Wednesday

April 16, 2014

April 16, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**calculus**

Use the principles of translating and reflecting to graph the function f(x)=(x-5)^3 +2
*Tuesday, February 4, 2014 at 1:27am*

**pre calculus**

determine the order of magnitude for the following: 1. a $1 bill and a dime 2. two products scored a PH level of 6.1 and 4.o I can't figure out #1… but I did get #2 to be 10^2.1 Please help with #1 and check to see I did #2 right?
*Monday, February 3, 2014 at 4:25pm*

**calculus**

y = (-5x+4)/(10-5x) = (4-5x)/(10-5x) At x=2 there is a vertical asymptote, since 10-5x=0 As x gets large, y -> 1 since y = (4/x - 5)/(10/x - 5) -> -5/-5 = 1
*Monday, February 3, 2014 at 5:47am*

**calculus**

y= -5x + 4 OVER 10 - 5x find the asymptotes of the function
*Monday, February 3, 2014 at 1:37am*

**Calculus II**

Solve the initial value problem using Taylor Series and the following conditions: y'(t) = y(t) + 2t y(0) = A
*Sunday, February 2, 2014 at 11:01pm*

**Calculus**

8.42
*Sunday, February 2, 2014 at 8:42pm*

**CALCULUS PLEASE HELP!!!**

as you know, the average velocity is the integral divided by the time interval. Since the position is the integral of the velocity, the average velocity in [a,b] = (s(b)-s(a))/(b-a) So, we have i) (s(2)-s(1))/(2-1) = (2sin2π+3)-(2sinπ+3) = 0 ii) (s(1.1)-s(1))/(1.1-1...
*Sunday, February 2, 2014 at 8:26pm*

**CALCULUS PLEASE HELP!!!**

SHOW WORK PLEASE!!! The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin πt + 3 cos πt, where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the ...
*Sunday, February 2, 2014 at 8:12pm*

**calculus**

g = m t + b m = (17-8)/(47-28) = .4737 so g = .4737 t + b 8 = .4737 (28) + b so b = -5.263 so g = 0.4737 t - 5.263
*Sunday, February 2, 2014 at 5:03am*

**calculus**

In a lab experiment 8 grams of acid were produced in 28 minutes and 17 grams in 47 minutes. Let g be the number of grams and m be the number of minutes. Find a linear equation that you could use to calculate g for any number of minutes.
*Sunday, February 2, 2014 at 3:28am*

**calculus**

Consider the function f(x) = 5 (x - 5)^(2//3) . For this function there are two important intervals: (-\infty, A) and (A,\infty) where A is a critical number. Find A
*Sunday, February 2, 2014 at 2:38am*

**CALCULUS**

I don't understand. Why are you adding 1/4? to both sides? I don't see how you are getting a quadratic formula from this. More importantly, thank you guys for the help!
*Sunday, February 2, 2014 at 12:29am*

**calculus**

I guess the foxes were factored in when calculating the surviving rabbit population, eh?
*Saturday, February 1, 2014 at 6:41pm*

**calculus**

I bet we want the maximum, but usually the foxes are included in these island problems. dP/dt = 0 at max = 120 -1.6 t^3 t^3 = 120 t = 4.93 months
*Saturday, February 1, 2014 at 6:23pm*

**Calculus & Vectors**

240 * 1.25 = 300 @ N20W = <-102.6,281.9> 240 * 2.00 = 480 @ N80E = <472.7,83.4> add 'em up to get <370.1,315.3> So, the distance back is 486.2 300+480+486.2 = 1266.2 km at 240 km/hr = 5.3 hours
*Saturday, February 1, 2014 at 6:18pm*

**Calculus & Vectors**

A search and rescue aircraft, travelling at a speed of 240 km/h starts out at a heading of N 20 degrees West. After travelling for 1h and 15 min, it turns to a heading of N 80 degrees E and continues for another two hours before returning to base. Find the total distance the ...
*Saturday, February 1, 2014 at 6:10pm*

**calculus - and?**

in the words of Larry King, "What's the question?"
*Saturday, February 1, 2014 at 5:32pm*

**calculus**

The rabbit population on a small island is observed to be given by the function P(t)=120t−0.4t4+700 where t is the time (in months) since observations of the island began.
*Saturday, February 1, 2014 at 5:28pm*

**CALCULUS**

Whoops !
*Saturday, February 1, 2014 at 11:47am*

**calculus**

Well if you only have five days total, hopefully it only takes the remaining one day to solve it! Otherwise, you're in trouble!
*Saturday, February 1, 2014 at 11:32am*

**CALCULUS**

noticed an error from line 2 to line 3 y^2 - y = x y^2 - y + 1/4 = x + 1/4 (y - 1/2)^2 =(4x+1)/4 y - 1/2 = ±√(4x+1)/2 y = 1/2 ± √(4x+1)/2 = (1 ± √(4x+1) )/2 , x ≥ -1/4
*Saturday, February 1, 2014 at 10:37am*

**CALCULUS**

x = y^2 - y y^2 - y + 1 = x + 1 (y-1)^2 = x + 1 y -1 = sqrt (x+1) y = 1 + sqrt (x+1)
*Saturday, February 1, 2014 at 9:46am*

**COURSE HELP PLEASE ms sue qq**

In high school, these courses are usually called Algebra I Algebra II Geometry Pre-calculus AP Calculus Biology AP Biology Chemistry AP Chemistry Physics AP Physics Work with your counselor to make sure you take the required courses for graduation AND as many of these as ...
*Saturday, February 1, 2014 at 9:41am*

**CALCULUS**

Find a formula for the inverse of the function. y=x^2-x, x>=(greater than or equal to) 1/2 Please give me a step by step explanation. I think my algebra is wrong... Ty
*Saturday, February 1, 2014 at 9:16am*

**calculus**

Four out of five days were used to create a problem, how many days will it take to solve it?
*Saturday, February 1, 2014 at 8:21am*

**calculus**

Four out of five days were used to create a problem, how many day will it take to solve it?
*Saturday, February 1, 2014 at 8:21am*

**integral calculus**

If you mean ∫[x,4] x/(x-4) dx that's a bit unusual to have the variable of integration as one of the limits. But ok, let's work it out as-is. ∫x/(x-4) dx = x + 4 log(x-4) Evaluating that at 4 and x, we have (4+4log(0))-(x+4log(x-4)) Looks like -∞ to ...
*Friday, January 31, 2014 at 10:56pm*

**integral calculus**

Evaluate the limit limit gose from x to 4 (x/x-4)integral from x to 4
*Friday, January 31, 2014 at 10:04pm*

**Calculus**

Thank you!
*Friday, January 31, 2014 at 10:00pm*

**Calculus**

point in first plane 0,0 something (0 , 0 , 10) d = | 10*0 -8*0 +2*10 -3 | /sqrt(10^2+8^2+4^2) = 17/13.4
*Friday, January 31, 2014 at 8:07pm*

**Calculus**

here is an example of how to do this: http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section5/718d65/718_65.html
*Friday, January 31, 2014 at 8:01pm*

**Calculus**

when x gets really big positive this is y = 2 (big) / sqrt(big ^2) which is y = 2 when x gets really big negative y = 2 (-big number) / big number which is y = - 2 well, let's look at the derivative since they say so [ (x^2+x+1)2 - 2x(.5)(x^2+x+1)^-.5(2x) ]/(x^2+x+1) yuuk...
*Friday, January 31, 2014 at 5:50pm*

**Calculus**

we have two points in that plane (8, 0, -2) and (6, 3, 3) a vector through those points has direction (6-8)i + 3 j + (2+2)k = -2 i + 3 j + 4 k so we have two vectors parallel to plane, their cross product is normal to the plane i j k -2 5 2 the given line direction -2 3 4 the ...
*Friday, January 31, 2014 at 5:35pm*

**Calculus**

Let f be the function given by f(x)= 2x/(sqrt(x^2 +x +1)) c. Write an equation for each horizontal asymptote of the graph of f. d. Find the range of f. Use f'(x) to justify your answer.
*Friday, January 31, 2014 at 5:26pm*

**Calculus**

vector normal to plane has direction 3 i + 2 j + 6 k line normal to plane through point is (1, -5 , 9) + (3, 2, 6) t where does that hit the plane? 3(1+3t) + 2(-5+2t) + 6(9+6t) = 5 solve for t go back and use that t to get x, y, z in plane x = 1+3t y = -5+2t z = 9+6t then d^2...
*Friday, January 31, 2014 at 5:09pm*

**Calculus**

pick a point in plane #1, say (2,3,1) Now just use the distance formula to get the distance to 6x-6y+6z=3 or, equivalently, 2x-2y+2z-1 = 0 d = |2*2-2*3+2*1-1|/√(2^2+2^2+1^2) = 1/√5
*Friday, January 31, 2014 at 4:47pm*

**Calculus**

see http://www.jiskha.com/display.cgi?id=1391198824 for the method
*Friday, January 31, 2014 at 4:36pm*

**Calculus**

(1+4t) +2 (4t) -(2-3t) = -1 solve for t then use that t to find x, y z
*Friday, January 31, 2014 at 4:35pm*

**Calculus**

going from the first point to the second dx = 3 dy = -2 dz = 4 so my line direction is 3 i -2 j + 4 k and my line in parametric form is (1,0,1) + (3, -2, 4) t at intersection x = 1+3t y = -2 t z = 1+4t and we know x + y + z = 10 2 + 5 t = 10 t = 8/5 so x = 1 + 24/5 = 29/5 y...
*Friday, January 31, 2014 at 4:32pm*

**Calculus**

perpendicular to the direction 1 i - 1 j + 2 k Vxi + Vyj + Vzk if perpendicular dot product is 0 Vx - Vy + 2 Vz = 0 also parallel to plane x+y+z = constant 2 so normal to the normal to that plane 1 i + 1 j + 1 k Vxi + Vyj + Vzk Vx + Vy + Vz = 0 so we have Vx - Vy + 2 Vz = 0 Vx...
*Friday, January 31, 2014 at 4:14pm*

**Calculus**

Find the distance between the given parallel planes. 4z = 4y − 4x, 6z = 3 − 6x + 6y
*Friday, January 31, 2014 at 3:15pm*

**Calculus**

Find the distance between the given parallel planes. 5x−4y+z=10, 10x−8y+2z=3
*Friday, January 31, 2014 at 3:14pm*

**Calculus**

Find the distance from the point to the given plane. (−3,8,7), x−2y−4z=8
*Friday, January 31, 2014 at 3:13pm*

**Calculus**

Find the distance from the point to the given plane. (1,−5,9), 3x+2y+6z=5
*Friday, January 31, 2014 at 3:12pm*

**Calculus**

Find parametric equations for the line through the point (0,2,2)that is parallel to the plane x+y+z = 2 and perpendicular to the line x=1+t, y=2−t, z=2t. (Use the parameter t.) (x(t), y(t), z(t)) =
*Friday, January 31, 2014 at 3:11pm*

**Calculus**

Where does the line through (1,0,1) and (3,−2,5)intersect the plane x+y+z=10? (x, y, z) =________?
*Friday, January 31, 2014 at 3:09pm*

**Calculus**

Find the point at which the line intersects the given plane. x = 1 + 4t, y = 4t, z = 2−3t ; x + 2y − z + 1 = 0 (x, y, z) = _____________?
*Friday, January 31, 2014 at 3:07pm*

**Calculus**

Find the point at which the line intersects the given plane. x = 4 − t, y = 5 + t, z = 2t; x − y + 3z = 3 (x, y, z) =____________?
*Friday, January 31, 2014 at 3:05pm*

**Calculus**

Find an equation of the plane. The plane that passes through the point (−1, 2, 1)and contains the line of intersection of the planes x + y − z = 4 and 4x − y + 5z = 4
*Friday, January 31, 2014 at 3:04pm*

**Calculus**

Find an equation of the plane. The plane that passes through (8, 0, −2)and contains the line x = 6 − 2t, y = 3 + 5t, z = 3 + 2t
*Friday, January 31, 2014 at 2:57pm*

**calculus**

22
*Friday, January 31, 2014 at 7:27am*

**CALCULUS**

y = 4-(5x-4) = 8-5x
*Friday, January 31, 2014 at 6:01am*

**CALCULUS**

Starting with the graph of f(x)=5x, write the equation of the graph that results from reflecting f(x) about the line y =4. y=_______? Thank you!
*Friday, January 31, 2014 at 12:27am*

**Math Pre-Cal**

If you do know calculus, find the derivative and set it equal to zero (the slope is 0 at a max or min) 4 x -12 = 0 so x = 3 then y = 2x^2-12x-6 = 2(9) -12(3) -6 =18 - 36 - 6 = -24 that vertex is a minimum because y gets big as x gets big
*Thursday, January 30, 2014 at 9:51pm*

**Math Pre-Cal**

If you do not know calculus you must find the vertex of that parabola.
*Thursday, January 30, 2014 at 9:45pm*

**Calculus**

l) (g)^2 + (h)^2=1 so 2 g g' + 2 h h' = 0 g g' = -h h' g' = -(h/g) h' 2)g' = h^2 and h' = -g g'/h h' = - g h^2/h = -gh that is part a then If h has a max, then h' has a zero - g h = 0 ? either g = 0 or h = 0 if h is max, look for ...
*Thursday, January 30, 2014 at 3:30pm*

**Calculus**

(a) using (i) and (ii) 2gg' + 2hh' = 0 2gh^2 + 2hh' = 0 gh + h' = 0 h' = -gh (b) from (a) and (iv), h'(0) = -g(0)*h(0) = 0 h" = -g'h-gh' so, h"(0) = -g'(0)h(0) Since h(0)>0, g'(0)>0, so h"(0) < 0 (c) from (ii) and...
*Thursday, January 30, 2014 at 3:15pm*

**Calculus**

Let g and h be any two twice-differentiable functions that are defined for all real numbers and that satisfy the following properties for all x: I) (g(x))^2 + (h(x))^2=1 ii) g'(x)= (h(x))^2 iii) h(x)>0 iv) g(0)=0 a)Justify that h'(x)=-g(x)h(x) for all x b) Justify ...
*Thursday, January 30, 2014 at 2:44pm*

**Calculus**

Thank you!
*Thursday, January 30, 2014 at 2:28pm*

**Calculus II**

you know it will be a plane normal to the vector <7,-3,-6> passing through (3/2,9/2,0). I'm sure you've covered that.
*Thursday, January 30, 2014 at 11:55am*

**Calculus II**

Find an equation of the set of all points equidistant from the points A(−2, 6, 3) and B(5, 3, −3)
*Thursday, January 30, 2014 at 11:33am*

**Pre-Calculus**

factor out 2x^5(x^2-9)^7 and you have 2x^5(x^2-9)^7 * (8x^2 + 3(x^2-9)) = 2x^5(x^2-9)^7 (8x^2+3x^2-27) and voila...
*Thursday, January 30, 2014 at 6:09am*

**Pre-Calculus**

Actually I figured out that you use both, and ended up with 16x^7(x^2-9)^7 + 6x^5(x^2-9)^8. but I don't understand how that simplifies to 2x^5(x^2-9)^7(11x^2-27)?
*Thursday, January 30, 2014 at 12:47am*

**Physics**

Problem 1 (80 points) Consider a spring of equilibrium length L, lying horizontally in a frictionless trough. The spring has cross sectional area S perpendicular to its length. The trough constrains the motion of the spring so that any wave propagating along the spring is a ...
*Wednesday, January 29, 2014 at 11:24pm*

**Pre-Calculus**

Find dy/dx if y= x^6[(x^2-9)^8]? So the answer is supposed to be 2x^5[(x^2-9)^7](11x^2-27), but I don't understand how to get it. Do you use product rule, chain rule or both? thanks
*Wednesday, January 29, 2014 at 9:44pm*

**Calculus**

first fall, 40 ft, then second bounce 2(40*3/5), third 2*(40*3/5)(3/5), etc Now, how can you determine what rest is?
*Wednesday, January 29, 2014 at 8:17pm*

**Calculus**

A superball is tossed vertically 40 feet and rebounds on each bounce 3/5 of the height from which it fell. How far will it travel before coming to rest?
*Wednesday, January 29, 2014 at 8:14pm*

**Calculus**

If you are going to integrate over y, the solid has two parts: a plain old cylinder of height 4 and thickness 2, and a variable-thickness shape of height 2. So, v = π(5^2-3^2)(4) + ∫[4,6] π(R^2-r^2) dy where R=5 and r=x-1=y-1 v = 64π + π∫[4,6] ...
*Wednesday, January 29, 2014 at 5:34pm*

**Calculus**

dy/dx = 2/2 = 1 so y = x + b 0 = 3 + b b = -3 so y = x - 3 if x(t) = k t for example then y(t) = x(t) - 3 y(t) = k t - 3
*Wednesday, January 29, 2014 at 5:03pm*

**Calculus**

c) as t --->oo v ---->1/t = 0 d) 500 = 5 + .5 ln (t^2+1) 495 (2) = ln(t^2+1) 990 = ln (t^2+1) e^(990) = t^2 + 1 oh, my, calculator overflow :)
*Wednesday, January 29, 2014 at 4:58pm*

**Calculus**

dx/dt = t/(1+t^2) that will be maximum when the derivative d^2x/dt^2 = 0 d^2x/dt^2 = [(1+t^2)1 -t(2t) ](1+t^2)^2 0 when numerator is zero 0 = 1 + t^2 - 2 t^2 0 = 1 - t^2 t^2 = 1 so max or min at t = 1 we know it is max because v decreases with big x a ) so max v = 1/2 at t = 1...
*Wednesday, January 29, 2014 at 4:53pm*

**Calculus**

A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given by v(t)= t/(1+t^2). a). Determine the maximum velocity of the particle. Justify your answer. b). Determine the position of the particle at t=6. c...
*Wednesday, January 29, 2014 at 4:38pm*

**Calculus**

write in parametric form the equation of the line that joins (1,-2) and (3,0)
*Wednesday, January 29, 2014 at 4:14pm*

**Physics**

Problem 1 (80 points) Consider a spring of equilibrium length L, lying horizontally in a frictionless trough. The spring has cross sectional area S perpendicular to its length. The trough constrains the motion of the spring so that any wave propagating along the spring is a ...
*Wednesday, January 29, 2014 at 3:50pm*

**Calculus **

Consider the solid obtained by rotating the region bounded by the following curves about the line x=1. y=x,y=0,x=4,x=6 Find the volume So it would be pi (integral from 3 to 6) of ((1-y)^2 -(1-0)^2) right? so then you integrate it and get pi(Y^3/3-y^2) from 3 to 6. ?
*Wednesday, January 29, 2014 at 3:15pm*

**Calculus**

If the dimensions are x wide and y high, then y = 160/x The printed area is a = (x-2)(y-1-7) = (x-2)(160/x - 8) = 176-8x-320/x da/dx = -8+320/x^2 da/dx=0 when x=2√10 Now figure y and you're done.
*Wednesday, January 29, 2014 at 6:16am*

**calculus**

I assumed you wanted the roots of your equation 3x^2+10x-8=7
*Wednesday, January 29, 2014 at 5:58am*

**calculus**

how did you get (-5 +/- sqrt(70))/3
*Wednesday, January 29, 2014 at 12:12am*

**Calculus**

h = t^(3/4) - 5t^(1/4) h' = 3/4 t^(-1/4) - 5/4 t^(-3/4) = (3√t-5) / 4t^(3/4) h' is undefined at t=0 h' = 0 at t = 25/9
*Wednesday, January 29, 2014 at 12:03am*

**calculus**

yes. the mean value theorem lets you pick any interval, but Rolle's theorem says f(a) = f(b). It's just a special case of the MVT where f'(c) = 0. Your interval here meets that condition. As for your exact value, forget about decimal approximations. (-5±&#...
*Tuesday, January 28, 2014 at 11:59pm*

**Calculus**

f = x^4 - 8x^2 + 6 f' = 4x^3 - 16x f" = 12x^2 - 16 f'=0 at x = 0,±2 at x = -2, f" > 0 so f has a local min at x = 0, f" = < 0 so f has a local max at x = 2, f" > 0 so f has a local min So, now just evaluate f(x) at 0,±2 and ...
*Tuesday, January 28, 2014 at 11:54pm*

**calculus**

i am on "rolles and the mean value theorem" and was just wondering, when i am doing rolles, do i really need to find the exact value of x where f'(c) = 0? for example: f(x) = (x+4)^2 (x-3) on [-4,3] i get to: 3x^2+10x-8=7 then i dont know if i then need to find ...
*Tuesday, January 28, 2014 at 11:51pm*

**Calculus**

Find the critical numbers of the function. h(t) = t3/4 − 5t1/4
*Tuesday, January 28, 2014 at 10:25pm*

**Calculus**

A poster is to have an area of 160 in2 with 1 inch margins at the bottom and sides and a 7 inch margin at the top. What dimensions will give the largest printed area? (Give your answers correct to one decimal place.)
*Tuesday, January 28, 2014 at 10:25pm*

**Calculus**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = (x2 - 1)3 [-1, 6]
*Tuesday, January 28, 2014 at 10:24pm*

**Calculus**

Find the absolute minimum and absolute maximum values of f on the interval below. f(x) = x4 - 8x2 + 6 [-3, 4]
*Tuesday, January 28, 2014 at 10:23pm*

**Calculus**

smaller number --- x larger number ----- x + 132 product=P=x(x+132) = x^2 + 132x dP/dx = 2x + 132 =0 for a min of P 2x = -132 x = -66 the smaller is -66 the larger -66+132 or +66 the two numbers are -66 and +66
*Tuesday, January 28, 2014 at 10:20pm*

**Calculus**

let the sides of the base be x cm each let the height be y V= x^2 y 4000/x^2 = y surface area = x^2 + 4xy = x^2 + 4x(4000/x^2 = x^2 + 16000/x d(surface area)/dx = 2x - 16000/x^2 = 0 for a min surface area 2x = 16000/x^2 x^3 = 16000 x = appr 25.2 cm y = 4000/25.2^2 = appr 6.3 cm
*Tuesday, January 28, 2014 at 10:17pm*

**Calculus**

Find two numbers whose difference is 132 and whose product is a minimum.
*Tuesday, January 28, 2014 at 10:04pm*

**Calculus**

Still not very understand. Where u get the 1/6?
*Tuesday, January 28, 2014 at 10:03pm*

**Calculus**

A box with a square base and open top must have a volume of 4,000 cm3. Find the dimensions of the box that minimize the amount of material used.
*Tuesday, January 28, 2014 at 10:03pm*

**Calculus**

A poster is to have an area of 160 in2 with 1 inch margins at the bottom and sides and a 7 inch margin at the top. What dimensions will give the largest printed area? (Give your answers correct to one decimal place.)
*Tuesday, January 28, 2014 at 10:02pm*

**pre-calculus**

S = 0.9p - 1500
*Tuesday, January 28, 2014 at 8:18pm*

**pre-calculus**

The suggested retail price for a new car is p dollars. The dealership advertises a 10% discount and a factory rebate of $1500. Write a function S in terms of p, representing the cost of the car after receiving the dealership discount.
*Tuesday, January 28, 2014 at 8:15pm*

**calculus**

I assume that Sn = sum(1..n) An, so Sn -> 3 Dunno where the x comes from, but if you mean when n > 1, An = Sn - s(n-1) = (3n-2)/(n+1) - (3(n-1)-2)/(n-1+1) = 5 / n(n+1)
*Tuesday, January 28, 2014 at 5:08am*

**calculus**

If the nth partial sum of an is sn=(3n-2)/(n+1) what is an when x>1? and what is the sum of an?
*Tuesday, January 28, 2014 at 1:06am*

**Calculus**

write and row reduce the augmented matrix to find the general solution. x + y - z = 7 2x - y - 5z = 2 -5x + 4y + 14z = 1 3x - y - 7z = 5
*Monday, January 27, 2014 at 6:51pm*

**Calculus HELP!!**

The 6 2 1 is supposed to be on the right hand side of the equal sign. I am so sorry again!!!
*Monday, January 27, 2014 at 5:38pm*

**Calculus HELP!!**

just add them up: 2u-3v-w = 12-6-6 8-6-2 10-9-1 = 0 0 0
*Monday, January 27, 2014 at 5:37pm*

**Calculus HELP!!**

Then after this, use that fact to find x1 and x2 that satisfies the equation: 6 2 6 4 2 2 5 3 {x1 and x2} = 1
*Monday, January 27, 2014 at 5:36pm*

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