Monday

April 21, 2014

April 21, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = xe^((−x^2)/98), [−6, 14]
*Tuesday, April 1, 2014 at 9:14pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (t square root of (64 − t^2)) ,[−1,8]
*Tuesday, April 1, 2014 at 9:11pm*

**Calculus Help!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = xe^(−x^(2)/98), [−6, 14]
*Tuesday, April 1, 2014 at 9:09pm*

**calculus**

5x^2
*Tuesday, April 1, 2014 at 8:19pm*

**Calculus Help Please!!!**

looks like we both messed up in the f(x) calculations should be f(4) = 2(64) - 3(16) - 72(4) + 7 = -201 f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 142 end-values: f(-4) = 2(-64) - 3(16)( - 72(-4) + 7 = 119 f(5) = 2(125) - 3(25) - 72(5) + 7 = -178
*Tuesday, April 1, 2014 at 6:34pm*

**Calculus**

f ' (t) = -8sin t + 2(4 cos t) = -8sint + 8cost = 0 for a max/min 8sint = 8cost sint/cost = 1 tant = 1 t = 45° or 225° or t = π/4 , t = 5π/4 -->(outside our domain) so evaluate f(0) f(π/4) f(π/2) and determine which is the largest and which ...
*Tuesday, April 1, 2014 at 6:24pm*

**Calculus Help Please!!!**

so it the same way you were just given in your previous post. Let us know what you got
*Tuesday, April 1, 2014 at 6:07pm*

**Calculus Help Please!!!**

f ' (x) = 6x^2 - 6x - 72 = 0 for a local max/min x^2 - x - 12 = 0 (x-4)(x+3) = 0 x = 4, or x = -3 f(4) = 2(64) - 3(16) - 72(4) + 7 = -201 f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 115 end-values: f(-4) = 2(-64) - 2(16)( - 72(-4) + 7 = 135 f(5) = 2(125) - 3(25) - 72(5) + 7 = -178 ...
*Tuesday, April 1, 2014 at 6:04pm*

**Calculus Help Please!!!**

f'(x) = 6x^2 -6x -72 f'(0) = 6(x^2 -x-12) 0 = 6(x+ 3) (x-4) x = -3, 4 f(-4) = 2(-4)^3 -3(-4)^2 -72(-4) +7= 139 f(-3) = 2(-3)^3 -3(-3)^2 -72(-3) +7=132 f(4) = 2(4)^3 -3(4)^2 -72(4) + 7= -145 f(5) = 2(5)^3 -3(5)^2 -72(5)+7= -178
*Tuesday, April 1, 2014 at 6:01pm*

**Calculus**

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 8 cos t + 4 sin 2t, [0, π/2]
*Tuesday, April 1, 2014 at 5:45pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (t square root of (64 − t^2)), [−1, 8]
*Tuesday, April 1, 2014 at 5:27pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x^3 − 3x^2 − 72x + 7 , [−4, 5]
*Tuesday, April 1, 2014 at 5:26pm*

**calculus and vectors**

Add the 2 equations ... 6x + y = 6 y = 6-6x let x = 0 , y = 6 2(0) + 4(6) + z = 5 --> z = -19) we have a point (0,6,-19) on the line of intersection let x = 1, y = 0 2(1) + 4(0) + z = 5 --> z = 3 and (1,0,3) is another point on that line So a direction vector of that ...
*Tuesday, April 1, 2014 at 5:18pm*

**calculus and vectors**

Determine an equation of the line of intersection of the planes 4x − 3y − z = 1 and 2x + 4y + z = 5.
*Tuesday, April 1, 2014 at 4:59pm*

**calculus**

max area for a given perimeter is a circle. Is the wire evenly divided? If so, c = p/2, so r = p/4π, and the total area is a = 2(πr^2) = 2π(p^2/16π^2) = p^2/8π If not, then if we have a tiny circle, (area effectively zero), then r = p/2π and a...
*Tuesday, April 1, 2014 at 12:30pm*

**calculus**

Twenty feet wire is used to make two figures? What is the maximum areas of enclosed figures.
*Tuesday, April 1, 2014 at 11:14am*

**Calculus**

Thanks Mr. Steve for the guidance.
*Tuesday, April 1, 2014 at 5:21am*

**Calculus**

Looks good to me. I get ∫[1,2] (3-x) - 2/x dx = 3x - 1/2 x^2 - 2logx [1,2] = 3/2 - 2log2 gotta be a typo in the answer. 2log2 is correct.
*Tuesday, April 1, 2014 at 4:52am*

**Calculus**

Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis ...
*Tuesday, April 1, 2014 at 3:07am*

**Calculus**

Good idea. However, in an attempt to use your math, and also to apply to a possibly more general problem later, try implicit differentiation: x^2 + y^2 = 18 2x + 2yy' = 0 y' = -x/y So, if -x/y = 1 and x^2+y^2 = 18, 2x^2 = 18 x = ±3 Now, figuring y should not be ...
*Monday, March 31, 2014 at 1:02pm*

**Calculus**

LOL, sketch a graph !
*Monday, March 31, 2014 at 11:46am*

**Calculus**

The circle defined by the equation x^2 + y^2 = 18 has two points where the slope of its tangent line is m=1. Find those points.
*Monday, March 31, 2014 at 11:41am*

**Calculus: Integral**

recall that sec^2 = 1+tan^2, so you have ∫sec^4(4x) dx = ∫sec^2(4x)(1 + tan^2(4x)) dx = ∫sec^2(4x) dx + ∫tan^2(4x) sec^2(4x) dx = 1/4 tan(4x) + (1/4)(1/3) tan^3(4x) and you can massage that in several ways.
*Monday, March 31, 2014 at 5:32am*

**Calculus: Integral**

I don't understand how to do this one integral problem that involves secant. I'm asked to find the integral of sec^4 (4x). I'm not really sure how to go about solving this problem.
*Monday, March 31, 2014 at 3:32am*

**Calculus Help Please!!!**

looking at a diagram, if A is a away from Q and B is b away from Q, then √(a^2+144) + √(b^2+144) = 39 a/√(a^2+144) da/dt + b/√(b^2+144) db/dt = 0 Now just plug in da/dt = 3.5 a = 5 b = 23.065 (from 1st equation when a=5) and solve for db/dt
*Monday, March 31, 2014 at 12:27am*

**Calculus Help**

v = 2/3 pi r^3 (half a sphere) dv = 2 pi r^2 dr now just plug in the given r and dr watch the units.
*Monday, March 31, 2014 at 12:07am*

**Calculus Help Please!!!**

v = 4/3 pi r^3 dv = 4 pi r^2 dr c = 2pi r dc = 2pi dr so, dr = dc/(2pi) meaning that dv = 4 pi r^2 dc/(2pi) = 2 r^2 dr so, using the given numbers, dv = 2*(80/2pi)^2 * 0.5 = 1600/pi^2
*Monday, March 31, 2014 at 12:05am*

**Calculus Help Please!!!**

looks good to me
*Monday, March 31, 2014 at 12:01am*

**pre calculus**

24
*Sunday, March 30, 2014 at 11:40pm*

**Calculus Help Please!!!**

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.) fourth root of (1 + 2x)≈ 1 + (1/2)x
*Sunday, March 30, 2014 at 11:21pm*

**Calculus Help Please!!!**

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. 1) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.) 2) What is the relative error? (Round your answer to three decimal ...
*Sunday, March 30, 2014 at 11:17pm*

**Calculus Help**

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.03 cm thick to a hemispherical dome with diameter 54 m. (Round your answer to two decimal places.)
*Sunday, March 30, 2014 at 11:16pm*

**Calculus Help Please**

f(x) = f(a) + (x-a) f'(a) f(x) = x^4 df/dx = 4x^3 let a = 2 then f(a) = 2^4 = 16 f'(a) = 4*8 = 32 f(x) = 16 + (x-a)(32) x-a = - .001 so f(1.999) = 16 -.001(32) = 16 - .032 f(1.999) = 15.968 with calculator it is 15.968 also
*Sunday, March 30, 2014 at 10:29pm*

**Calculus Help Please!!!**

at x = 3 y = 3*3 - 9 = 0 at x = 2.4 y = 3(2.4) - 2.4^2 = 1.44 delta y = 1.44 -0 = 1.44 dy/dx = 3 - 2 x at x = 3 dy/dx = 3 - 6 = -3 if dx = -.6 dy = -3 (-.6) = 1.8
*Sunday, March 30, 2014 at 10:21pm*

**Calculus**

Thanks Damon, that really clears it up for me
*Sunday, March 30, 2014 at 10:17pm*

**Calculus Help Please**

Use a linear approximation (or differentials) to estimate the given number. (1.999)^4
*Sunday, March 30, 2014 at 10:14pm*

**Calculus Help Please!!!**

Oh, I see f(x) = ln (1+x) df/dx = 1/(1+x) d^2f/dx^2 = -1/(1+x)^2 f(x) = f(0) + [1/(1+0)] x - x^2/2! ... f(x) = 0 + x - x^2/2 + ..... well at a first cut when is x^2/2 =.1 x? x/2 = .1 x = .2
*Sunday, March 30, 2014 at 10:13pm*

**Calculus Help Please!!!**

Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) y = 3x − x^2, x = 3, Δx = −0.6 Δy=???
*Sunday, March 30, 2014 at 9:57pm*

**Calculus Help Please!!!**

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.) ln(1 + x) ≈ x xE
*Sunday, March 30, 2014 at 8:28pm*

**Calculus Help Please!!!**

Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 3.5 ft/s. How fast is cart B moving toward Q at...
*Sunday, March 30, 2014 at 7:16pm*

**Calculus Help Please!!!**

LOL - Guess which of us is the mathematician and which is the Engineer :)
*Sunday, March 30, 2014 at 7:15pm*

**Calculus Help Please!!!**

when the water has depth x, the cross-section is a trapezoid with bases 30 and 30+x. So the volume of water at depth x is v = (60+x)/2 * x * 500 cm^3 = 250x^2 + 15000x so, knowing that dv/dt = (500x + 15000) dx/dt just solve that for dx/dt when x=20
*Sunday, March 30, 2014 at 7:12pm*

**Calculus Help Please!!!**

Q = incoming flow rate = .1 m^3/min dh/ dt = Q A where A = surface area = length * width at 20 cm depth which depth is (1/2) height width = 30 + 1/2(70-30) = 30+20 = 50 cm = .5 m wide water surface so A = 5 * .50 =2.5 m^3 so finally dh/dt = .1 * 2.5 = .25 m/min = 25 cm/min
*Sunday, March 30, 2014 at 7:11pm*

**Calculus Help Please!!!**

A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when ...
*Sunday, March 30, 2014 at 6:43pm*

**physics**

find the difference in PE in the two locations. Hold one charge stationary. PEtotal= kQQ/.1 PE(new total)=KQQ/.06 subtract the first from the second, that must equal the work done. There are more difficult ways to work this, involving finding work in an integral calculus ...
*Sunday, March 30, 2014 at 10:11am*

**Calculus**

when dy/du = u^½ what does y = ? Just a simple power rule substitution. dy/du = u^n y = u^(n+1) / (n+1) + C
*Saturday, March 29, 2014 at 4:38pm*

**Calculus**

Thanks Steve! Finally got this one correct on Math Lab! I am eternally grateful, and have made an account here so I can get help and I have already tried helping others as well. I am very pleased with the service of this site and am glad to have found it :) Hooray!!!
*Saturday, March 29, 2014 at 4:34pm*

**Calculus**

When dy/dx=(x-6)^½ what does y equal?
*Saturday, March 29, 2014 at 4:11pm*

**Calculus**

the rate of change of volume is the surface area times the rate of change of height 450 ft^3/min = surface area * dh/dt surface area = pi r^2 450 = pi (30^2) dh/dt dh/dt = .159 ft/min You could do this by saying V = pi r^2 h dV/dh = pi r^2 dV/dh*dh/dt = pi r^2 dh/dt chain rule...
*Friday, March 28, 2014 at 11:48pm*

**Calculus**

Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h ...
*Friday, March 28, 2014 at 11:39pm*

**math**

assume no calculus allowed thus complete he square to find the vertex of this parabola 16 t^2 - 20 t - 2 = -h 16 t^2 -20 t = - h + 2 t^2 - 5/4 t = - h/16 + 1/8 t^2- 5/4 t+ 25/64 = -h/16 + 8/64 + 25/64 (t - 5/8)^2 = -(1/16)(h - 33/4) so in 5/8 of a second it reaches the vertex ...
*Friday, March 28, 2014 at 10:37pm*

**Calculus**

so it is positive for x>.316 and for x <.316
*Friday, March 28, 2014 at 8:12pm*

**Calculus**

Note that you did the second derivative correctly. It is easier to write it their way y" = 10e^(-5x^2)(10x^2-1) It is zero at x = +/- .316
*Friday, March 28, 2014 at 8:11pm*

**Calculus**

first http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29 Then this for second derivative and graph https://www.wolframalpha.com/input/?i=second+derivative+of+e^%28-5x^2%29
*Friday, March 28, 2014 at 7:57pm*

**Calculus**

first http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29
*Friday, March 28, 2014 at 7:52pm*

**Calculus**

At what interval is e^(-5x^2) concave up? I know the second derivative is 100x^2*e^-5x^2-10*e^-5x^2 but I just can not figure this one out. Thank you for your help!
*Friday, March 28, 2014 at 7:38pm*

**Calculus**

Thanks for the advice. I checked the problem statment and answer several times and got the same result. I also suspect it to be a print mistake in the book.
*Friday, March 28, 2014 at 7:46am*

**Calculus**

As a first check, I went to http://www.wolframalpha.com/input/?i=2%E2%88%AB[3%2C4]+%282%2F3+%E2%88%9A%28x^2-9%29%29+dx and saw that they show the area as 2.28 So, I suspect there is an error in the problem or the answer. Your calculation appears to be correct, ...
*Friday, March 28, 2014 at 5:43am*

**Calculus**

Find the area cut off by x=4 from the hyperbola x^2/9-y^2/4=1. Answer is 4.982 in the book. I have proceeded as under: Y=2/3*sqrt(x^2-9) and rhe reqd. area is double of integral 2/3*sqrt(x^2-9) from 3 to 4. Int= 2/3*[xsqrt(x^2-9)/2 – 9/2*log{x+sqrt(x^2-9)}] from 3 to 4 =x...
*Friday, March 28, 2014 at 2:17am*

**Calculus Help Please!!!**

One of us made an arithmetic mistake. It is up to Tanya to get it right :)
*Thursday, March 27, 2014 at 9:06pm*

**Calculus Help Please!!!**

According to Newton's Law of Cooling T(t) = roomtemp + (37 - 20)e^(-kt) , where t is the time in hours and k is a constant so we get two equations: 32.5 = 20 + 17e^(-kt) ---> 12.5 = 17e^(-kt) and 30.3 = 20 + 17e^(-k(t+1)) ---> 10.3 = 17e^(-kt - k) divide them: 115/...
*Thursday, March 27, 2014 at 9:02pm*

**Calculus Help Please!!!**

rate of change of temp proportional to temp above room temp which is 20 to make it easy on the arithmetic define T' = real T - 20 dT'/dt = k (T') dT/T' = k dt ln T' = k t T' = C e^kt let's call t = 0 at 1:30 T = 32.5 so T' = 12.5 12.5 = C e^0 = ...
*Thursday, March 27, 2014 at 8:54pm*

**Calculus Help Please!!!**

I am going to assume it is algebra first take t = 0 at 1:30 T = Ti - k t Ti = 32.5 so T = 32.5 - k t 30.3 = 32.5 - k (1 hour) k = 2.2 so T = 32.5 - 2.2 t 37 = 32.5 - 2.2 t t = - 2.04 call it 2 hours so by that linear model 11:30 am now I will work on the more realistic ...
*Thursday, March 27, 2014 at 8:38pm*

**Calculus Help Please!!!**

Are you sure this is calculus? You want exponential decay to room temp? Or is it algebra and you want a linear function?
*Thursday, March 27, 2014 at 8:30pm*

**Calculus Help Please!!!**

In a murder investigation, the temperature of the corpse was 32.5 C at 1:30pm and 30.3 C an hour later. Normal body temperature is 37.0 C and the temperature of the surrounding was 20.0 C. When did the murder take place? PLEASE SHOW STEP BY STEP
*Thursday, March 27, 2014 at 8:16pm*

**Calculus**

best use radians, pi/2
*Thursday, March 27, 2014 at 3:20pm*

**Calculus**

when t=90?
*Thursday, March 27, 2014 at 2:59pm*

**Calculus**

y" = 5cos(t) so, when is that zero?
*Thursday, March 27, 2014 at 2:52pm*

**Calculus**

A weight oscillates in a vertical motion according to the position function y(t)=-5 cos(t). Assuming t≥0, when will the acceleration if the weight be zero for the first time?
*Thursday, March 27, 2014 at 2:51pm*

**Calculus**

-9.8 m/s^2
*Thursday, March 27, 2014 at 2:45pm*

**Calculus**

An object in free fall has its distance from the ground measured by the function d(t)=-4.9t^2 +50, where d is in meters and t is in seconds. If gravity is the only acceleration affecting the object, what is gravity's constant value?
*Thursday, March 27, 2014 at 2:37pm*

**Calculus**

Ahh. I see that I was interpreting 243^3/5 as (243^3)/5
*Thursday, March 27, 2014 at 12:05pm*

**Calculus**

just using ln a^b = b ln a ln (243^3/5 *32^4/5) = ln ( (3^5)^(3/5) * (2^5)^(4/5) ) = ln ( 3^3 * 2^4) = ln (27*16) = ln(432) 1/5 ln (243^3 * 32^4) = ln [ (243^3 * 32^4) ^(1/5) ] = ln (243^(3/5) * 32^(4/5) ) = .... = ln(432)
*Thursday, March 27, 2014 at 12:01pm*

**Calculus**

No one is bothered by the fact that 5 does not divide powers of 2 and 3?
*Thursday, March 27, 2014 at 11:37am*

**Calculus**

same as y": -cosx
*Thursday, March 27, 2014 at 11:25am*

**Calculus**

If y=cos x, what is y^(6) (x)?
*Thursday, March 27, 2014 at 11:18am*

**Calculus**

I agree with Damon's "huh" since ln (243^3/5 *32^4/5) = 1/5 ln (243^3 * 32^4)
*Thursday, March 27, 2014 at 10:30am*

**CALCULUS problem**

int x^-3 dx = -.5 x^-2 + c at x = 3 = -.5/9 at x = 1 = -.5 so A = .5 - .5/9 = .5(8/9) = 4/9 B at x = h int = -.5/h^2 right half = -.5/9 +.5/h^2 left half = -.5 h^2 +.5 so -.5 h^2 + .5 = -5/9 +.5/h^2 1/h^2 = .5 + 5/9 = 4.5/9 + 5/9 = 9.5/9 = 19/18 int 1 to 3 of pi (x^-6)dx = -pi...
*Thursday, March 27, 2014 at 9:53am*

**CALCULUS problem**

There are four parts to this one question, and would really appreciate if you could show and explain how you get to the answer, because I tried looking up how to find the answer myself, but nothing made sense. Thank you! 11. The region R is bounded by the x-axis, x = 1, x = 3...
*Thursday, March 27, 2014 at 8:51am*

**Calculus**

huh?
*Thursday, March 27, 2014 at 7:51am*

**Calculus**

good except for this step: ln (243^3/5 *32^4/5) should be 1/5 ln (243^3 * 32^4)
*Thursday, March 27, 2014 at 5:36am*

**Calculus**

The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00 at time t hours after 12:00, (at t=0) the minute hand is at 4.5 sin(2pi*t) the hour hand is at 2.0sin(2pi*t/12) at 9:00, the distance d is d^2...
*Thursday, March 27, 2014 at 5:34am*

**Calculus**

x = ln 243 y = ln 32 LET Z = e^((3x + 4y)/5) ln [z ]= (3x+4y)/5 ln z = (1/5)( 3 ln 243 + 4 ln 32) = ln (243^3/5 *32^4/5) = ln (27*16) = ln(432) if ln z = ln 432 then z = 432
*Thursday, March 27, 2014 at 3:54am*

**Calculus**

Posted by MG on Wednesday, March 26, 2014 at 6:54pm. The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines) The distance between the tips of the hands is changing at ...
*Thursday, March 27, 2014 at 3:36am*

**Calculus**

If e^x = 243 and e^y = 32 then e^((3x + 4y)/5) =? The answer is 432, but I don't understand why.
*Thursday, March 27, 2014 at 3:34am*

**College Calculus**

Thank you for the response, i tried the method mentioned above three times and was incorrect each time, i double checked all my work to match the method above. The correct answer is always just .3 under the answer All of your arithmetic is right as well, so it is not that. ...
*Thursday, March 27, 2014 at 3:31am*

**pre calculus**

C(x) = 2.00 for 0 < x <= 1 2.00 + .20(10x) for 1 < x < 2 since there are 10 charging units per mile.
*Thursday, March 27, 2014 at 12:10am*

**pre calculus**

A taxi company charges $2.00 for the first mile (or part of a mile) and 20 cents for each succeeding tenth of a mile (or part). Express the cost C (in dollars) of a ride as a piecewise-defined function of the distance x traveled (in miles) for 0 < x < 2
*Thursday, March 27, 2014 at 12:07am*

**Calculus**

Thank you very much!
*Wednesday, March 26, 2014 at 11:46pm*

**Pre calculus**

yes. The reason the law of sines can give two triangles is because sin(x) is positive all the way from 0 to 180. cos(x) becomes negative for x>90, so the formula takes that into account, always leaving only one possible answer. I mean, think about it geometrically. If you ...
*Wednesday, March 26, 2014 at 11:46pm*

**Calculus**

Since velocity is the derivative of position, v(t) = -32t + 160 now just solve -32t+160 = 32 -32t+128 = 0 t = 4
*Wednesday, March 26, 2014 at 11:39pm*

**Calculus**

The height in feet above the ground of a ball thrown upwards from the top of a building is given by s=-16t^2 + 160t + 200, where t is the time in seconds. If the maximum height is 600 feet, what is v^-1(32)? The answer is supposed to be 4 seconds, but I don't understand ...
*Wednesday, March 26, 2014 at 11:07pm*

**argggh - Calculus**

messed up in my expansion volume should have been 4x^3- 120x^2 + 800x and V' = 12x^2 - 240x + 800 = 0 3x^2 - 60x + 200=0 to get x = 4.23
*Wednesday, March 26, 2014 at 10:11pm*

**Calculus**

First things first: width --- s length ---- 2s 2s^2 = 800 s^2 = 400 s = 20 so the piece of metal is 20 by 40 let the side of the square to be cut out be x so the width is 20-2x the length is 40-2x the height is x Volume = x(20-2x)(40-2x) = 2x^3 - 120x^2 + 800x d(Volume)/dx = ...
*Wednesday, March 26, 2014 at 10:05pm*

**Calculus **

You are given a piece of sheet metal that is twice as long as it is wide an has an area of 800m^2. Find the dimensions of the rectangular box that would contain a max volume if it were constructed from this piece of metal by cutting out squares of equal area at all four ...
*Wednesday, March 26, 2014 at 9:49pm*

**Calculus**

you should have recalled that sin (-x) = -sinx so that sin(-.1) could not have been positive. did you mean -.1 ?
*Wednesday, March 26, 2014 at 9:14pm*

**Calculus**

Use a tangent line approximation at x=0 to estimate the value of sin(-0.1). I got 0.1
*Wednesday, March 26, 2014 at 8:56pm*

**Calculus**

yes v(t) = h ' (t) = -16t + 5 so v(0) = -16(0) + 5 = 5
*Wednesday, March 26, 2014 at 8:54pm*

**Calculus**

The vertical position of an object is modeled by the function h(t)=-16t^2 +5t+7, where h is measured in feet and t is measured in seconds. Find the object's initial velocity (that is, the velocity at t=0). Is it 5 feet per second?
*Wednesday, March 26, 2014 at 8:46pm*

**Pre calculus**

For the most part, will a law of cosines always be one triangle? As in one triangle to solve?
*Wednesday, March 26, 2014 at 8:35pm*

Pages: <<Prev | 1 | 2 | 3 | **4** | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>

Post a New Question | Current Questions