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April 20, 2014

April 20, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**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*

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

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*

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