Related Rates Question?
918095
The distance between the streetlight and the girl is x\nThe distance between the streetlight adn the shadow is y.\nWe know that y varies with time : y = 4ft * t \nwhere t is the time in seconds.\nThe ratio of y to the height of the streetlight is the same as\nthe ration of x and the height difference between the head of the girl and the streetlight (triangles with same angles)\nSo\nx / (16ft-4ft) = y / 16ft\nx / 12ft = 4ft * t / 16ft = t / 4\nx =  3ft * t\nSo the girl is walking at 3ft/s
795870
5) W = M + 6, W + M = 30\n\n6) x + (x + 1) + (x + 2) = 153\n\n7) 5x = 3(x + 2) + 14\n\n8) x + (x + 2) + (x + 4) = 54\n\n9) Rate = 356, Distance = 1424\n1424 = 356 * Time
965285
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....
806819
A.\n\nF=ma, so, a= F/m\n\nIn this case, a=7/5ms^2\n\nB.\n\nv=u+at, u=0 (initial velovity), t = 2 sec, so\n\nv=0+7/5 x 2 metres per second = 14/5 m/s\n\nC.\n\nThe time to reach Sasha will be:\n\nt = d/v = 11.2/(14/5) seconds\n\nI will let you use your calculator!
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
888670
First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.\nNow, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.\nWe want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).\nNow, we need to differentiate both sides with respect to time.\nThus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).\nWe need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t.  So the overall equation becomes:\nd/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.\nNow, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.\n\nI'm likely to have made a mistake, but hopefully that gives you the idea of what to do.
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
932092
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.
887499
avg speed = total distance / total time: \n\ntime 5 min = 1/12 hrs  and \ndistance covered is 140 miles/hr x 1/12 hrs \n= 11 2/3 miles\n\nRemaining distance would take \n[300-11 2/3] / 160 hrs \n=288 1/3 dived ed by 160 \n= 865/480 hrs\n\nTotal time is 1/12 hrs + 865/480 hrs \n= 905/480 hrs giving \n\navg speed as 300 x 480 /905 \n= 159.13 miles/hrs (approx) Ans\nRest of the information is given to 'entertain' the students!
271582
3.84*10^5 km ( 1000m /  1km ) =  384000000m\n\n\n  distance / rate = time:\n\n   384000000m/3.0*10^8ms = 1.28sec
759319
acceleration=rate of change of velocity\n=80*5280/3600*6ft/sec^2\n=19.6 ft/sec^2\n\n2.0=(70*5280/3600)^2-2f(157)\n314s=67.1ft/s^2
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
1129113
It is 9.8 m/sec , some they have made it 10 :)
577594
12.66 mph\n\nor 12.65558195 to be (more) exact!
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
806280
There has been one careless conversion of 50 mph to not quite the correct number of ft/sec.  Remember that 60 mph = 88 ft/sec exactly.\n\nS = V^2 / 2A, and V = 88 * 50 / 60, so S = 74.691. . . feet.
365841
To solve this question, you need to set up a basic proportion:\n\n2.5/18 = x/60 (to find the rate, you must solve for x)\n\ncross-multipy so that 2.5(60) = 18x\n\nnow you have 150 = 18x\n\ndivide both sides by 18 to get x by itself\n\nx = 8.3...\n\nx = 8 1/3 mph.