Monday
March 17, 2014

Homework Help: Math: Calculus

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

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

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

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

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