Friday

April 18, 2014

April 18, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Calculus Help **

In the following x,y and (a) are all variables. Show steps please! Thank you! 1) y^2 = x^2+a^2 2) y^2+ay = x^2+ax+a^2
*Thursday, March 13, 2014 at 8:19pm*

**Pre Calculus**

rcosθ = 3 r = 3secθ
*Thursday, March 13, 2014 at 5:04pm*

**Pre-Calculus**

r = 5sinθ r^2 = 5r sinθ x^2+y^2 = 5y x^2 + (y - 5/2)^2 = 25/4
*Thursday, March 13, 2014 at 5:03pm*

**CALCULUS ECONOMICS**

Consider the problem of a competitive firm which has fixed costs of $1000, semi-fixed-costs of $1000, and variable costs given by q^2. QUESTION: What is the maximum market price at which the firm decides to supply zero?
*Thursday, March 13, 2014 at 4:10pm*

**CALCULUS ECONOMICS**

Consider the problem of a competitive firm which has fixed costs of $1000, semi-fixed-costs of $1000, and variable costs given by q2. QUESTION: What is the maximum market price at which the firm decides to supply zero?
*Thursday, March 13, 2014 at 4:09pm*

**CALCULUS ECONOMICS**

Consider a market in which consumption of the good being traded generates a positive externality. There are 100 identical consumers, each with a utility function given by (1/2)*(q^(1/2))+m +(G^(1/2)) where G denotes the total level of consumption in the market. The good is ...
*Thursday, March 13, 2014 at 4:08pm*

**CALCULUS ECONOMICS**

Consider a market in which consumption of the good being traded generates a positive externality. There are 100 identical consumers, each with a utility function given by (1/2)*(q^(1/2))+m +(G^(1/2)) where G denotes the total level of consumption in the market. The good is ...
*Thursday, March 13, 2014 at 4:06pm*

**CALCULUS ECONOMICS**

Consider an oligopolistic market with two firms. Each of them produces using a cost function given by c(q)=q^2. The aggregate demand in the market is given by 1000−p. Suppose that, in order to increase production, the government gives the firms a $100 per-unit produced ...
*Thursday, March 13, 2014 at 3:58pm*

**CALCULUS ECONOMICS**

Consider the same setting as in the previous question. Suppose that firms are NOT owned by consumers. Let s denote the size of the per-unit subsidy/tax given to the firms. Let positive values of s denote subsidies, and negative values of s denote taxes. QUESTION: What is the ...
*Thursday, March 13, 2014 at 3:54pm*

**CALCULUS ECONOMICS**

Consider an oligopolistic market with two firms. Each of them produces using a cost function given by c(q)=q^2. The aggregate demand in the market is given by 1000−p. Suppose that, in order to increase production, the government gives the firms a $100 per-unit produced ...
*Thursday, March 13, 2014 at 3:53pm*

**CALCULUS ECONOMICS**

Consider an economy in which a monopolistic firm serves two identical, but separate markets, called A and B. The aggregate inverse demand in each market is given by 1000−q. The cost function for the monopolist is given by (qA+qB)^2, where qA andqB denotes the amount sold...
*Thursday, March 13, 2014 at 3:52pm*

**CALCULUS ECONOMICS**

Consider a market in which aggregate demand is given by 1000−10p, and aggregate supply is given by 10p, where p denotes the market price. QUESTION: What is the maximum amount of revenue that the government can raise using a per-unit sales tax on consumers?
*Thursday, March 13, 2014 at 3:50pm*

**CALCULUS ECONOMICS**

Consider the problem of a rational consumer with an experienced utility function given by 8*x^(1/2)+m. Let p=$1 p/unit denote the market price of good x. Suppose that, initially, the firm selling the good matches his purchases as follows: for every x units that he buys, he ...
*Thursday, March 13, 2014 at 3:49pm*

**CALCULUS ECONOMICS**

Consider the problem of a rational consumer with an experienced utility function given by 8x√+m. Let p=$1 p/unit denote the market price of good x. Suppose that, initially, the firm selling the good matches his purchases as follows: for every x units that he buys, he ...
*Thursday, March 13, 2014 at 3:47pm*

**Pre-Calculus**

r = sqrt (4^2+3^2) which we know is 5 because of 3 4 5 right triangle so we do not need c^2=a^2+b^2 -4 = 5 cos theta cos theta = -4/5 theta = 143 if you use your calculator but that would be wrong because we are in QUADRANT 3 where sin and cos BOTH negative cos^-1 4/5 = 36.9 ...
*Thursday, March 13, 2014 at 3:38pm*

**Pre Calculus**

r = 3, theta = 0
*Thursday, March 13, 2014 at 3:29pm*

**Pre-Calculus**

Write the polar equation in rectangular form... r=5sintheta
*Thursday, March 13, 2014 at 3:28pm*

**Pre-Calculus**

Find the polar coordinates of each point with the given rectangular coordinates. Use degrees. (-4,-3)
*Thursday, March 13, 2014 at 3:25pm*

**Pre-Calculus**

r cos theta = x = -2.598 = - 3(sqrt3/2) r sin theta = y = +1.5 (-2.6 , 1.5 ) or exact: (-3 sqrt 3 /2 , 3/2 )
*Thursday, March 13, 2014 at 3:23pm*

**Pre Calculus**

Write the rectangular equation in polar form... x=3
*Thursday, March 13, 2014 at 3:17pm*

**Pre-Calculus**

Find the rectangular coordinates of the point with the given polar coordinates. (3, 150°)
*Thursday, March 13, 2014 at 3:15pm*

**Calculus**

well, x^2/(2logx) = x^2/log(x^2), so if you can figure out the limit of x/logx, you're ok. Since x->0 and logx->-∞, x/logx -> 0/-∞ = 0 -------------- e^-x -> 0 as x->∞, so the upper limit -> 0 e^0 = 1, so the lower limit -> 2
*Thursday, March 13, 2014 at 12:45pm*

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. ...
*Thursday, March 13, 2014 at 2:45am*

**calculus**

nn
*Wednesday, March 12, 2014 at 6:15pm*

**CALCULUS HELP**

If we let u = 8x-3 v = 6x+7 then we can make the expression a little easier to read during the computational steps: f' = (5u^4*8*v^12-u^5*12v^11*6)/v^24 Now, you can cancel out v^11 top and bottom to get f' = (40u^4*v - 72u^5)/v^13 Now factor out 4u^4 to get f' = ...
*Wednesday, March 12, 2014 at 5:37pm*

**AP Calculus AB**

5
*Wednesday, March 12, 2014 at 3:48pm*

**CALCULUS HELP**

Find the derivative of the following function showing your work and fully simplifying your answer. STEP BY STEP PLEASE!!! f(x)=(8x-3)^5/(6x+7)^12 THANK YOU SO MUCH!!!
*Wednesday, March 12, 2014 at 2:34pm*

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. ...
*Wednesday, March 12, 2014 at 6:13am*

**Calculus**

limit as x->2- = 5 limit as x->2+ = 5 since one-sided limits are the same, and f(2)=5, f is continuous Do you have to do the delta-epsilon limit proof?
*Wednesday, March 12, 2014 at 5:29am*

**Calculus**

Show, using limits, that f(x) = x2 – x + 3, is continuous at x = 2.
*Wednesday, March 12, 2014 at 2:15am*

**Calculus**

product rule and chain rule -- if y = uv y' = u'v + uv' Here u=csc^2 x and v = arccos(1-x^2) y' = -2 cscx cscx cotx arccos(1-x^2) + csc^2 x 1/√(1-(1-x^2)^2) * -2x kind of messy, no? well, recall that if cos u = 1-x^2, sin u = x, so we really have y = -csc...
*Wednesday, March 12, 2014 at 12:12am*

**Calculus- typo**

64/6 + 8 + 20 - ( -64/6 + 8 -20 ) 64/3 + 0 + 40 184/3
*Wednesday, March 12, 2014 at 12:06am*

**Calculus**

what you have tried is the volume when rotating the region about the x-axis. But, of course, the volume of a disk is pi r^2 dx, not pi r^3 dx In addition, your radius is 3-x, not y, since you are rotating about the line x=3, not y=0. So, if you want to use discs, you need to ...
*Wednesday, March 12, 2014 at 12:01am*

**Calculus**

At time t, using the law of cosines, d^2 = (7t)^2 + (5t)^2 - 2(5t)(7t)cosπ/4 = 74t^2 - 35√2 t^2 = (74-35√2)t^2 so, 2d dd/dt = 2(74-35√2)t Now, plug in your numbers to get d(15), and solve for dd/dt at t=15 Remember to use appropriate units
*Tuesday, March 11, 2014 at 11:47pm*

**Calculus**

as usual, draw a diagram and you will see that if θ=0 when x=0, tanθ = x/4 so, sec^2θ dθ/dt = 1/4 dx/dt Now, you have θ and dθ/dt, so just solve for dx/dt
*Tuesday, March 11, 2014 at 11:43pm*

**Calculus**

A lighthouse is located on a small island 4 km away from the nearest point P on a straight shoreline and its light makes three revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P? (Round your answer to one decimal place.)
*Tuesday, March 11, 2014 at 10:12pm*

**Calculus**

Two people start from the same point. One walks east at 5 mi/h and the other walks northeast at 7 mi/h. How fast is the distance between the people changing after 15 minutes? (Round your answer to three decimal places.)
*Tuesday, March 11, 2014 at 10:11pm*

**Calculus Check my answer**

integral (x^2/2 + x + 5)dx x^3 /6 + x^2/2 + 5 x at 4 minus at -4 64/6 + 8 + 20 - ( -64/6 + 8 -20 ) 64/3 + 20 + 40 64/3 + 180/3 = 244/3 = 81 1/3
*Tuesday, March 11, 2014 at 10:00pm*

**Calculus**

Can you tell me if im doing this right? Find the volume of the region obtained by revolving the area below about the line x=3. y=x3, x=2, y=0 v=pi[2,0] (x^3)^3dx= pi[2,0] x^9 v=pi/10 x^10(2,0)= 1024/10 pi
*Tuesday, March 11, 2014 at 9:46pm*

**Calculus Check my answer **

Find the area between the curves y= x^2/2 +2 and y = –x – 3 on the interval –4 ≤ x ≤ 4. I got 40/3 is that correct?
*Tuesday, March 11, 2014 at 9:43pm*

**Calculus**

Can someone help me with finding the derivative of this function? y = -( csc x)2 (cos-1(1-x2))
*Tuesday, March 11, 2014 at 9:42pm*

**Calculus Help**

just use the quotient rule and chain rule. If f = u/v f' = (u'v-uv')/v^2 f' = 5(8x-3)^4(8)(6x+7)^12 - (8x-3)^5(12)(6x+7)^11(6)]/(6x+7)^24 = -16(8x-3)^4(21x-31)/(6x+7)^13
*Tuesday, March 11, 2014 at 4:30pm*

**Calculus Help**

Find the derivative of the following function showing your work and fully simplifying your answer. f(x)=(8x-3)^5/(6x+7)^12 Thank you!!!
*Tuesday, March 11, 2014 at 3:36pm*

**FREE BODY DIAGRAM calculus**

12
*Tuesday, March 11, 2014 at 12:46pm*

**calculus**

The circle in the x-y plane is x^2 + (y - a/2)^2 = a^2/4 x^2 = a^2/4 - (4y^2-4ay+a^2)/4 = (y^2-ay)/4 The limits of integration in the x-y plane are 0 < y < a 0 < x < (1/2)√(y^2-ay) then use symmetry and multiply by 4
*Tuesday, March 11, 2014 at 12:06am*

**calculus**

Find the surface area of the part of the sphere x^2+y^2+z^2=a^2 inside the circular cylinder x^2+y^2=ay (r=a*sin(θ) in polar coordinates), with a>0. First time posting on this website, sorry for the lack of details on my attempts but I am really not sure where to start...
*Monday, March 10, 2014 at 10:22pm*

**Calculus**

the answers are the same √[(x-3)/2] = √[(x-3)/2 * 2/2] = √(2x-6)/2
*Monday, March 10, 2014 at 8:16pm*

**Calculus**

I think you did it correctly. Of course the domain is limited because the answer is imaginary if x <3
*Monday, March 10, 2014 at 6:39pm*

**Calculus**

Determine the equation of the inverse function if f(x) = 2x^2+3, and x≥0. The answer is supposed to be f^-1(x)=[√(2x-6)]/2. This is what I did: x=2y^2+3 2y^2=x-3 y^2=(x-3)/2 y=√[(x-3)/2] Did I do something wrong? Thanks!
*Monday, March 10, 2014 at 5:37pm*

**Calculus**

If f(x)= -4x^2+7 and x≤0, what is the equation of the inverse function? The answer is supposed to be f^-1(x)= -[√(7-x)]/2, but this is what I did: x=-4y^2+7 -4y^2=x-7 y^2=(x-7)/-4 y= √[(x-7)/-4] Did I do something wrong? By the way, for the original function...
*Monday, March 10, 2014 at 2:12pm*

**Calculus Help**

3/31: t=90 4/21: t=111 L'(t) = 2.8cos[(2π/365)(t − 80)](2π/365) now just plug in t=90 and t=111 The answer, of course, will be in hr/day.
*Monday, March 10, 2014 at 5:11am*

**Calculus**

Since the derivative is the slope of the tangent line, we need -2sinx - 2cosx*sinx = 0 -2sinx(1+cosx) = 0 sinx = 0 means x = nπ cosx = -1 means x = (2n+1)π so, f(x) has a horizontal tangent at x=nπ
*Monday, March 10, 2014 at 5:03am*

**Calculus **

Find all points on the graph of the function f(x) = 2 cos x + cos^2 x at which the tangent line is horizontal. (Use n as your arbitrary integer.) smaller y-value (x,y)= larger y-value (x,y)=
*Sunday, March 9, 2014 at 10:40pm*

**Calculus Help**

A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is L(t) = 12 + 2.8 sin[(2π/365)(t − 80)]. Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and April 21. (Assume there are...
*Sunday, March 9, 2014 at 9:56pm*

**calculus**

r' = 2.4/(3t+6) r = 0.8 log(3t+6)+c solve for c at t=0 in 0.8 log6 + c = 3.8 now just plug in t=16 for the final answer.
*Saturday, March 8, 2014 at 10:02pm*

**calculus**

r' = 2.1/(t+5) r = 2.1 log(t+5) + C at t=0, 3.8 = 2.1 log5 + C C = 3.8-2.1log5 = 0.42 so, r = 2.1 log(t+5)+0.42 now just plug in t=27
*Saturday, March 8, 2014 at 9:59pm*

**calculus**

a = 2t-4, not 2t^2-4
*Saturday, March 8, 2014 at 9:55pm*

**Calculus**

It requires 8 inch pounds of work to stretch a certain spring 2 inches from its rest position. Assuming that the spring follows hooke's law, what is k? Please answer this question. Thanks for your answers in advance.
*Saturday, March 8, 2014 at 8:13pm*

**calculus**

A. Integrate the velocity x = (t^3)/3-2t+3t substitute 4 and 6 [(4^3)/3 - 2(4)+3(4)] - [(6^3)/3 - 2(6)+3(6)] = m? B. get the derivatives of velocity a = 2t^2 - 4 substitute t=6 2(6)^2-4 = m/s^2?
*Saturday, March 8, 2014 at 8:00pm*

**Integral calculus**

It requires 8 inch pounds of work to stretch a certain spring 2 inches from its rest position. Assuming that the spring follows hooke's law, what is k? Thanks for your answers! :)
*Saturday, March 8, 2014 at 7:36pm*

**calculus**

a=2.675543 b=7.23334
*Saturday, March 8, 2014 at 6:04pm*

**calculus**

The velocity of a skateboard is v(t) = t^2 - 4 t + 3 m/s when moving in a straight line. A. Find the the change in displacement of the skateboard between 4 seconds and 6 seconds. (Note this may or may not be negative, meaning it goes in the opposite direction, if so then be ...
*Saturday, March 8, 2014 at 6:03pm*

**calculus**

\int_{4}^{13} f(x) \,dx - \int_{4}^{11} f(x) \,dx = \int_{a}^{b} f(x) \,dx where a= and b= .
*Saturday, March 8, 2014 at 6:02pm*

**calculus**

pretty easy upper limit of b is 25.15 lower limit of a is 16.23
*Saturday, March 8, 2014 at 6:01pm*

**calculus**

A long narrow piece of land gets flooded each year by a river. The flooded area is in the shape of the area under the curve y = 2.3 x^3 and above the x-axis, for 0 \le x \le 3.2. All the distances are in metres.
*Saturday, March 8, 2014 at 6:01pm*

**calculus**

a. The value of \displaystyle \int_{-2}^{-1} \frac{14}{ 4 x } dx is b. The value of \displaystyle \int_{1}^{2} \frac{14}{ 4 x } dx is
*Saturday, March 8, 2014 at 6:01pm*

**calculus**

The following definite integral can be evaluated by subtracting F(B) - F(A), where F(B) and F(A) are found from substituting the limits of integration. \int_{0}^{4} \frac{1600 x +1200 }{(2 x^2 +3 x +1)^5}dx After substitution, the upper limit of integration (B) is : and the ...
*Saturday, March 8, 2014 at 6:00pm*

**calculus**

At a summer campfire, the radius of a marshmallow on a stick expands at the rate of \ {r ' (t)} = \frac{2.4 }{3 t + 6} mm/s where t is the time of heating in seconds. Initially the radius was 3.8 mm. Find the radius after 16 seconds using the following steps: When ...
*Saturday, March 8, 2014 at 6:00pm*

**calculus**

Ice cream drips out of the bottom of an ice cream cone on a hot day at a rate of r(t) mL per second, as a child eats it slowly, where t is in seconds. If r(t) = 10 e^{-k t}, complete the definite integral expressing the quantity of ice cream lost in the first 3 minutes(s). (...
*Saturday, March 8, 2014 at 5:59pm*

**calculus**

At a summer campfire, the radius of a marshmallow on a stick expands at the rate of \ {r ' (t)} = \frac{2.1 }{1 t + 5} mm/s where t is the time of heating in seconds. Initially the radius was 3.8 mm. Find the radius after 27 seconds using the following steps: When ...
*Saturday, March 8, 2014 at 5:59pm*

**calculus**

Find the area of the region under the curve y = 16 e ^{4 x} between x = -1.4 to x =1.4 .
*Saturday, March 8, 2014 at 5:58pm*

**Calculus**

what's the trouble? Straightforward integration: a = ∫[-2.6,2.6] 12e^(3x) dx = 4e^3x [2.6,-2.6] = 9762.41
*Saturday, March 8, 2014 at 5:47pm*

**typo, gravity wrong**

a) Hey, you know what h is when t = zero! b ) h = -4(1) + 16(1) + 9 = 21 c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola t^2 - 4 t = -h/4 + 9/4 t^2 - 4 t + 4 = -h/4...
*Saturday, March 8, 2014 at 1:54pm*

**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*

**Calculus Help Please!!!**

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*

**Calculus Help Please!!! **

--- 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*

**Calculus Help Please!!!**

(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*

**Calculus Help Please!!! **

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*

**Calculus Please help!**

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*

**Calculus...URGENT..show steps please**

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*

**Calculus...URGENT..show steps please**

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*

**Calculus Please help!**

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*

**Calculus Please help!**

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*

**Calculus Please help!**

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*

**Calculus Please help!**

$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*

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