These are the questions and use the formula(F=x2-5x) to answer the question.\n\n1.In the last six months, Anne has received tickets for driving 85 mph, 102 mph, 78 mph, and 99 mph. What are the domain and range for this relation?\n\n2. What speed would a person have been traveling on the interstate if he or she had to pay a fine of $176?
138501
for 1), it is acceptable to name your domain as being 85 <= x <= 102, where x is a set of natural numbers (numbers greater than 0). If you want to be more specific, u can say the domain consists of all 4 speeds.\n\nYour range is indicated by the lowest fine paid up to the highest fine paid. In this case, sub in the lowest and highest speeds into the equation. The range is bounded by these values.\n\n2) Sub it into the equation as F, and solve for x. Use quadratic formula or factoring :)
534544
1) \nSB takes 3 hours to do the job.\n\nfraction of job done by SB in 1 hr = 1/3\n\nSimilarly Fraction of job done by SW in 1 hr = 1/2.5 = 2/5\n\nFraction of job done by both SB and SW in 1 hr = 1/3+2/5 = 11/15\n\nNow Both SB and SW can finish the job together in : 1 / (11/15) = 15/11 = 1.36 hrs.\n\n Shortcut:\n\n1 / (X and Y together) = 1/ X + 1 / Y\n\n2)\n\nNow when Mark and his agent meet they would have covered the 500 Kms together as they \n\ntravel the same road to meet each other.\n\nsuppose after x hrs they meet.\n\n\nMark has travelled : 55x km\nAgent has travelled : 65x km\n\nTotal distance : 55x + 65x = 500\n\n120x = 500 => x = 500/120 = 4.167 hrs\n\n3) Use the same approach as in 1) above.\n\nUse the shortcut too if u need.\n\n1/ ans = 1/12 + 1/9\n\nand find the ans!\n\n4)\n\nLet the cost of the pencil be x.\ncost of pen = 3x\n\n5 * x + 4 * 3x = 4.25\n17x = 4.25\nx = 0.25\n\nSo pencil costs 25 cents!\n\n\nHope the above answers help! Do let me know if they dont! :-)
368251
1.  The key here is to understand how far the two trains have traveled in total during the first 3 hours.  We are told that the two trains started 500 miles apart, and 3 hours later are now 80 miles apart.  Thus they traveled 420 miles in total. \n\nIf you let x represent the rate of speed that the Freight train is traveling, and x+20 represent the rate of speed that the Passenger train is traveling, you can set the following algebric equation:\n\n3 hrs * x mi/hr  + 3 hrs * ( 20 + x ) mi/hr = 420\n 3x + 60 + 3x = 420\n6x = 360\nx =60 mi/hr.\nx+20 = 80 mi/hr.\n\n2.  The key is too figure out the total number of employees needed to break even.  Regardless of how many employees you have, we are told that it costs $2000 per day to run the factory.  In addition to that  we are told that for each employee, we must pay him or her $300.  These are the total costs for the operation.\n\nOn the other hand, we are told that each employee can produce $500 in revenue each day.\n\nThus, we need to create an algebric equation where the total costs = the total revenue\n\nLet x represent the number of workers, and set up the following algebric equation:\n\nCosts = Revenue\n\n$2000 (Fixed Costs) + $300 * x (Variable Costs) = $500 * x\n\n2000 + 300x = 500x\n2000 = 200x\n10 = x\n\nYou will need 10 workers to break even, and the total gross value of ball bearings produced will be 10 * $500 or $5000.\n\nHope that I have cleared things up for you!
552029
1. Bob's favorite ice cream used to cost $1.50.... Then the price was raised to $1.75. Find the percent increase.\n(175/150)-1)*100%=16 2/3% increase\n\n2. Jenny paid $12,000 for her car. Five yrs later she sold it for $8500. What percent decrease was this?\n(8500/12000-1)*100%= -29 1/6% 29 1/6% decrease\n\n3. The Music Box marked up their CD's 75%. If the music box charges $20 for a special cd, how much did it cost before the mark-up? (hint: the answer is not $15)\n$20=1.75x\nx=20/1.75=$11.43\n\n4. Quinn deposits $2000 in a savings account that pays 4% compounded semi-annuallly. HOw much money will Quinn have in her account after two years? Give the expression. (you do not have to simplify) \n\n$2000(1.02)^4=$2164.86
514801
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.
415867
They travelled a total of 476 miles in 4 hours.  So together they are travelling (476/4) mph or 119 mph.\n\nSo you have two equations:\nJ + S = 119\nJ - S = 5\n\n2J = 124\nJ = 62 mph\nS = 57 mph
634072
The term "victim" refers to her injuries, not her legal standing.  Sounds like your g.f. made a left turn in front of a car that struck her car.\n\nIf that's the case, then the other driver had the right of way and your g.f. is at fault.\n\n~~~~~~~~~~\n\nWicked Magnet below wrote: "BUT...if neither saw the other, and the other party was "clearly going faster..."\n\nIf neither party saw the other, then how do you determine which party was "clearly going faster"?  As far as these two drivers were concerned, the other driver was invisible.  How can you say an invisible driver was speeding?
412936
67.6 miles per hour\nif it takes the officer five minutes to catch someone a quarter mile away from him, the driver of the first car is going 2.4 miles per hour slower than the cop's 70 miles per hour.\n\n12/5=2.4\n70-2.4=67.6\n\npretty simple if you just figure it the same way you calculate how long it will take to go 30 miles at 55mph on a highway...same math applies.  it just seems trickier because it is asked in a scenario of third person.
82860
Let r = the rate on the country roads.  Then r-9 is the freeway rate since he travels 9 mph slower there.  The formula needed is d=rt, or in another form t = d/r where t = time, d = distance and r = rate.  Since both have the same t, \n\n20/r = 15 (r-9)\n\ncross multiply to get 20r - 180 = 15r\n\nadd 180 to both sides and subtract 15r from both sides to get\n\n5r = 180   Divide by 5 to get\n\nr = 36.  So he goes 36 mph on the side roads.
484841
well if the car needs to go around twice and average 60 mph then it would need to travel 120 mph total, 60*2.  If it went once around at 30 mph, 120-30=90. Therefore it would have to go 90mph for the second lap.\n\nAnother way to look at it would be to use the D=RT formula for distance = rate times time.  Solve it for the time meaning T=D/R so the time it would take for two laps to average 60mph would be 10miles/60mph or 1/6 of an hour, or 10 mintues.  Now one lap of 5 miles at 30mph would take T=5m/30mph or 1/6 hour = 10 minutes.  This leaves no time for the second lap to take place.\n\nThis is a paradoxical question.  If the second lap was at 90 mph then the average would be 60mph (30mph+90mph)/2=\n120mph/2=60mph.
236673
C. Evaluate the factorial\n1.5,4,3,2,1\n2.9,8,7,6,5,4,3,2,1\n3.7,6,5,4,3,2,1\n4.12,11,10,9,8,7,6,5,4,3,2,1\n\nB. Determine how many different license plates are possible if (a) digits and letters can be repeated, and (b) digits and letters cannot be repeated \n1.(b)01 abcd, 23 efgh, 45 ijkl, 67 mnop, 89 qrst (a)01 abcd, 23 efgh, 45 ijkl, 67 mnop, 89 qrst, 12 uvwx, 34 yzab, 56 cdef, 78 ghij, 90 klmn\n2.(b)abcde 0, fghij 1, klmno 2, pqrst 3, uvwxy 4 (a)abcde 0, fghij 1, klmno 2, pqrst 3, uvwxy 4, zabcd 5, efghi 6, jklmn 7, opqrs 8, tuvwx 9\n\nA. Each event can occur in the given number of ways. Find the number of ways all of the events can occur.\n1. Event 1:3 ways can occur 1 way, Event 2:5 ways can occur 1 way\n2. Event 1:4 ways can occur 1 way, Event 2:6 ways can occur 2 ways, Event 3:9 ways can occur 2 ways, Event 4:7 ways can occur 1 way. 6 total ways\n\nGlad I could help or hopefully I could help
673313
Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
447360
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
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
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
297281
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
305207
d = vt\ntP = 5tR\ndP = 300\ndR = 160\nvP + 50 = vR\n\n300 = vP * tP\n\n160 = vR * tR\n160 = (vP + 50) * tP/5\n160 = (300/tP + 50) * tP/5\n800 = 300 + 50tP\n500 = 50tP\n\n-----------------------------\ntP = 10 hours\ntR = 10/5 = 2 hours\nvP = 300/10 = 30 mph\nvR = 160/2 = 80 mph\n-----------------------------