? How would I solve for x? ( never taken calculus)
*Friday, November 15, 2013 at 7:58pm*

**Calculus**

c(x) = (10 - x) + 1.4√(x^2 + 25) dc/dx = -1 + 1.4(2x)/2√(x^2+25) = 1.4x/√(x^2+25) - 1 because d/dx √u = 1/2√u du/dx
*Friday, November 15, 2013 at 5:41pm*

**Calculus**

What would the derivative of c(x) = (10 - x) + 1.4*sqrt((x^2) + 25)
*Friday, November 15, 2013 at 5:29pm*

**Calculus**

8yy' = x 8y dy = x dx 4y^2 = 1/2 x^2 + c 4(16) = 1/2 (64) + c c = 32 4y^2 = 1/2 x^2 + 32 8y^2 - x^2 = 64 y^2/8 - x^2/64 = 1
*Friday, November 15, 2013 at 1:43pm*

**Calculus**

y' = (x-7)e^(-2y) e^2y y' = x-7 e^2y dy = x-7 dx 1/2 e^2y = 1/2 x^2 - 7x + c e^2y = x^2 - 14x + c 2y = log(x^2-14x+c) y = 1/2 log(x^2-14x+c) Pick a c that fits a particular solution
*Friday, November 15, 2013 at 1:40pm*

**Calculus**

Solve the seperable differential equation 8yyŒ =x. Use the following initial condition: y(8)=4. Express x^2 in terms of y.
*Friday, November 15, 2013 at 1:37pm*

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