Wednesday

April 16, 2014

April 16, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**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*

**Calculus**

Find the particular solution of the differential equation particular solution of the differential equation dy/dx=(x−7)e^(−2y).
*Friday, November 15, 2013 at 1:36pm*

**calculus**

outside perimeter is 2(4x+y) = 8x+2y inside sections use 3y cost is thus c = 23(8x+2y) + 12(3y) = 184x + 82y Now, if the pens each occupy 14m^2, then y = 14/x, so c = 184x + 82(14/x) = 184x + 1148/x minimum cost where dc/dx=0, or 184 - 1148/x^2 = 0 x = √(287/46) = 2.497 ...
*Friday, November 15, 2013 at 11:52am*

**calculus**

So, we have 3 square sections and 4 sides. If the sides are of length x, and the height is y, then the cardboard used is 3x^2 + 4xy = 12 so, y = (12-3x^2)/4x The volume is v = x^2y = x^2(12-3x^2)/4x = (12x - 3x^3)/4 dv/dx = 3 - 9/4 x^2 max volume when dv/dx=0, or x=2/√3 ...
*Friday, November 15, 2013 at 11:46am*

**calculus**

A closed cardboard box is made with a square top and bottom, and a square horizontal shelf inside that divides the interior in half. A total of 12 square meters of cardboard is used to make the top, sides, bottom, and shelf of the box. What should the dimensions of the box be ...
*Friday, November 15, 2013 at 11:35am*

**calculus**

A dog kennel with four pens is to be constructed. The pens will be surrounded by rectangular fence that costs $23 per meter. The rectangle is partioned into four pens of equal size with three partitions made of fence that costs $12 per meter. Each pen measures x meters wide by...
*Friday, November 15, 2013 at 11:34am*

**Pre calculus**

cos(Ø - 42°) = -1 I know cos 180° = -1 Ø - 42 = 180 Ø = 222° , only one solution between 0° and 360° (look at the graph of cos Ø ) The period of cos(Ø-42) is 360° so a general solution is Ø = 222° + 360k°...
*Thursday, November 14, 2013 at 10:23pm*

**Pre calculus**

Solve for all values between zero and 360 degrees. cos(theta-42 degrees)=-1 What is the general solution and what is the solution????
*Thursday, November 14, 2013 at 9:48pm*

**Calculus Inegral**

This does not have an answer in elementary functions. Ei(2lnx)
*Thursday, November 14, 2013 at 6:34pm*

**Calculus Inegral**

integral of x/lnx
*Thursday, November 14, 2013 at 6:19pm*

**Pre-Calculus**

√(1x+2y) + √(1xy) = 8.24 changing the radicals to exponents is really the only way, explicitly or implicitly, but √(1x+2y) ≠ √(1x) + √(2y) !!! Anyway, we have, using implicit differentiation (and deleting those useless 1 coefficients), &#...
*Thursday, November 14, 2013 at 1:01pm*

**Pre-Calculus**

no it's actually 1x and 1xy. It's not division.
*Thursday, November 14, 2013 at 12:48pm*

**Pre-Calculus**

Before I attempt this ..... I am curious why you would put a coefficient of 1 in front of the variables, such as in (1x + 2y) and (1xy) At this stage of Calculus, they would certainly be understood and not needed. I am tempted to guess you meant (1/x + 2y) and √(1/xy) ...
*Thursday, November 14, 2013 at 10:18am*

**Pre-Calculus**

Find the slope of the tangent line to the curve √(1x+2y) + √(1xy) = 8.24 at the point (2,8)? I know you have to use implicit differentiation, but the radicals keep making me mess up algebraically. Is the changing the radicals to exponents the fastest way? please ...
*Thursday, November 14, 2013 at 9:20am*

**calculus**

1500
*Wednesday, November 13, 2013 at 8:59pm*

**Pre-Calculus**

nevermind I made a mistake it's p(x)=-1x/10 + 465
*Wednesday, November 13, 2013 at 8:39pm*

**Pre-Calculus**

A manufacture has been selling 1050 television sets a week at 360 dollars each. A market survey indicates that for each 20 -dollar rebate offered to a buyer, the number of sets sold will increase by 200 per week. p(x)=1x/10 + 465 How large of a rebate should the company offer ...
*Wednesday, November 13, 2013 at 4:06pm*

**Pre-Calculus**

Thank you all so much!
*Wednesday, November 13, 2013 at 12:18am*

**Pre-Calculus !!**

Nice one, Reiny. Went right to the heart of the matter of implicit derivatives.
*Tuesday, November 12, 2013 at 11:58pm*

**or - Pre-Calculus**

D^2 = (x^2 + 8x + 12) 2D dD/dt = 2x dx/dt + 8 dx/dt dD/dt = dx/dt (2x+8)/2D dD/dt = dx/dt (x+4) / √(x^2 + 8x + 12)
*Tuesday, November 12, 2013 at 11:54pm*

**Pre-Calculus**

wikipedia has a simple proof
*Tuesday, November 12, 2013 at 11:53pm*

**Pre-Calculus**

D = √(x^2 + 8x + 12) We can also rewrite this as D = (x^2 + 8x + 12)^(1/2) Now we have to get dD/dt. To get the derivative, recall that if x is raised to a certain constant, the derivative is x^n = n*x^(n-1) * dx We multiply the exponent to x, which is jow raised to 1 ...
*Tuesday, November 12, 2013 at 11:52pm*

**Pre-Calculus**

It's really just the chain rule in disguise: D = √u where u is a function of x dD/dx = 1/(2√u) du/dx du/dx = 2x+8 Now, since x is a function of t, dD/dt = dD/dx dx/dt and voila
*Tuesday, November 12, 2013 at 11:46pm*

**Pre-Calculus**

When you do implicit differentiation, how does D = √(x^2 + 8x + 12) turn into dD/dt = [(x + 4)(dx/dt)]/√(x^2 + 8x + 12)? please explain...I don't even understand where the (x+4) comes from. D means distance, but that's irrelevant
*Tuesday, November 12, 2013 at 11:35pm*

**College Level Calculus**

number of extra trees per acre --- n number of trees per acre = 20+n yield per tree = 720 - 15n number of oranges = N = (20+n)(720-15n) = 14400 + 420n - 15n^2 N ' = 420 - 30n = 0 for a max of N 30n = 420 n = 14 for the best yield there should be 20+14 or 34 trees per acre
*Tuesday, November 12, 2013 at 11:18pm*

**College Level Calculus**

Each orange tree grown in California produces 720 oranges per year if not more than 20 trees are planted per acre. For each additional tree planted per acre, the yield per tree decreases by 15 oranges. How many trees per acre should be planted to obtain the greatest number of ...
*Tuesday, November 12, 2013 at 10:31pm*

**calculus**

just carry tha too yA crazy kiddo!
*Tuesday, November 12, 2013 at 9:52pm*

**Pre-Calculus**

Can someone please help me with this problem? De Moivre’s theorem states, “If z = r(cos u + i sin u), then zn = rn(cos nu + i sin nu).” • Verify de Moivre’s theorem for n = 2. a. Provide a correct proof that includes written justification for each step.
*Tuesday, November 12, 2013 at 9:38pm*

**calculus**

assuming you swim along a chord that subtends an angle θ, you swim 2rsin(θ/2) walk πr-rθ total time is thus t = 1/10 r sinθ/2 + 1/50 (πr-rθ) = r/50 (π - θ + 5sin(θ/2)) for minimum time, we need dt/dθ=0 5/2 cos(θ/2) - ...
*Tuesday, November 12, 2013 at 9:19pm*

**calculus **

you stand on the shore of a circular lake and you wish to reach the exact opposite your current position. you can swim 20 feet per minute and run 50 feet per minute. what path should you take to reach your destination as quickly as possible?
*Tuesday, November 12, 2013 at 4:41pm*

**calculus**

34234234
*Monday, November 11, 2013 at 11:01pm*

**Calculus**

let the ladder be y ft above the ground x^2 + y^2 = 15^2 when foot is x ft from wall y^2 = 225-x^2 y = (225-x^2)^(1/2) 2x dx/dt + 2y dy/dt = 0 dy/dt = -2x dx/dt/(2y) = (-x dx/dt)/y = -2x(1/2)x/√(225 - x^2) = -x^2/√(225-x^2)
*Monday, November 11, 2013 at 10:39pm*

**Calculus**

A ladder 15 feet long leans against a vertical wall. Supppose that when the bottom of the ladder is x feet from the wall, the bottom is being pushed towards the wall at the rate of 1/2x feet per second. How fast is the top of the ladder rising at the moment the top is 5 feet ...
*Monday, November 11, 2013 at 10:30pm*

**Calculus**

You need to factor out the e^-x as well, giving you R" = -15e^-x (x-2) R" >0 for x<2 You are correct. Since R" < 0 for x>2, the graph is concave down on (2,7)
*Monday, November 11, 2013 at 7:30pm*

**Calculus**

27 ft of wire is to be used to form an isosceles right triangle and a circle. Determine how much of the wire should be used for the circle if the total area enclosed is to be a maximum? I can only find the minimum because the parabola is opening upward
*Monday, November 11, 2013 at 6:11pm*

**Calculus**

For R"(x) = -15[(x-1)(e^-x)-(e^-x)] with 0 < x < 7 What interval is the graph concave and and concave down? I know concave up is (0,2). Is that right? Does concave down exist?
*Monday, November 11, 2013 at 6:06pm*

**calculus**

if the triangle has side b, and the circle has circumference c, then 2b+b√2 + c = 27 so, b = (27-c)/(2+√2) the area is a = 1/2 b^2 + πr^2 = 1/2 ((27-c)/(2+√2))^2 + π(c/(2π))^2 This a parabola with vertex (minimum area) at c = 27(1 - 1/(1+(3-2&#...
*Monday, November 11, 2013 at 5:18pm*

**calculus**

27 ft of wire is to be used to form an isosceles right triangle and a circle. Determine how much of the wire should be used for the circle if the total area enclosed is to be a minimum? Maximum?
*Monday, November 11, 2013 at 4:34pm*

**Check my CALCULUS work, please! :)**

#3 is 0. exponentials grow much faster than any power of x. l'Hospital's Rule shows that given enough iterations, the derivatives in the numerator go to zero while that pesky e^x remains in the bottom. #4. Note that g(x) changes sign, so the limit does not exist. The ...
*Monday, November 11, 2013 at 12:23pm*

**Check my CALCULUS work, please! :)**

#1 and #2 are correct #3 --- here is a neat trick that works for most limit questions. Use your calculator and try a number close to your approach value e.g. for #1 I used h = .001 and then evaluated the expression for #3, try a "large" number. However for this one ...
*Monday, November 11, 2013 at 12:18pm*

**Check my CALCULUS work, please! :)**

Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 7.01 7.1 7.2 g(x) 9.44 10.21 10.92 -11.08...
*Monday, November 11, 2013 at 9:24am*

**calculus**

since Euler's method is used for differential equations, do you mean y' = x+y^2 y(0) = 0 ?
*Sunday, November 10, 2013 at 11:47pm*

**calculus**

Use Euler's method with step size .2 to estimate y(.4), where y(x) is the solution of the initial value problem y=x+y^2, y=0. Repeat part a with step size .1
*Sunday, November 10, 2013 at 9:45pm*

**Really need help in Calculus Problem?!**

Use Euler's method with step size .2 to estimate y(.4), where y(x) is the solution of the initial value problem y=x+y^2, y=0. Repeat part a with step size .1
*Sunday, November 10, 2013 at 8:27pm*

**Calculus**

critical points are where P'=0 or is undefined. P' = 12/(x-1)^2 Since P' is always positive, P is always increasing P'=0 never, so there are no min/max P' is undefined at x=1, the vertical asymptote. Enter this function at wolframalpha.com to see its graph ...
*Sunday, November 10, 2013 at 8:21pm*

**Calculus**

P(x) = 12x / (1-x) What are the critical points and where is P(x) increasing and decreasing?
*Sunday, November 10, 2013 at 8:17pm*

**Algebra**

ok. sorry. I forgot this was algebra, not calculus. recall that the vertex of ax^2+bx+c=0 is at x = -b/2a R(x) is a parabola, with vertex at x = 360/0.1 R(3600) = 648,000
*Sunday, November 10, 2013 at 7:34pm*

**CALCULUS**

You have the equation -- just plug in the numbers. 1/p + 1/7 = 1/5 1/p = 1/5 - 1/7 1/p = 2/35 p = 35/2 1/p + 1/q = 1/5 1/q = 1/5 - 1/p 1/q = (p-5)/5p q = 5p/(p-5) The above is just algebra I, you know. any online graphing site can do this. Try wolframalpha.com limit as p->...
*Sunday, November 10, 2013 at 7:20pm*

**Calculus**

yeah, that's what I got, too, if you factor out the 15 e^-x
*Sunday, November 10, 2013 at 7:08pm*

**CALCULUS**

If you could give an explenation with the answers, that'd be wonderful so I actually know how to solve similar problems in the future. :) Thank you so much! 1. A convex lens with focal length f centimeters will project the image of an object on a point behind the lens. If ...
*Sunday, November 10, 2013 at 6:27pm*

**Calculus**

When I try to find the first derivative, get R = 15e^-x - 15xe^-x..
*Sunday, November 10, 2013 at 5:16pm*

**Calculus**

R = 15x e^-x R' = 15(1-x) e^-x R'=0 when x=1 min or max? R" = 15(x-2) e^-x R"(1) < 0, so R(1)=15 is a maximum
*Sunday, November 10, 2013 at 4:33pm*

**Calculus Sigma Notation**

7 ∑ 2^k k=3 It appears that a=3,n=5 but that's hard to say, since you provide so little data on their relevance.
*Sunday, November 10, 2013 at 4:30pm*

**Math**

critical numbers have nothing to do with concavity. They are used for determining min/max, or where f(x) is undefined. For your f(x) above, f"(x) = 6, which is always positive. Makes sense, since the curve is a parabola which opens upward, and is thus always concave up. ...
*Sunday, November 10, 2013 at 4:26pm*

**Calculus**

Sorry, find the local extrema for the revenue function: R = xp
*Sunday, November 10, 2013 at 4:16pm*

**Calculus**

For p = 15e^-x, 0 < x < 7, find the local extrema
*Sunday, November 10, 2013 at 4:10pm*

**Calculus Sigma Notation **

Find a and n such that: 8+16+32+64+128 (k=a) a=? n=?
*Sunday, November 10, 2013 at 4:08pm*

**CALCULUS**

A. the numerator is always positive, so since there is a vertical asymptote at x=2, (i) -infinity (ii) +infinity (iii) not exist B. f(x) is very conveniently defined, so that we have (i) sin(pi/6) = 1/2 (ii) cos(pi/6) = √3/2 (iii) not exist C. again a convenient ...
*Sunday, November 10, 2013 at 3:33pm*

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