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April 20, 2014

April 20, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**pre-calculus**

C = .30*pi r^2 + .1* 2 pi r h pi r^2 h = 24 pi r^2 h = 24 h = 24/r^2 C = .30 pi r^2 + .2 pi r (24)/r^2 C = pi (.3 r^2 + 4.8/r)
*Thursday, January 23, 2014 at 7:40pm*

**pre-calculus**

A company plans to manufacture a container having the shape of a right circular cylinder, open at the top, and having a capacity of 24pi cubic inches. If the cost of the material for the bottom is $.30 per square inch and that for the curved sides is $.10 per square inch, ...
*Thursday, January 23, 2014 at 7:33pm*

**calculus**

t = 0 in 1990 t = 5 in 1995 s = A e^kt when t = 0, s = 68 so 68 = A when t = 5 427 = 68 e^5k 6.28 = e^5k ln 6.28 = 1.84 = 5 k so k = .367 so s = 68 e^.367 t
*Thursday, January 23, 2014 at 1:49am*

**calculus**

Suppose that the sales at Borders bookstores went from 68 million dollars in 1990 to 427 million dollars in 1995 . Find an exponential function to model the sales (in millions of dollars) as a function of years,t, since 1990.
*Thursday, January 23, 2014 at 1:37am*

**Math- Pre calculus**

idfk
*Wednesday, January 22, 2014 at 11:59pm*

**Calculus and vectors**

A cylindrical tank is to have a capacity of 1000 m³. It is to fit into a foundry that is 12 m wide with a height of 11 m. The base of the tank will cost half as much as the top. The metal for the side of the tank will cost four fifths as much as the top. An equation that ...
*Wednesday, January 22, 2014 at 10:16pm*

**Math**

Easy if you know calculus. h ' (t) = -10t + 22 = 0 for a max/min of h(t) 10t = 22 t = 2.2 h(2.2) = -5(2.2)^2 + 22(2.2) = appr 24.2 m If you don't know Calculus, the x of the vertex of your parabola is -b/(2a) = -22/-10 = 2.2 sub in 2.2 into h(t) , same as above If you ...
*Wednesday, January 22, 2014 at 9:30pm*

**calculus**

multiply both numerator and denominator by 1/n then Lim (-1)^n (5+7/n)/(6+5/n) so as n gets large, the second part becomes (5/6) the first part is +-1, depending if n is odd or even. there is not a limit. however, the funcion is +- 5/6 and is bounded by those two numbers.
*Wednesday, January 22, 2014 at 9:24pm*

**calculus**

limit as n approaches infinity for the function ((-1)^n)((5n+7)/(6n+5))
*Wednesday, January 22, 2014 at 9:19pm*

**Calculus**

Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer is supposed to be positive or negative. ...
*Wednesday, January 22, 2014 at 1:20pm*

**Calculus**

V= (l)(w)(h) since the length and width do not change, only V and h. dV/dt = (lw) dh/dt We must have the same units, 20 cubic feet = 20(12^3) or 34560 cubic inches when V = 34560 and dV/dt = -4 34560 = (lw) h lw = 34560/h -4 = 34560/h dh/dt dh/dt = -h/8640 inches/second Are ...
*Wednesday, January 22, 2014 at 7:53am*

**Calculus**

Water is leaking out of a pool at a rate of 4 cubic inches per second, what is the rate of change of the height of the water at the instant the volume equals 20ft cubic feet
*Wednesday, January 22, 2014 at 12:25am*

**Calculus**

f'(x) = 6x - 3 f'(a) = 6a - 3
*Tuesday, January 21, 2014 at 8:32am*

**calculus**

I would solve it for y first 1-xy = x-y y - xy = x-1 y(1 - x) = x-1 y = (x-1)/(1-x) = -1 , x ≠1 Well, isn't that special !!! so y' = 0 and of course y '' = 0 Wolfram took it as a pair of intersecting lines. Notice the horizontal line y = -1 The vertical ...
*Tuesday, January 21, 2014 at 8:31am*

**Calculus**

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*Tuesday, January 21, 2014 at 12:11am*

**calculus**

what is the 2nd derivative of: 1-xy = x-y i get (-y-1)/(x-1) for the first the second i know is (-y-1)/(x-1)^2 but i keep getting (0)/(x-1)^2 please show me the steps i should be taking and if possible what i did wrong
*Monday, January 20, 2014 at 10:57pm*

**Calculus**

I've been having a hard time with these lately... could someone explain it to me? Find f '(a). f(x) = (3x^2 − 3x + 4)
*Monday, January 20, 2014 at 10:50pm*

**calculus**

Thanx!!
*Monday, January 20, 2014 at 10:27pm*

**Calculus, Limits**

Thank you! That was a huge help!
*Monday, January 20, 2014 at 10:25pm*

**calculus**

top ---> 2 |x^3| because sqrt(x^6) is + but sqrt(x^3) is - bottom ----- x^3 which is - so 2 |x|^3 /-|x|^3 = -2
*Monday, January 20, 2014 at 7:15pm*

**Calculus, Limits**

when t --oo top --> t^2 bottom --> -t^2 result --> -1 try it with some big numbers :)
*Monday, January 20, 2014 at 7:11pm*

**calculus **

Find the limit lim x→−∞ sqrt(4x^6 − x)/(x^3 + 6)
*Monday, January 20, 2014 at 7:07pm*

**Calculus, Limits**

Find the limit. lim t → ∞ (sqrt t + t^2)/(6t − t^2)
*Monday, January 20, 2014 at 7:04pm*

**calculus**

express P and A as a function of L and W Then just use the chain rule to find the derivative (d/dt)
*Monday, January 20, 2014 at 12:34am*

**calculus**

a rectangle is 10 inches by 6 inches whose sides are changing. write a formula for both the perimeter and how fast each is changing in terms of L and W how do i do this, an explanation instead of an answer would be most appreciated, thanks in advance.
*Monday, January 20, 2014 at 12:26am*

**Calculus**

x = -3 is also excluded from the domain, eh? since x^2-9 = 0
*Sunday, January 19, 2014 at 6:53am*

**Calculus**

domain of f(x) : any real number, x ≠ 3 lim (x-3)/(x^2 - 9) , x --->3 = lim (x-3)((x-3)(x+3)) = lim 1/(x+3) , as x -->3 = 1/6
*Sunday, January 19, 2014 at 12:02am*

**Calculus**

What is the domain of f(x)= (x-3)/(x^2-9)? lim f(x) x->3
*Saturday, January 18, 2014 at 10:18pm*

**calculus**

Oddly, the answers agree, since 8 = y(x+1), so -8/(x+1)^2 = -(8/(x+1))/(x+1) = -y/(x+1)
*Saturday, January 18, 2014 at 6:32am*

**calculus**

yx+y = 8 y dx/dx + x dy/dx + dy/dx = 0 (x+1)dy/dx = -y dy/dx = -y/(x+1)
*Friday, January 17, 2014 at 6:17pm*

**calculus**

i need to use implicit differentiation to find the derivative: yx+y = 8 i keep getting y'=-1-y. i already know the derivative of this, by using explicit by simplification: y(x+1)=8 y=(8)/(x+1) y'= -(8)/(x+1)^2 how do i go about doing this the right way?
*Friday, January 17, 2014 at 5:49pm*

**science/math**

Using Calculus a = -4 v = -4t + c when t=0 , v = 30 30 = 0 + c v = -4t + 30 s = -2t^2 + 30t + k when t = 0 , s = 0, so k = 0 s = -2t^2 + 30t when the car stops, v = 0 4t = 30 t = 30/4 = 7.5 seconds when t = 7.5 s = -2(7.5)^2 + 30(7.5) = 112.5 m
*Thursday, January 16, 2014 at 11:25pm*

**physics**

I wonder if you have had calculus... velocity=dx/dt=35-21t^2. so velocity is zero when t^2=35/32 solve for t. acceleration= dv/dt=-42t so acceleation is zero only when t=0
*Tuesday, January 14, 2014 at 9:34pm*

**calculus**

Of course you could do it the sophisticated way, where the derivative is zero f' = 2 x + 4 = 0 when x = -2 then y = -5
*Tuesday, January 14, 2014 at 4:01pm*

**calculus**

That is a parabola. It is horizontal at the vertex.
*Tuesday, January 14, 2014 at 4:00pm*

**calculus**

at what point do the graph of the function f(x)=(x^2)+4x-1 have a horizontal tangent?
*Tuesday, January 14, 2014 at 3:04pm*

**Calculus**

Are you sure x is not approaching zero? then you have cos x = 1 - x^2/2 ...... and sin (x^2/2)/x = x/2 = 0
*Tuesday, January 14, 2014 at 1:54pm*

**Calculus**

lim x to 1 of (sin(1-cosx))/x
*Tuesday, January 14, 2014 at 1:34pm*

**Calculus**

To be continuous, the two graph must be "linked" at x = 3 that is, x^2 - 1 = 2ax for x = 3 9-1 = 6a a = 8/6 = 4/3 so the two functions are f(x) = x^2 - 1 for x < 3 and f(x) = (8/3)x for x≥ 3 (note that (3,8) would be on both graphs, ignoring the restrictions)
*Monday, January 13, 2014 at 11:27pm*

**Calculus**

correction its f(x)=(x^2)-1 not f(x)=(x^2)-a
*Monday, January 13, 2014 at 10:24pm*

**Calculus**

For what value of a is f(x)=(x^2)-a x<3 f(x)=2ax x(> or =)3 continuous at every x?
*Monday, January 13, 2014 at 10:23pm*

**calculus**

as you know, the slope of the tangent line is just the derivative. So, y'(x) = -4x giving us y'(2) = -8 and now we have a point and a slope, so the line is y-1 = -8(x-2)
*Monday, January 13, 2014 at 8:43pm*

**calculus**

a.Find the slope of the tangent line to the curve y=9-2x^2 at the point (2,1) b.Find an equation of this tangent line.
*Monday, January 13, 2014 at 4:39pm*

**physics**

T = 2 pi sqrt(L/g) I could compute T for the two values of g but the change in g is small so better to use calculus for sport let's make L = 1 meter T = 2 pi sqrt L (g^-.5) =2 pi sqrt(1/9.81) = 2 Seconds for L = 1 and g = 9.81 dT/dg = 2 pi sqrt L(-.5)(g^-1.5) so dT = - pi ...
*Monday, January 13, 2014 at 2:21pm*

**calculus**

well, maybe they're as accurate as your spelling... Your very first line is in error: should be -3x^2 toward the end. Fix that, and everything's ok. If you visit wolframalpha.com and enter derivative (x^2 + 2x - 6)^2 (1-x^3)^2 you will get several different ways of ...
*Monday, January 13, 2014 at 5:10am*

**calculus**

i need to find the derivative using chain rule: (x^2 + 2x - 6)^2 (1-x^3)^2 i got the answer from a site but the problem is i cannot get my work to match up with the answer, i dont know what im doing wrong. answer: 10x^9 + 36x^8 − 64x^7 − 182x^6 + 168x^5 + 80x^4 + ...
*Monday, January 13, 2014 at 3:00am*

**Integral Calculus**

s = integral v dt s = t^3 - 5 t^2 + 15 t + c when t = 1, s = 8 8 = 1 - 5 + 15 + c c = -3 so s = t^3 - 5 t^2 + 15 t - 3 then when t = 0, s = -3 when t = 5 s = 125-125 + 75 -3 = 72 so the jeepney (whatever that is) went from -3 to + 72 or 75 ft
*Sunday, January 12, 2014 at 7:08pm*

**Integral Calculus**

A jeepney's velocity over time in ft/sec as it moves along the x-axis is governed by the function v(t)=3t^2-10t+15. If the jeep's position at t= 1 second is 8ft, what is the total displacement made by the jeepney from t=0 to t=5sec? please I need help, i just need the ...
*Sunday, January 12, 2014 at 7:01pm*

**pre-calculus**

thank you :)
*Sunday, January 12, 2014 at 6:45pm*

**pre-calculus**

whew :)
*Sunday, January 12, 2014 at 6:40pm*

**pre-calculus - typo**

Scratch all that, since s = d/√3 v = s^3 = d^3/√27 a = 6s^2 = 2d^2
*Sunday, January 12, 2014 at 6:38pm*

**pre-calculus**

edge length s diagonal of a side = s sqrt 2 then right triangle with legs s and s sqrt 2 d^2 = s^2 + 2 s^2 = 3 s^2 d = s sqrt 3 s = d/ sqrt 3 = d (3)^-.5 area = 6s^2 = 6d^2 (3)-1 = 6d^2/3 =2 d^2 vol=s^3=[d(3^-.5)]^3 = d^3 /(3 sqrt 3) = (d^3 sqrt 3)/9
*Sunday, January 12, 2014 at 6:36pm*

**pre-calculus**

d^2 = 3s^2 so, s = (d/3)^(1/2) v = s^3 = (d/3)^(3/2) A = 6s^2 = 2d
*Sunday, January 12, 2014 at 6:31pm*

**pre-calculus**

Express the edge length of a cube as a function of the cube's diagonal length d. Then express the surface area and volume of the cube as a function of the diagonal length.
*Sunday, January 12, 2014 at 6:21pm*

**Calculus**

Find the position s as a function of time t from the given velocity v=ds/dv. Then evaluate the constant of integration so that s = s0 when, t = 0. 1. v = 3t2 , s0 = 4 I don't quite understand what the question is asking or what is s0 mean. Than
*Sunday, January 12, 2014 at 11:24am*

**Calculus**

V = π r ² h = 1ft³ h = 1/π r ² SA = π r ² + 2 π r h = π r ² + 2 π r * 1/π r ² = π r ² + 2 / r SA ' = 2π r - 2 / r² = 0 2π r = 2 / r² r³ = 1/π r = 1/ ³√π...
*Sunday, January 12, 2014 at 8:01am*

**calculus**

Make a diagram let the foot of the ladder be x ft from the fence let the ladder reach y ft above the ground I see similar triangle so set up a ratio 8/x = y/(x+1) xy = 8x+8 y = (8x+8)/x let the length of the ladder be L L^2 = (x+1)^2 + y^2 = (x+1)^2 + [(8x+8)/x]^2 = x^2+2x+1...
*Sunday, January 12, 2014 at 7:35am*

**Pre-Calculus**

I know that sin (π/6) or sin 30° = .5 so (π/4)(x-6) = π/6 or (π/4)(x-6) = π - π/6 = 5π/6 (in II) case 1 times 12 and divide by π 3(x-6) = 2 3x - 18 = 2 3x=20 x = 20/3 case 2 times 12 , and divide by π 3(x-6) = 10 3x - 18=10 3x...
*Saturday, January 11, 2014 at 9:02pm*

**Pre-Calculus**

Here is correct question: Solve sin (pi/4*(x-6))=0.5, 0<x<360
*Saturday, January 11, 2014 at 8:10pm*

**Pre-Calculus**

Find an equation of the line that satisfies the given conditions Through (-2,4); slope -1 How would you solve that by using the point slope form and writing it in the general equation of a line? ax +by + c =0
*Saturday, January 11, 2014 at 7:30pm*

**Pre-Calculus**

y = -1 x + b 4 = -1(-2) + b 2 = b so y = -x + 2
*Saturday, January 11, 2014 at 7:21pm*

**Pre-Calculus**

Find an equation of the line that satisfies the given conditions Through (-2,4); slope -1
*Saturday, January 11, 2014 at 7:10pm*

**Pre-Calculus**

maybe you mean sin [pi(x-6)/4] sin (pi/6) = .5 (x-6)/4 = 1/6 6 x - 36 = 4 x = 40/6 = 20/3
*Saturday, January 11, 2014 at 5:43pm*

**Pre-Calculus**

Solve sin (pi/4(x-6))=0.5, 0<x<360
*Saturday, January 11, 2014 at 4:18pm*

**Calculus**

The following table displays the number of HIV diagnoses per year in a particular country. Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 Diagnoses 2512 2343 2230 2113 2178 2495 2496 2538 2518 a) Using Curve Expert or another curve modelling program, determine an equation ...
*Saturday, January 11, 2014 at 12:17am*

**calculus**

y= (x) / ((x^2)-1)^(-1/2) do you have a typo? That is the same as: y = x (x^2-1)^(+1/2) dy/dx = x(1/2)(x^2-1)^-.5(2x) +(x^2-1)^.5 = x^2(x^2-1)^-.5 + (x^2-1)^.5
*Friday, January 10, 2014 at 5:58pm*

**calculus**

there are tutoring chatrooms, but not here. To play around with derivatives and much more, visit wolframalpha.com. For this problem, enter derivative (x) / ((x^2)-1)^(-1/2) and it will show you (2x^2-1)/√(x^2-1) and you can click on the "step-by-step solution" ...
*Friday, January 10, 2014 at 5:49pm*

**calculus**

i need to find the derivative using chain rule and show steps: y= (x) / ((x^2)-1)^(-1/2) and is it possible to have free tutoring online like through a chatroom? i am in need of help
*Friday, January 10, 2014 at 5:44pm*

**Pre calculus**

since distance = speed * time, 50x + 40(1-x) = 48
*Thursday, January 9, 2014 at 6:16pm*

**4 CALCULUS study problem**

Never mind! I solved them all and got them right. :)
*Thursday, January 9, 2014 at 2:14pm*

**4 CALCULUS study problem**

1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/...
*Thursday, January 9, 2014 at 1:23pm*

**Pre calculus**

A woman went for a 48km drive in her car, partly in town in the country. She drove at an average of 40km/hr in town and an average speed of 50km/hr in the country. If the drive took her one hour altogether then how long did she drive in the country?
*Thursday, January 9, 2014 at 12:53pm*

**calculus**

f'(x)=[-5* 1/2]*[1/sqrt(x^2-4x+1)] * (2x-4)
*Wednesday, January 8, 2014 at 10:22pm*

**algebra**

Since you have no calculus (the easy way), lets do it this way. Obviously, it is a parabola going up, then down. So look for the half way point between both x intercepts. So find the x intercepts. 0=-480(F-3.25)^2 + 5270 F^2+6.5F+3.25^2-5070/480=0 check that, then put it in ...
*Wednesday, January 8, 2014 at 10:21pm*

**calculus**

updating previous question so easier to understand im using chain rule to answer this question: f(x) = -5 sqrt(x^2 - 4x + 1) what is f'(x)? tried to use calculator to get the answer to base my answer off of but this is a nonreal number. does this mean i will have to use ...
*Wednesday, January 8, 2014 at 9:57pm*

**calculus**

For the life of me, I dont see a question. Am I missing something? I assume the power of the quadratic is 1/2, it is hard to tell from your grouping symbols.
*Wednesday, January 8, 2014 at 9:30pm*

**calculus**

im using chain rule to answer this question: -5 ((x^2) - 4x + 1)^(1)/(2) tried to use calc to get answer to base off of but this is a nonreal number. does this mean i will have to use imaginary numbers? please help
*Wednesday, January 8, 2014 at 9:07pm*

**Pre Calculus**

Been there, done that a few minutes ago http://www.jiskha.com/display.cgi?id=1389226155
*Wednesday, January 8, 2014 at 8:47pm*

**Pre Calculus**

Find the exact value of sin(1/2arcsin(-7/25))
*Wednesday, January 8, 2014 at 8:23pm*

**Pre-Calculus**

sin A = x/sqrt(x^2+4) draw that right triangle side adjacent to A is sqrt (x^2+4-x^2) = sqrt 4 = 2 so cos A = 2/sqrt(x^2+4) sec A = 1/cos A = sqrt(x^2+4)/2
*Wednesday, January 8, 2014 at 7:47pm*

**Calculus**

I bet you mean y = x/(x+3) The denominator is 0 when x = -3 so x = -3 is not in our domain. Otherwise the domain is all real numbers. Inverse x = y/(y+3) x y + 3 x = y y(x-1) = -3 x y = -3x/(x-1) domain is all real numbers except x = 1 f[f^(-1)] = f[-3x/(x-1) ] = [-3x/(x-1...
*Wednesday, January 8, 2014 at 7:41pm*

**Calculus**

Let f(x)=(x/x+3). Find Domain Find f^(-1) and its domain Verify f*f^(-1)=f^(-1)*f=x Find (f(x+h)-f(x))/h.
*Wednesday, January 8, 2014 at 7:15pm*

**Pre-Calculus**

Write an algebraic expression in x for sec(arcsin(x/sqrt(x^(2) +4)))
*Wednesday, January 8, 2014 at 7:10pm*

**Pre-Calculus**

y = a x^2 + b x + c if x = 1 3 = a + b + c if x = 2 5/2 = 4 a + 2 b + c if x = -1 1 = a - b + c we have 3 linear equations with 3 unknowns. There are many ways to solve but lets try to eliminate c first 1 a +b + c = 3 4 a+2b + c = 5/2 ------------------ subtract - 3 a - 1 b = ...
*Wednesday, January 8, 2014 at 1:55pm*

**Pre-Calculus**

Find a quadratic function f such that f(1)=3, f(2)=5/2, f(-1)=1 I came out with 3 a+b+c=3, 4a+2b+c=5/2, a-b+c=1. And I know that I should do synthetic division but I am stuck.
*Wednesday, January 8, 2014 at 1:21pm*

**Calculus**

You have to specify your domain. For log(x-1) you need x>1 For log(1-x) you need x<1 May times you will find it written that ∫ dx/x = log |x| + C just for this reason.
*Wednesday, January 8, 2014 at 5:51am*

**Calculus**

Integrate x dx/(1-x). I have proceeded thus- Int xdx/(1-x)=int -(x-1+1)/(x-1) =-Int[1+ 1/(x-1)]dx =-Int dx-Int dx/(x-1) =-x-log(x-1). On differentiating, we get original expression- d/dx[-x-log(x-1)]=-1-1/(x-1)=-x/(x-1)=x/(1-x). However, the answer in the book is -x-log...
*Wednesday, January 8, 2014 at 4:04am*

**AP Calculus AB**

y´=3(cosx)²(-sinx) -3(sinx)²cosx= = - 3 sinx cosx(cosx+sinx)
*Friday, January 3, 2014 at 3:13pm*

**AP Calculus AB**

Find the derivative of: y=(cosx)^3-(sinx)^3
*Friday, January 3, 2014 at 12:23pm*

**AP Calculus AB**

1. assign x and y values to the length and width of each box within the pens, and use the formula 600=9x+8y to isolate the y value (i got the 8 and 9 by counting the number of length and width values existed in the diagram) you should get y=(600-9x)/8 2. then, since they asked...
*Thursday, January 2, 2014 at 5:09pm*

**math**

I have a hunch you actually mean: g(x)=x/ (x-7) because that is much more exciting than what you typed. Now clearly if x = 7 we have a serious problem with this function. It has a zero for its denominator and is therefore undefined at x = 7. Now what if x = 8 ? g(8) = 8/1 = 8 ...
*Friday, December 27, 2013 at 2:15pm*

**Pre calculus**

First step, add/subtract multiples of 180° until you have an angle between -180° and 180° If θ is in QI, you're done. If θ is in QIII, reflect through the origin. That is, add 180°. Done. If θ is in QIV, y<0, so reflect through the x-axis...
*Thursday, December 26, 2013 at 12:12am*

**Pre calculus**

1. Reference angle for -280 degrees? I do not understand why you add 360???? 2. Theta=? If reference angle =60 and theta terminates in quadrant 3. What does terminate mean? Please show work and explain why. My priority is to understand the concept ... 3. Exact value for sec ...
*Wednesday, December 25, 2013 at 10:30pm*

**calculus-changing voltage**

V = IR V is increasing at the rate of 1 volt/sec while I is decreasing at the rate of 1/3 amp/sec. Let t = time in sec What is dV/dt? What is dI/dt? Find the rate at which R is changing when V = 12 volts and I = 2 amp
*Monday, December 23, 2013 at 12:45pm*

**calculus**

if these are logs base ten 10^log x = x = 10^5.3027 = 10^5 * 10^.3027 = 10^5 * 2.007705 or 2.008 * 10^5
*Saturday, December 21, 2013 at 5:25pm*

**calculus**

Evaluate to four significant digits log x = 5.3027
*Saturday, December 21, 2013 at 3:48pm*

**CALCULUS - Check my answers :)**

#5 At a time of t seconds after the boy passed under the balloon, distance covered by boy = 15t height of balloon = 5 + 5t let the distance between them be d ft d^2 = (5t+5)^2 + (15t)^2 2d dd/dt = 2(5t+5)(5) + 2(15t)(15) dd/dt =( 5(5t+5) + 15(15t) )/d when t = 3, d^2 = 20^2 + ...
*Thursday, December 19, 2013 at 10:30pm*

**CALCULUS - Check my answers :)**

#4, the way you typed it, ... y' = 2x/3 when x = 8, y' = 16/3, which is not found in any of your choices as the slope if you meant , f(x) = x^(2/3) then y' = (2/3)x^(-1/3) = (2/3)(1/x^(1/3)) when x = 8 y' = (2/3)(1/2) = 1/3 the only one that has a slope of 1/2 ...
*Thursday, December 19, 2013 at 10:18pm*

**CALCULUS - Check my answers :)**

And 5 would be 25?
*Thursday, December 19, 2013 at 9:28pm*

**CALCULUS - Check my answers :)**

Is number 4 the last one?
*Thursday, December 19, 2013 at 9:24pm*

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