You are on a trip to one of the Cartesian islands. In the cellar of your cabin you found an oldmap that claims to hold the secret location of a buried treasure. The map shows 4 trees at location A, B, C, and D. The trees at A and D are along an east-west road that runs through the island. the tree at B is directly north of the tree at A, and the tree at C is directly north of the tree at D.\nThe map's directions say to trace a path from the tree at B to the tree at D, and to do the same for the trees at A and C. The treasure is buried halfway between the intersection of these paths (point E) and the road. Because three small lakes dot the island, you are only able to measure the distances between B and D, A and C, and A and D:\nB to D=300 paces\nA to C=260 paces\nA to D=240 paces\nTo find the location of the treasure, you must determine the distance from E to the road and then walk half the distance from E. How far is E from the road and how far from E is the buried treasure?
202534
Hi! The name of the island suggests we should find the treasure using graphs! :)\n\nLet us assume first that point A lies on the west of point D. Consider the point A to be the origin. Thus point D will lie on the X-axis and the straight lines A-B and D-C would be parallel to Y-axis. Draw the graph on paper choosing an appropriate scale, say 1 cm=10 paces. In the solution, I'm finding the points as your REAL coordinates if you are on the island itself.\n\nWe've been given the distances BD, AC, AD; find the distances AB and DC using Pythagoras theorem. They come out to be 180 and 100 paces respectively. Since we took A to be the origin, the coordinates of point B are (0,180) and for C are (240,100). The coordinates of A and D are (0,0) and (240,0).\n\nHaving found the points B, D, A, C, we can now find the equation of the lines BD and AC using the Two-Point formula. They come out to be:\n\nline BD: y=(-3/4)x + 180\nline AC: y=(5/12)x\n\nThese two lines intersect at the point E. Find the point E by equating y in both cases. Then find 'x', and put it in any one equation to find 'y'. E comes out to: (1080/7,540/7).\n\nSince the road is the X-axis itself, the distance of E from the road is its' x-coordinate: 1080/7 paces ~= 154.28.\n\nHalf the distance from E to the road is 540/7 paces, approx 77.14 paces. So the treasure is buried at a distance of a little more than 77 paces and little less than 78 paces from point E towards the road.
113789
Here is a solution.  Let x be the distance the motorist traveled at 20km/h on city road, and let y be the distance the motorist traveled at 50km/h on the highway.  We are looking for y.\n\nFirst, we know that the total distance is 155km\n\nx+y=155     (eq1)\n\nSecond we know that if you travel x km at 20km/h, it will take you x/20 hours.  Since the total hours is 4, then:\n\nx/20 + y/50 = 4      (eq2)\n\nThat is two equations, two unknowns, so we can solve the problem.  We'll rearrange eq1 to be:\n\ny=155-x     (eq3)\n\nThen substituting this into eq4 yields\n\nx/20 + (155-x)/50 = 4 \n\nor\n\n5x/100 + 2(155-x)/100 = 4\n5x+2(155)-2x=400\n3x=90\n\nx=30\nThat gives us x, and we can solve for y from (eq 3)\n\ny=155-30\n\ny=125\n\nAnswer: 125 km
368219
I'm going to use minutes instead of hours.\nWe know that on the first trip, D = 120S (D=distance, S=Speed of first trip).\nWe know that one the second trip, D = 90(S+6).\n\nTherefore 120S = 90S + 540, or 30S = 540, and S = 18kph.\nTherefore the distance is 36 kilometers.\nChecking, 24(1.5) for the second trip also = 36.
565632
You need to g to the log cabin and look at the books on the shelf, check out the one with the same thing on the spine as the box and that is where you will get your answer. After you cross the river you click on the tree to find your destintion then click on thetruck. then you will aget a map and it will take you to the log cabin.
10839
pier_haringsma is correct.  You do not need any additional information.  Let \nD = total distance of the trip  \nD/2 = the distance the driver has gone so far\nT1 = amount of time he took to travel the first half of the trip\nT2 = the complete length of time of the whole trip (both halves).\n\n   Now, he wants the average ( D / T2 ) to equal 30.  His average so far is ( (D/2) / T1 ) = 15.\n\nD/T2 = 30, so D = 30*T2\n(D/2) / T1 = 15, so D = 30*T1\n\nD = 30*T1 = 30*T2, so T1 = T2.\n\n  So the driver wants the total time for the trip (BOTH halves), T2, to equal the time for the first half of the trip, T1.  This means he would have to travel the second half of the trip in no time, which is impossible.
258112
OK, I drew the picture and this is what I came up with.\n\nThe anchor point or bottom of the bridge touches the ground at the edge of the shoulder.  Lets call the middle of the road on the ground (0,0).  So, the anchor point is at (8,0).\n\nIt also says that when you get to the edge of the shoulder and the road that the clearance must be 2.5 meters.  This point would be (6,2.5).  Symmetry gives us (-8,0) and (-6,2.5).  And just for fun, note that the parabola is open down, just in case that comes up.\n\nThe generic quadratic is ax^2 + bx + c.\n\nLets put some of our points in and solve for a,b, and c.\n\na(8)^2 + b(8) + c = 0.  So, c = -64a - 8b  using (8,0)\n\na(-8)^2 + b(-8) +c =0.  So, c = -64a + 8b using (-8,0)\n\nSetting these 2 values for c equal to each other I get that b=0. (We should have known this already because we put the vertex directly on the y-axis.  This will always give you a b value of zero because the quadratic has to be able to factor into the form \n(sqrt(a) - sqrt(c))(sqrt(a) + sqrt(c))\n)\n\nSo, y = ax^2 + c.  Lets put a third point in: (6,2.5)\n\na*36 + c = 2.5 or c = 2.5 - 36a.  From directly above we have that c = -64a - 8b, but b turned out to be 0, so,\nc = 2.5 - 36a = -64a  --> 2.5 = -28a  \n--> a = -2.5/28 = -5/56\n\nc = 2.5 - 36(-5/56) = 2.5 + 45/19  = 95/19 + 45/19 = 140/19.\n\nI come up with y = -5/56*x^2 + 140/19 as the equation for the bridge.  The highest point will be when x = 0, so, the maximum clearance is at (0,140/19).
672883
When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
460256
QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
99597
its 127.98 ft high above josh's head. It needs to be explained diagramatically which is not possible in this text space but I will try to explain you here.\n\nDraw one straight line perpendicular(vertical) to ground(meaning a horizontal line sometimes referred as base in trigonometry) and say at x distance draw another line parallel to the first line and perpendicular to the ground and this line should be shorter than the first line, consider the second vertical line as height of josh. Now join the base it should be a straight horizontal line and also join the upper end of both the line the upper end will lead to elevation of 47 degree. now draw a perpendicular from josh's head to the first vertical line that you draw. \nthe line which joins the upper end of both vertical lines measures 175 ft.and the small section in the first vertical line is the distance to be found.\nso it leads to \nsine 47 = h/175\nh= sine47 * 175\nh = 127.98 ft\n\nI hope I was able to explain you well, in case you ned more help send me a mail I will explain you. :)
251348
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
14546
N.O is wrong. The right answer is below:\n\n1st-you have to find the area of the plywood:\n   4.5*8.25=37.125 feet squared\n2nd-find the area of the circle he cut out:\n   A=pi*r^2\n   A=3.14*(1.5)^2\n   A=7.069\n3rd-subtract the area of the circle from the plywood's area:\n   A=30.056\n4th-mutiply this leftover area by 1.75 to get the answer:\n   1.75*30.056=$52.60\n\nSo his friend payed $52.60 for the leftover plywood!
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
368981
The fact that he took his return trip later just complicates the problem. The question can be restated as one person leaves from the top of the mountain and one person leaves from the bottom of the mountain. They leave at the same time. Explain why, if they follow the same path, they will run into each other each other at some point. Stated this way, it is obvious, right?\n\nIf you want a more rigorous solution, here you go. This is a continuity question. Let f(t) denote the distance *from the bottom of the mountain* at time t on the trip up the mountain and let g(t) denote the distance *from the bottom of the mountain* at time t. Change your units so that your time starts at zero (so 8AM is t=0) and you make it up the mountain in time 1 (so 1PM is t=1) and your distance from the bottom to the top of the mountain is 1. With this set up, g(x) lies in the interval [0,1] for every x in the interval [0,1], and the same holds for f(x). Consider the function g(t)-f(t). The physical interpretation makes it clear that f(t) and g(t), hence g(t)-f(t), are continuous functions of t; that is, you don't just disappear at one point and reappear at another point without having travelled through every point in between. g(0)-f(0)=1-0=1>0 and g(1)-f(1)=g(1)-1<=0. The intermediate value theorem tells you that there is some time t_0 such that g(t_0)-f(t_0)=0. That is, there is some time t_0 such that g(t_0)) = f(t_0). Your deja vu moment happens at time t_0.
517032
for the first hour:\nJose travels 5 miles\nMario travels 7 miles\nMelanie travels 9 miles\n\nfor the second hour:\nJose travels 10 miles\nMario travels 14 miles\nMelanie travels 18 miles\n\nJose   (1,5) and (2, 10) are your points\n(10 - 5)/(2-1) = 5 mph\nMario  (1, 7) and (2, 14)\n(14-7)/(2-1) = 7 mph\nMelanie (1,9) and (2, 18)\n(18-9)/(2-1) = 9 mph\n\nANSWERS\nJose\n    5 mph\nMario\n     7 mph\nMelanie\n    9 mph
373483
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
401110
The distance travel is going to be 1/2 mile.   How far away are you are from the beginining of the first 1/4 mile could vary depending on how straight of a line you traveled and which angle you turned once you stopped but the maxium distance can not be greater than 1/2 mile (perfectly striaght line).  So C and D can not be correct.  A and B can be correct.  Being a perfectly straight road is not very possible then A is correct.  But if this is for your math class then most likely B is what the teacher is looking for.
445431
I used to get tripped up with these too!\n\nSo, it's all about speed, distance, time (s = d/t) and finding the 2 things you know, and getting an equation for the third unknown and solving it. Here, the speeds are given, and the difference in times are given.\n\nSay d is the distance Matt has to go, and t1 & t2 are the times he took to get there the first and second time.\n\nt1 = d/75\nt2 = d/45\n\nalso, he arrived one hour early the first time, and one hour late the second ==> the difference between these two times is 2 hours.\n\nt2 - t1 = 2\n=> d/45 - d/75 = 2\n=> (75-45)d/45*75 = 2\n=> d = 2*45*75/30\n=> d = 225 miles.