Tuesday
March 11, 2014

# Homework Help: Math: Calculus

Calculus
what's your name. i think i'm in your class. I'm having the same problem. Hopital isn't the way to solve this problem though.
Friday, March 7, 2014 at 8:54pm

calculus
3 * 2 pi = 6 pi radians/minute pi/6 = 30 degrees by the way I call your angle theta A dA/dt = 6 pi rad/min tan A = x/7 x = 7 tan A dx/ dt = 7 d/dt(tan A ) = (7/cos^2A) dA/dt cos^2 (30) = .75 so dx/dt = (7 miles/.75)(6 pi rad/min) dx/dt = 176 miles/min * 60 = 1055 miles/hr so ...
Friday, March 7, 2014 at 7:26pm

calculus
by the way, d/dt (tan A) = sec^2 A * dA/dt = (1/cos A)^2 * dA/dt
Friday, March 7, 2014 at 7:14pm

calculus
A searchlight rotates at a rate of 3 revolutions per minute. The beam hits a wall located 7 miles away and produces a dot of light that moves horizontally along the wall. How fast (in miles per hour) is this dot moving when the angle \theta between the beam and the line ...
Friday, March 7, 2014 at 7:13pm

calculus
Hey, I just did one very much like this. Your turn.
Friday, March 7, 2014 at 7:03pm

calculus
A hot air balloon rising vertically is tracked by an observer located 4 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is \frac{\pi}{3} , and it is changing at a rate of 0.1 rad/min. How fast is the ...
Friday, March 7, 2014 at 7:00pm

calculus
Oh, it is pointed at the bottom? area = pi r^2 radius = (1/3) h for 1 at the top and 0 at the bottom so surface area A = pi (1/9)h^2 for h = 2.6, A = 2.36 meters^2 d V = A dh dV/dt = A dh/dt or dh/dt = (1/A)dV/dt dV/dt given as 1.4 m^3/min so dh/dt = 1.4/2.36 = .593 m/min
Friday, March 7, 2014 at 7:00pm

Calculus
y = 1 e^(rx) y' = r e^(rx) y" = r^2 e^(rx) r^2 - 4 r + 1 = 0 r = [4 +/- sqrt (16 -4) ] /2 r = [ 4 +/- 2 sqrt 3 ]2 r = 2 +/- sqrt 3
Friday, March 7, 2014 at 6:49pm

Calculus
for what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0 Show steps please! Thank you!
Friday, March 7, 2014 at 6:43pm

calculus
A man of height 1.5 meters walk away from a 5-meter lamppost at a speed of 1.8 m/s. Find the rate at which his shadow is increasing in length.
Friday, March 7, 2014 at 6:30pm

calculus
A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.4 \text{m}^3\text{/min}. How fast is the water level rising when it is 2.6 m?
Friday, March 7, 2014 at 5:53pm

y' = 8x-6x^2 so, at x=a, y' = 8a-6a^2 = m Use the above info to plug into the point-slope form of the lines. No sweat.
Friday, March 7, 2014 at 4:44pm

--- Find the slope m of the tangent to the curve y = 4 + 4x^2 − 2x^3 at the point where x = a. ---- Find equations of the tangent lines at the points (1,6) and (2,4). (1,6) Y(x)= (2,4) Y(x)=
Friday, March 7, 2014 at 4:40pm

(c) (15005-10237)/2 = 2384 (d) (16684-12435)/2 = 2124.5 (e) Looks like growth is slowing down
Friday, March 7, 2014 at 4:22pm

c) Estimate the instantaneous rate of growth in 2006 by measuring the slope of the tangent line through (2005, 10237) and (2007, 15005). d) Estimate the instantaneous rate of growth in 2007 by measuring the slope of the tangent line through (2006, 12435) and (2008, 16684). e)...
Friday, March 7, 2014 at 4:07pm

Calculus Help
dy/dx = 2ax + bx when x = 1, 2a + b = 6 when x = -1 -2a +b = -14 add them 2b = -8 b= -4, then a = 5 to find c, sub in (2,17) into the original: 17 = 4a + 2b + c 17 = 20 -8 + c c = 5 y = 5x^2 - 4x + 5 check: for (2,17) 17 = 20 - 8 + 5 ---> true dy/dx = 10x - 4 at x = 1, dy/...
Friday, March 7, 2014 at 3:33pm

Calculus Help
Find a parabola with equation y = ax^2 + bx + c that has slope 6 at x = 1, slope −14 at x = −1, and passes through the point (2, 17).
Friday, March 7, 2014 at 2:56pm

Criminal justice
1. Which of the following, according to Carl Klockars, is NOT an important consideration in determining whether the good ends of police work justify immoral means in a given scenario? A. Are there other, non-dirty, means that may be effective but that we may be overlooking? B...
Friday, March 7, 2014 at 2:21pm

f(-3) = 54 + 18 + 9 + 2 = 83 f(-1) = -2 + 2 + 3 + 2 = 5 slope = (5-83)/(-1+3) = -39 f ' (x) = -6x^2 + 4x - 3 then -6x^2 + 4x - 3 = -39 6x^2 - 4x -36 = 0 3x^2 - 2x - 18 = 0 x = (2 ± √220)/6 = appr 2.8054 or appr -2.1388
Friday, March 7, 2014 at 8:20am

The "solving" becomes a bit easier if you change the original equation to (x-y)^2 - 6x + 2y + 17 = 0 sub in : y = x - 3 (x - (x-3))^2 - 6x + 2(x-3) + 17 = 0 9 - 6x + 2x - 6 + 17 = 0 -4x = -20 x = 5 then y = 2 the horizontal asymptote touches at (5,2) I will do the ...
Friday, March 7, 2014 at 8:03am

2xdx-2y dx-2xdy + 2y dy -6dx+2dy=0 for horizontal lines, dy/dx=0 2x-2y-6)/(-2x+2y+2)=0 or y=x-3 for vertical tangent, dy/dx=undifined or -2x+2y+2=0 y=x-2 now solve the points on the parabola. for horizontal lines, substutute x-2 for y in the given equation, solve. Then, do the...
Friday, March 7, 2014 at 6:02am

average slope=(f(5)-f(2))/(5-2) set average slope above to equal f=-2/sqrt(x) then solve for x.
Friday, March 7, 2014 at 5:56am

Consider the function f(x)=4sqrt(x)+4 on the interval [2,5] . Find the average or mean slope of the function on this interval _______ <---A By the Mean Value Theorem, we know there exists a c in the open interval (2,5) such that f'(c) is equal to this mean slope. For ...
Friday, March 7, 2014 at 3:25am

f(x) -2x^3+2x^2-3x+2 Find the average slope of this function on the interval (–3–1) ________ <--A By the Mean Value Theorem, we know there exists a c in the open interval (–3–1) such that f'(c) is equal to this mean slope. Find the value of c in the ...
Friday, March 7, 2014 at 3:23am

calculus
A certain radioactive isotope decays at a rate of 2% per 100 years. If t represents time in years and y represents the amount of the isotope left then the equation for the situation is y= y0e-0.0002t. In how many years will there be 93% of the isotope left?
Friday, March 7, 2014 at 3:08am

calculus
$4000 is invested at 9% compounded quarterly. In how many years will the account have grown to$14,500? Round your answer to the nearest tenth of a year
Friday, March 7, 2014 at 2:53am

$4000 is invested at 9% compounded quarterly. In how many years will the account have grown to$14,500? Round your answer to the nearest tenth of a year
Friday, March 7, 2014 at 2:52am

Use implicit differentiation to find the points where the parabola defined by x^{2}-2xy+y^{2}-6x+2y+17 = 0 has horizontal and vertical tangent lines. List your answers as points in the form (a,b).
Friday, March 7, 2014 at 2:29am

Calculus
122.7761
Thursday, March 6, 2014 at 10:33pm

pre calculus
I mean to where I have to graph the following function using transformations. And have to be sure to graph all of the stages on 1 graph. Then I have to state the domain and range.
Thursday, March 6, 2014 at 8:28pm

calculus
3.2ft/s
Thursday, March 6, 2014 at 10:28am

Calculus
.00092
Thursday, March 6, 2014 at 1:09am

pre-calculus
147b^3-35b^2-19b+3 = 0 Since 147 = 7*7*3, I'd try using those coefficients first, since the inside coefficients are relatively small. A little playing around shows that T(b) = (7b-1)(7b-3)(3b+1)
Thursday, March 6, 2014 at 12:27am

pre-calculus
Find the real and imaginary zeros for the following polynomial function. T(b)= 147b^3-35b^2-19b+3
Thursday, March 6, 2014 at 12:22am

Math- NOT CALCULUS
Hey, I told you use completing the square to find the vertex, boring. 2 n^2 - 84 n = - p - 45 n^2 - 42 n = -p/2 - 22.5 n^2 - 42 n + (21)^2 = -p/2 -22.5 + 441 (n-21)^2 = - (p/2 - 418.5) (n-21)^2 = - (1/2) (p - 837) so 21 taxis making 837
Wednesday, March 5, 2014 at 9:06pm

Math- NOT CALCULUS
The hourly profit ($P) obtained from operating a fleet of n taxis is given by P=-2n^2+84n-45 What is the profit if 20 taxis are on the road? What is the maximum hourly profit? What number of taxis gives the max hourly profit? How much money is lost per hour if no taxis are on ... Wednesday, March 5, 2014 at 8:58pm Math P(20) = -2(400) + 84(20) - 45 if you do not know calculus, find vertex of parabola by completing square. I am assuming calculus. 0 = -4 n + 84 n = 21 at maximum so Pmax = -2(21^2) + 84(21) - 45 21, been there, did that -45 when n = 0 Wednesday, March 5, 2014 at 8:30pm Pre - calculus what happens to the cosine on the right side, when its in this step cos^3x/sinx + cosxsinx^2/sinx Wednesday, March 5, 2014 at 8:21pm Pre - calculus (cos/sin)/(1/ cos^2)+(cos/sin)/(1/sin^2) cos^3/sin + cos sin cos^3/sin + cos sin^2/sin cos (cos^2 + sin^2)/sin but cos^2 + sin^2 = 1 so cos/sin or cot x Wednesday, March 5, 2014 at 8:10pm Pre - calculus Can someone please explain how to simplify this proiblem: cotx/sec^2 + cotx/csc^2 Wednesday, March 5, 2014 at 7:59pm calculus Hmmm. I get y = -x^3+bx^2+cx+d y' = -3x^2+2bx+c y" = -6x+2b inflection at x = -2 means -6x+2b = 0 b = -6 y = -x^3 - 6x^2 + cx + d extrema at x = -5,1 y' = (x+5)(x-1) = x^2+4x-5 we need -3x^2, so y' = -3x^2-12x+15 so c = 15 y = -x^3 - 6x^2 + 15x + d f(2) = 11, ... Wednesday, March 5, 2014 at 11:48am calculus I assume you mean y or f(x) = -x^3 +bx^2 +cx + d if so y' = -3 x^2 + 2 b x + c and y" = -6 x +2 b at x = -5 , y' = 0 -75 -10 b + c = 0 at x = 1 , y' = 0 -3 + 2 b + c = 0 so 2 b + c = 3 -10 b + c = 75 -----------------subtract 12 b = -72 b = -6 then -12 + c = 3... Wednesday, March 5, 2014 at 11:28am calculus Find the correct values for the equation -x^3 +bx^2 +cx + d, using this information, local min x= -5, local max (1,11). Point of inflection = -2. Wednesday, March 5, 2014 at 11:14am Pre-Calculus Vo = 32Ft/s[35o] Xo = 32*cos35 = 26.21 Ft/s. Yo = 32*sin35 = 18.35 Ft/s. a. Y = Yo + g*Tr = 0 @ max. ht. Tr=(Y-Yo)/g = (o-18.35)/-32.4=0.566 s = Rise time or time to reach max. ht. hmax = ho + Yo*t + 0.5g*Tr^2 hmax = 4.5 + 18.35*0.566 - 16.2*0.566^2 = 9.70 Ft above gnd. b. ... Tuesday, March 4, 2014 at 8:39pm pre calculus 1000x + 2000(x+.005) = 190 x = 6% Tuesday, March 4, 2014 at 5:34am pre calculus assuming a proper allocation of papers and distance, then 1/x = 1/70 + 1/80 x = 112/3 minutes Tuesday, March 4, 2014 at 5:32am pre calculus Candy and Tim share a paper route. It takes Candy 70 min to deliver all the papers, and it takes Tim 80 min. How long does it take the two when they work together? Tuesday, March 4, 2014 at 12:46am pre calculus Jack invests$1000 at a certain annual interest rate, and he invests another $2000 at an annual rate that is one-half percent higher. If he receives a total of$190 interest in 1 year, at what rate is the $1000 invested? Tuesday, March 4, 2014 at 12:46am Pre-Calculus Solve the following inequality in terms of natural logarithms (ln). (e^6x)+2 is less than or equal to 3. Monday, March 3, 2014 at 6:56pm Pre-Calculus Go back and check your volleyball question. I just saw it and you missed the angle of launch in your equations. Monday, March 3, 2014 at 6:35pm Pre-Calculus Hey Alexis, between the two of us we have been doing an awful lot of problems for you. Many of them are the same old, same old. It is time for you to try and post what you did. Monday, March 3, 2014 at 6:30pm Pre-Calculus I got ( +36 , 24 , 39 ) Monday, March 3, 2014 at 6:27pm Pre-Calculus That is the resultant, but you need the opposite to that so move it around 180 to cancel it Monday, March 3, 2014 at 6:21pm Pre-Calculus Oh, I see from your other questions you are not using vector notation x = 0 + t = t y = 4 + 6 t Monday, March 3, 2014 at 6:18pm Pre-Calculus <-36,24,15> Monday, March 3, 2014 at 6:18pm Pre-Calculus y = 70 sin41° t - 16t^2 x = 70 cos41° t solve for t when y = 10 use that t to get x. Monday, March 3, 2014 at 6:16pm Pre-Calculus Find vector a<3,2,-4> X vector b<-6, 9,0> Monday, March 3, 2014 at 6:15pm Pre-Calculus y = 6 x + 4 through point (0,4) slope = 6 well (0,4) + (1,6) t should work Monday, March 3, 2014 at 6:14pm Pre-Calculus Phil Dawson is a professional place kicker for the Cleveland Browns. On average, he kicks the ball at a 41 degree angle and with an initial speed of 70 feet per second. For future reference, goal posts are 10 feet high in the NFL. a) Write parametric equations to model Dawson... Monday, March 3, 2014 at 6:12pm Pre-Calculus t = x - 6 y = 2 (x-6) - 4 y = 2 x -16 Monday, March 3, 2014 at 5:56pm Pre-Calculus Write an equation in slope-intercept form of the line with the given parametric equations. x=t+6 y=2t-4 Monday, March 3, 2014 at 5:48pm Pre-Calculus Write Parametric equations of -3x+1/2y=2 Monday, March 3, 2014 at 5:46pm Pre-Calculus A 15N force acting at 15 degrees north of east and a 18N force acting at 79 degrees north of west act concurrently on an object. What is the magnitude and direction of a third force that produces equilibrium on the object? Show sketch and work. ------ x-component:: 15*cos(15 ... Monday, March 3, 2014 at 5:44pm pre calculus x^4 > x^2 x^2 > 1 x > 1 or x < -1 Monday, March 3, 2014 at 5:20am calculus 2x Monday, March 3, 2014 at 2:30am pre calculus solve nonlinear inequality x^4 > x^2 Monday, March 3, 2014 at 12:38am calculus the fraction remaining after t years is (1/2)^(t/5730) So, you want (1/2)^(t/5730) = .7 t/5730 = log(.7)/log(.5) ... Monday, March 3, 2014 at 12:25am calculus Skeletal remains had lost 70% of the C-14 they originally contained. Determine the approximate age of the bones. (Assume the half life of carbon-14 is 5730 years. Round your answer to the nearest whole number.) Sunday, March 2, 2014 at 9:21pm Calculus Help Let P(t) be the percentage of Americans under the age of 18 at time t. The table gives values of this function in census years from 1950 to 2000. (I didnt put the table here because it will not display well so the table has basically t as time from 1950 to 2000 and P(t) values... Sunday, March 2, 2014 at 6:57pm Pre-Calculus Find N if log base 6 (6^7.8)=N Sunday, March 2, 2014 at 5:51pm Physics 110 km/hr(1000/3600) = 30.6 m/s 119 km/hr = 33.1 m/s x position of truck = -211.7+ 30.6 t v car = 2.3 t until v = 33.1 at t = 14.4 s and x = (1/2)(2.3)(14.4^2) = 238 m after that xcar = 238 + 33.1 (t-14.4) so during car acceleration period: d = Xcar - Xtruck = .5(2.3)t^2 - 211... Sunday, March 2, 2014 at 3:47pm Calculus Please help! the linearization is just finding a straight line that is close to the curve at the given point. That is just the tangent line. So, since when y = √(1+2x), y' = 1/√(1+2x) y(0) = 1 y'(0) = 1 and the point-slope form for the line is y-1 = 1(x-0) y = x+1 A=1 ... Saturday, March 1, 2014 at 6:28am Pre-Calculus Reiny gave you good polar coordinates, but if all you want is parametric equations, try x = t/3 y = 1 + t/2 or x = -(2+t)/3 y = -2t Saturday, March 1, 2014 at 6:20am Pre-Calculus y = v sinθ t - 16t^2 x = v cosθ t You have θ and v, so plug those in and solve for x when y=10 to get the max goal distance. or, y = x tanθ - 16/(v cosθ)^2 x^2 Saturday, March 1, 2014 at 6:15am Pre-Calculus we know x =rcosØ and y = rsinØ -3rcosØ + (1/2)rsinØ = 2 -6rcosØ + rsinØ = 2 r(sinØ - 6cosØ) = 2 or r = 2/(sinØ - 6cosØ) looks good: http://www.wolframalpha.com/input/?i=pol​ar+r+%3D+2%2F%28sinx+-+6cosx%29 Friday, February 28, 2014 at 9:19pm Pre-Calculus Phil Dawson is a professional place kicker for the Cleveland Browns. On average, he kicks the ball at a 41 degree angle with an initial speed of 70 feet per second. For future reference, goal posts are 10 feet high in the NFL. a) Write parametric equations to model Dawson'... Friday, February 28, 2014 at 8:56pm Pre-Calculus Write Parametric equations of -3x+1/2y=2 Friday, February 28, 2014 at 8:49pm Pre-Calculus Jake serves a volleyball with an initial velocity of 32 feet per second from 4.5 feet above the ground at an angle 0f 35 degrees. a) Write parametric equations to model the situation. b) How far will the ball travel( if it hits the ground)show work Friday, February 28, 2014 at 8:47pm Calculus Please help! dy/dx = 3(x^2 + 6)^2 dy = 3(x^2 + 6)^2 dx when x = 2 and dx = .05 dy = ...... just plug in the above values. Friday, February 28, 2014 at 8:41pm Calculus Please help! The linearization at a=0 to sqrt(1+2x) is A+Bx where A is____ and where B is _____? A=? B=? Ty Friday, February 28, 2014 at 8:20pm Calculus Please help! The differential of the function y=(x^2+6)^3 is dy=______dx. When x=2 and dx=0.05, the differential dy=_______? A) dy=______dx? B) the differential dy=_______? Friday, February 28, 2014 at 8:19pm Calculus Please help! darn them brackets 23t^2 vs (23t)^2 Friday, February 28, 2014 at 5:26pm Calculus Please help! let the time be t hours past noon. I see a right-angled triangle , with sides 23t and 50 + 18t, with D as the distance between them D^2 = (23t)^2 + (50+18t)^2 D^2 = 529t^2 + 2500 + 1800t + 324t^2 = 853t^2 + 1800t + 2500 2D dD/dt = 1706t + 1800 dD/t = (853t + 900)/D , #1 at 3:... Friday, February 28, 2014 at 5:23pm Calculus Please help! so, did you draw a diagram as suggested? a = 1/2 bh so, b = 2a/h da/dt = 1/2 (db/dt * h + b * dh/dt) Now just plug in your values: 3500 = 1/2 (db/dt * 7000 + 2*87/7 * 2500) db/dt = -7.88 Friday, February 28, 2014 at 4:55pm Calculus Please help! As usual, draw a diagram. The distance between the ships at time t hours is d^2 = (50+18t)^2 + 23t^2 At 3 pm, t=3, so d = √(104^2+69^2) = 124.8 2d dd/dt = 36(50+18t)+46t so, dd/dt = 15.6 knots Friday, February 28, 2014 at 4:45pm Calculus Please help! Let the radius be r, so the diameter is 2r then the height is 2r V = (1/3)π r^2 h = (1/3) π r^2 (2r) = (2/3)π r^3 dV/dt = 2π r^2 dr/dt when h = 24 2r = 24 r = 12 and dV/dt = 30 30 = 2π (144) dr/dt dr/dt = 30/(288π) = 5/(48π) ft/min or appr .... Friday, February 28, 2014 at 4:00pm Calculus Please help! Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 24 feet high? Recall that... Friday, February 28, 2014 at 3:10pm Calculus Please help! At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.) Friday, February 28, 2014 at 3:02pm Calculus Please help! The altitude of a triangle is increasing at a rate of 2500 centimeters/minute while the area of the triangle is increasing at a rate of 3500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7000 centimeters and the area is 87000... Friday, February 28, 2014 at 2:51pm Calculus Help f(x) = 1/[ x^.5 +1 ] f(z) = 1/[ z^.5 + 1 } f(x)-f(z) = 1/[ x^.5 +1 ] - 1/[ z^.5 + 1] = ([z^.5+1]-[x^.5+1])/(x^.5z^.5 +x^.5+z^.5+1) = (z^.5-x^.5) /(x^.5z^.5 +x^.5+z^.5+1) Divide by (x-z) which is (x^.5-z^.5)(x^.5+z^.5) and get -(x^.5+z^.5) / (x^.5z^.5 +x^.5+z^.5+1) let z --->... Friday, February 28, 2014 at 1:10pm Calculus Help Use f(x)= 1/(square root(x)+1) to answer the following; -- Use the difference quotient f'(x)=lim z->x (f(x)-f(z))/ (x-z) to find f'(x)! Please show steps!!! Thank you!!! Friday, February 28, 2014 at 12:31pm calculus I think we went through this before. note I think the somewhat unusual exponent notation may mean | x^2 + x^-2| = | x^2 + 1/x^2 | http://www.wolframalpha.com/input/?i=plo​t+y%3D|+x^2+%2B+1%2Fx^2+| Friday, February 28, 2014 at 9:53am calculus the graph can be seen here: http://www.wolframalpha.com/input/?i=plo​t+y%3D|x^2+%2Bx-2|+for+-4+%3C%3D+x+%3C%3​D+3 That should make it easy to answer the other questions. If not, come on back and show where you get stuck. Friday, February 28, 2014 at 6:40am calculus Find the asymptote, interval of monotonicity, critical points, the local extreme points, intervals of concavity and inflection point of the following functions. Sketch the graph. a)f(x)=|x2 +x-2| Friday, February 28, 2014 at 3:32am Calculus by the time you get to calculus, you certainly should know how to plug in a value and evaluate a function! Where do you get stuck? (a) q(5) = 10300*e^(-0.34*5) = 1882 (b) as you know, if y = e^u y' = e^u u' And, as you know, the derivative is the rate of change. So, q... Thursday, February 27, 2014 at 11:33pm Calculus q = f(p) = 10300e^−0.34p a) Find the number of products sold when the price of the product is$5. (Round your answer to the nearest whole number.) Number of products sold: b) Find a formula for the rate of change in the number of products sold when the price is p dollars...
Thursday, February 27, 2014 at 10:11pm

Calculus Help
get a better table
Thursday, February 27, 2014 at 6:18pm

Calculus Help
Let P(t) be the percentage of Americans under the age of 18 at time t. The table gives values of this function in census years from 1950 to 2000. How would it be possible to get more accurate values for P'(t).
Thursday, February 27, 2014 at 5:06pm

calculus
given the student's use of strange exponent notation it might be |x^2 + 1/x^2 | http://www.wolframalpha.com/input/?i=|x^​2+%2Bx^-2|
Thursday, February 27, 2014 at 3:20pm

calculus
Or, post your results and we can verify them.
Thursday, February 27, 2014 at 2:50pm

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