Gloria started at teh beginning of a jogging trail and walked at the rate of 3 miles per hour.  One-half hour later, Denise started at the beginning of the same trail and ran at the rate of 5 miles per hour. How many miles had each of them traveled when Denise overtook Gloria???\nA.  5/3\nB.  9/4\nC.  15/4\nD.  4\n\n....LAST ONE!!!\n\nThe area of a starfish is 10 square centimeters.  This starfish is enlarged so that the perimeter of the starfish is three times the original perimeter.  What will be the area, in square centimeters, of the starfish in the enlarged form???\na. 30\nb. 90\nc. 150\nd. 300
137884
Let t be the time it takes for Denise to catch Gloria expressed in hours.\n\nGloria will have been walking for (t + 1/2) hours, at 3 mph\nso she will have gone (t+1/2) * 3 miles.\n\nDenise will have been jogging t hours at 5 mph, so she will have gone 5t milse.\n\nSince they are at the same place, the distances will have to be the same: 3(t+1/2) = 5t\n\nSolve for t:\n3t + 3/2 = 5t\n\n3/2 = 2t\n\n3/4t = t\n\nCheck Denise goes 3/4 * 5 = 15/4 miles\n\nGloria goes (3/4 + 1/2) * 3 = (5/4) * 3 = 15/4 miles\n\n\n\nIf you have a square, and you triple the length of the side,  the resulting square has an area that is 9 times that of the original square.  The same thing works for all figures in the plane. If you enlarge them by a factor of 3, the area grows by a factor of 9.   \n\nIf you were to break the star down into a collection of triangles, and analyzed the area of each one, you would see that you were multiplying the base and altitude of each triangle by a factor of three, so you are multiplying the area of every one of the pieces by 9. So 90 is your correct answer.
561400
20 min.\n\njim goes 1 1/3 miles in 20 min\njohn goes 1 2/3 miles in 20 min\ntogether = 3 miles
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a) The disk increased from 10 rev/s to 15 rev/s. Acceleration was constant so the average speed was 12.5 rev/s. At that average speed, how many seconds did it take to make 60 revs?\n60 revs * 1 s/12.5 revs = 4.8 seconds. Round to 5 seconds.\n\nTherefore the acceleration was \n(delta rev/s) / (delta time) = (5 rev/s) / 5 s = 1.0 rev/s^2\n\nb) See above, that was 5 seconds\n\nc) An acceleration of 1 rev/s^2 means each second the speed increases 1 rev/s. So it took 10 seconds to reach 10 rev/s.\n\nd) The average speed from start to the time speed reached 10 rev/s was 5 rev/s. In 10 seconds at that speed it would make 50 revolutions.
439200
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
268244
Call Bonnie's speed x in miles per hour and her driving time in hours h\nSo 540 = xh\n\ncall Buffy's speed y in miles per hour and her driving time in hours t\nSo 135 = yt\n\nWe know that h = 3t\nWe also know y = x-15\nThat will let us reduce the number of variables.  Write both the distance equations using only x and t\n540 = xh = 3xt\n135 = yt = (x-15)t\n\n3xt = 540\nxt -15t = 135\n\nSolve\nt= 540/3x = 180/x\nAnd replace\n\nx(180/x) -15(180/x) = 135\n180 - 135 = 15(180)/x\n45x = 15(180)\nx = 60\nMeans Bonnie drove 60 mph for 9 hours\nBuffy drove 45 mph for 3 hours
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Specify the ratio of Paul's speed to Jennifer's speed.\n60\n---\n80\n\nNow reduce this fraction by dividing the numerator and denominator by 20:\n3 \n--\n4\n\nSo Paul can type 3/4 as fast as Jennifer.\n\nC) 3/4 as fast
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a)\n\nby dropping 3.2 meters tarzan lost \n\nPE=mgh=688x3.2= 2201.6 J  ,\n\nthis is converted to KE\n\n1/2 * m * V^2 , m=688/9.81= 70.13251784\n\nv=7.92 m/s, \n\nnow we can find centri acceleration\n\na=v^2/r = (7.923635529^2)/18 = 3.488 m/s2\n\nF=70.13 * 3.488 = 244.6 N\n\nso total force is 244.61344+688 = 932.61344 N  \n\nthis is less than 950 N\n\nthe vine doesn't break\n\nb) already found it in part a, max force was \n\n932.61344 N
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Answer to question A :\n420 in 1001\nin percent = 41.958%\n\nThe way I worked this out :\n(8x7)/(2x1) multiplied by (6x5)/(2x1) to get the 420\nand (14x13x12x11)/(4x3x2x1) to get the 1001\n\nI assume the second part (B) you are asking how many you would have to take out to guarantee that you get at least one guppy. The answer would be 9\n\nYou have not made the second part of the question very clear, so I don`t really know what you are after?
218974
Sandy, and here's why:\n\nlet's say the distance is 30 miles (you can plug any number into this and you will get the same result).\n\nJack's total trip takes him 4 hours (1 from x to y, 3 from y back to x).\n\nSandy's total trip takes her 3 hours (1.5 from x to y, 1.5 from y back to x).\n\nRegardless of distance, Sandy wins.
515038
Ok. 45 minutes is equal to 3/4 of an hour. To find the distance Travis has travelled at that point in time, you will need to multiply 55 (his rate) by 3/4. That is equal to 41.25 miles. Now we need to find if Roy will ever catch up to him. We are trying to find when their distances are equal, so time can be our variable. \n\n41.25 + 55t = 65 t \n41.25 = 10t\nt = 4.125 hours\nNow, if we plug that in to 41.25 + 55 t and it is greater than 450, then Roy can't reach Travis before Travis reaches Pecos. \n41.25 + 55 (4.125) = 268.125\n4.125 hours after Roy leaves Dallas, he will catch up to Travis 268.125 miles from Dallas.
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d= v .t\n> d = 50 km/hr (18 min)\n>d = 50 km/60 min (18min)\nd= 50 km (18)/60\nd =900 km/60\nd = 15 km >>the distance travelled\n\nIf you walk at an average of 3 miles per hour (approximately 4.8 km/hr) at a distance of 15 km:\nt= d/v\n=15 km/4.8 km/hr\nt = 3.125 hrs or (3 hours and 7.5 minutes)>>the time you need to walk on 15 km at the speed of 4.8 km/hr.♥
409143
She lands 0.7332 miles from the point on the shore which is 2 miles from the boat(call this point A) Note: the point where she should land (call it B) does not depend on the distance to the village but only upon the relative speeds of walking and rowing.\nLet x be the angle from the perpendicular she heads towards the shore and d be the distance in miles that the boat will travel at this angle, then 2/d = cosx and so d = 2/cosx miles.The amount of time taken to go this distance is d/2mph =1/cosx hrs.\nShe will land at point B that is y miles from point A and y/d = sinx so y = d*sinx = 2sinx/cosx miles.\nIf the distance to the village is D then she will have to walk D-2sins/cosx miles to the village. This will take her (D-2sinx/cosx)/5 hrs\nThe total time,t , taken is therefore\n\nt = (1/cosx +(D-2sinx/cosx)/5) hrs \n\ndt/dx = (secx*tanx) -(2/5)sec^2(x) \n\n= secx(tanx-(2/5)secx)\n\nSince secx is never 0, the derivative will be 0 when tanx = 2/5*secx, which is true when sinx = 2/5 and cosx=0.917\n\nSo y = 2sinx/cosx = 0.7332 miles from point on the shore closest to the boat.
344633
Let\nx = speed of Sachiko\nx+5 = speed of Jose\n\nSince they travel a distance of 476 miles in opposite direction\nthen,\n\nd(Sachiko) + d(Jose) = 476\ns(Sachiko)t + s(Jose)t = 476\n(x)(4) + (X+5)(4) = 476\n4x + 4x + 20 = 476\n8x = 456\nx = 57 \nx+5= 62\n\nTherefore,\nspeed of Sachiko = 57 miles per hour\nspeed of Jose = 62 miles per hour
387156
Equate distances and solve for time.\n\ngraces distance: 5*10 + 5*t    \nwill's distance: 15*t\n\nNote that I've started the clock when will starts.  That's because the question asks how long will must ride.\n\nNow, just equate the distances and solve for t.\n\nI get 5 minutes.\n\ngrace = 50+25 = 75\nwill = 5*15 = 75\n\nBtw, I graphed these on my TI-83 and looked for the intersection.  It's helpful to draw the lines.
390184
a) 5000*,4 = 2000 feet, so a = 1000 feet\n\nb) x^2/a2 -y2/b^2 =1, where;\n   where the two foci are located at (-c,0) and (c, 0) so c= 4000 ft\n   2a = difference in distances between foci and whale\n   b^2 = c^2-a^2\n\nc) Closest when y = 0 which give x= abs(a)\n\nd) y=+ or - b/a[sqrt ( x^2-a^2)], so if x < a, then y is imaginary. Therefore the region between x=a and x=-a must be excluded, that is, the whale can never be in that region. The whale will always be on the hyperbola branch that is associated with the mike that first receives the signal. The whale can be on either branch of the hyperbola, but never (obviously) on both at the same time. The distance from one foci to the whale will always be < or > than the distance from the other foci to the whale. The distances can never be equal. If they were, this technology would not work.\n\nI believe I included all the info you need to answer the question completely. It would be nice if a diagram could be included.
352035
Distance = Rate * Time\n\nYour first question does not make sense.  you are telling us that they will streak around a 1.5 mile track at 15 mph for 15 minutes and then asking us how long it would take.  You just told us 15 minutes.\n\nif you want to know "how far would they have traveled in 15 minutes at a rate of 15 mph on a 1.5 mile track" you can find out by:\n\nthe answer is 3.75 miles.\n\nD=RT\nD = 15 miles per hour * 15 minutes\nfirst you have make sure you are using the same units for time, either use minutes or hour for both the rate and time.\n\nI chose to change the 15 minutes into hours which = 1/4 of an hour.  use a ratio:  of minutes to hours and reduce 15/60 = 1/4\n\nthen multiply  15 m/h * 1/4h and you end up with 3.75 miles.\n\nwhen you say 15 miles per hour it = 15miles/1 hour in fraction form.\n\nwhen you multiply 15 miles / 1 hour * 1 hour /4  you cancel out the hours units and you are left with only miles.\n\nIf you want to know how long it would take them to travel one time around the track:\n\nD = RT\nT= D/R\n\nTime = 1.5 miles / (15 miles / 1 hour)\n\nRemember when you divide fractions you invert the second one and them multiply them:\n\nTime = 1.5 miles * 1 hour / 15 miles\nmultiply and cnacel out the miles unit and you get:\n(1.5 * 1) / 15\n1.5/15\n= .1 hours\n\nchange .1 hours into minutes:\n1 hour = 60 minutes so:\n.1 * 60 = 6 minutes\n\nit takes them 6 minutes to do a lap, (which is crazy for someone that size)
366360
im curious how this helpful to both of us??\n\nlynns distance (neglect directions )is 20 km....but her displacement (a vector) is 2 km east....because going north 5 and them south 3 and south 2 later cancels out n/s distance and going 6 east but 4 west leaves with 2 east..draw it out\n\n\nthe UFO moved 10 km North, it moved 20 east but them 30 west, subtract to leave 10 West...so the TOTAL displacement is 20 km\n\nBUT use pythagorean for 2\n\n10^2+10^2=200, sqrt 200, leaves 14.14 km NW