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 Matrices-and-determiminant/105148: An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. 1 solutions Answer 76528 by ankor@dixie-net.com(15652)   on 2007-10-19 16:12:08 (Show Source): You can put this solution on YOUR website!An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. : Let s = speed in still air : (s+25) = speed with wind (s-25) = speed against the wind : The trip out and the trip back are equal distance. Write a distance equation. Distance = time * speed : 5(s-25) = 4(s+25) 5s - 125 = 4s + 100 5s - 4s = 100 + 125 s = 225 mph in still air : : Check the solution by finding the distances to be equal: Against: 5(200) = 1000 mi With: 4(250) = 1000 mi confirms our solution
 Trigonometry-basics/105033: Use the properties of addition and multiplication to complete the statement. 18 • 4 = 4 • ____ Not sure how this is done at all, help please.1 solutions Answer 76514 by ankor@dixie-net.com(15652)   on 2007-10-19 14:44:11 (Show Source): You can put this solution on YOUR website!Use the properties of addition and multiplication to complete the statement. 18 • 4 = 4 • ____ : 18 * 4 = 4 * 18 : They are just trying to remind you that 18*4 is the same as 4*18 It's the same with addition. the order of the number does not matter. That's the "property of addition and mult" : This is not the case with dividing and subtracting, however. : It's probably obvious to you anyway, but they just want to make a rule about it>
 Linear_Algebra/105059: A change machine contains nickels, dimes, and quarters. There are 75 coins in the machine, and the value of the coins is \$7.25. There are 5 times as many nickels as dimes. Find the number of coins of each type in the machine.1 solutions Answer 76513 by ankor@dixie-net.com(15652)   on 2007-10-19 14:35:57 (Show Source): You can put this solution on YOUR website!: A change machine contains nickels, dimes, and quarters. : Just write an equation for each statement: : "There are 75 coins in the machine," n + d + q = 75 : "the value of the coins is \$7.25." .05n + .10n + .25q = 7.25 : " There are 5 times as many nickels as dimes." n = 5d : Find the number of coins of each type in the machine. : Substitute 5d for n in the 1st equation: 5d + d + q = 75 6d + q = 75 q = (75-6d); use this for substitution later : Substitute 5d for n in the 2nd equation: .05(5d) + .10d + .25q = 7.25 .25d + .10d + .25q = 7.25 .35d + .25q = 7.25 : In the above equation substitute (75-6d) for q: .35d + .25(75-6d) = 7.25 .35d + 18.75 - 1.5d = 7.25 .35d - 1.5d = 7.25 - 18.75 -1.15d = -11.5 d = -11.5/-1.15 d = +10 dimes : Find the nickels from n = 5d n = 5(10) n = 50 nickels : Find q using the 1st equation: 50 + 10 + q = 75 60 + q = 75 q = 75 - 60 q = 15 quarters : : Check our solutions on the 2nd equation:. .05(50) + .10(10) + .25(15) = 2.50 + 1.00 + 3.75 = 7.25; confirms our solutions : Did this makes sense to you? A lot of steps but each one is logical and simple. Any questions?
 Functions/105056: This question is from textbook Algebra 1 Solve for b2 A=1/2h(b1+b2)1 solutions Answer 76509 by ankor@dixie-net.com(15652)   on 2007-10-19 13:40:46 (Show Source): You can put this solution on YOUR website!Solve for b2 A=1/2h(b1+b2) : A = h(b1 + b2) : Multiply both side by 2, and you have: 2A = h(b1 + b2) : Divide both sides by h, resulting in: = b1 + b2 : Subtract b1 from both sides: - b1 = b2
 Polynomials-and-rational-expressions/105028: Simplify the expression. Write the answer with positive exponents only. (m^4)-3/(m^-2)41 solutions Answer 76507 by ankor@dixie-net.com(15652)   on 2007-10-19 13:31:40 (Show Source): You can put this solution on YOUR website!(m^4)-3/(m^-2)4 : Assume you mean: = ; exponents are multiplied : = ; combine the exponents where they result in a positive : Do you understand that you subtract the exponents to divide, but subtracting a neg exponents results in adding it?
 Equations/105032: A straight line with a slope of 5 contains the points(1,2) and (3,k) find the value of k1 solutions Answer 76506 by ankor@dixie-net.com(15652)   on 2007-10-19 13:14:29 (Show Source): You can put this solution on YOUR website!A straight line with a slope of 5 contains the points(1,2) and (3,k) find the value of k : Remember our old friend the "slope formula" m = : In your problem: x1 = 1; y1 = 2 x2 = 3; y2 = k m = 5 : So we have: = 5 : = 5 : Multiply both sides by 2, gets rid of the denominator k - 2 = 2(5) k = 10 + 2 k = 12, (y2 = 12) : : Check that in the original equation: = 5; confirms our solution : Did this help you?
 Trigonometry-basics/104987: Solve. x/3 – 4 = 1 I have no idea how this even works can some one help. the x/3 is a fraction.1 solutions Answer 76503 by ankor@dixie-net.com(15652)   on 2007-10-19 12:54:46 (Show Source): You can put this solution on YOUR website!Solve. x/3 – 4 = 1 : - 4 = 1 : The first thing you want to do, is get rid of that annoying fraction. Multiply equation by 3 to do that (you mult each term by 3): : 3* - 3(4) = 3(1) : Cancel out the denominator and you have x - 12 = 3 : x = 3 + 12; added 12 to both sides : x = 15 ; : Confirm our solution by substituting 15 for x in the original equation: - 4 = 1 5 - 4 = 1 : Did this help you?
 Radicals/104967: Simplify. Assume all variables represents positive numbers. sq rt x^4y^31 solutions Answer 76502 by ankor@dixie-net.com(15652)   on 2007-10-19 12:45:54 (Show Source): You can put this solution on YOUR website!Simplify. Assume all variables represents positive numbers. sq rt x^4y^3 : : Factor it into perfect squares: : Extract the squares : Which is: : Did this help?
 Linear-systems/105124: I keep trying over and over again but cannot find the solution to this: (3m-2)/5 = (4-m)/3 everytime I do it i keep getting m=6.5 . 1 solutions Answer 76490 by ankor@dixie-net.com(15652)   on 2007-10-19 10:38:09 (Show Source): You can put this solution on YOUR website! = : Cross multiply and you have: 3(3m-2) = 5(4-m) : 9m - 6 = 20 - 5m; multiplied what's inside the brackets : 9m + 5m = 20 + 6 : 14m = 26 : m = 26/14 : m = 13/7 : : Check on a calc using decimals: m = 1.857 = .714 = .714; confirms our solution : Did this help?
 Trigonometry-basics/104986: Which of the ordered pairs (0, 0), (–2, 10), (–1, –5), (–3, 9) are solutions for the equation y = 5x? Please help me I am so stuck on this. 1 solutions Answer 76488 by ankor@dixie-net.com(15652)   on 2007-10-19 10:05:08 (Show Source): You can put this solution on YOUR website!Which of the ordered pairs (0, 0), (–2, 10), (–1, –5), (–3, 9) are solutions for the equation y = 5x? : you can do this, not at all hard. The 1st value in the pair is x, the 2nd is y : Take the pair (0,0), substitute 0 for x in the equation y = 5x y = 5(0) y = 0, so (0,0) is a solution: : The next pair: (-2,10), substitute -2 for x in the same equation y = 5x y = 5(-2) y = -10, not +10 so it is not a solution : The next pair (-1,-5) y = 5(-1) y = -5, so this is a solution : The next pair (-3,9) y = 5(-3) y = -15 so this is not a solution : How about this, can you do this now?
 Linear-equations/105069: This question is from textbook Algebra 2 and Trig At 2 am snow started falling at a constant rate. At 10:30 am Harry began clearing his driveway at a constant rate. At 11 am when it stopped snowing the average depth of the snow on the driveway was 2 in. By 11:24 the driveway was clear. At what rate did the snow fall in in./hour. I am really confused at this problem and don't even know where to start.1 solutions Answer 76484 by ankor@dixie-net.com(15652)   on 2007-10-19 08:16:22 (Show Source): You can put this solution on YOUR website!At 2 am snow started falling at a constant rate. At 10:30 am Harry began clearing his driveway at a constant rate. At 11 am when it stopped snowing the average depth of the snow on the driveway was 2 in. By 11:24 the driveway was clear. At what rate did the snow fall in in./hour. : Get some facts in our mind, and write them down: : No. hrs the snow fell: 11:00 - 2:00 = 9 hrs No. hrs shoveling snow: 11:24 - 10:30 = :54 min; 54/60 = .9 hrs No. hrs shoveling after the snowing stopped: 11:24 - 11:00 = :24 min = 24/60 = .4 hr : At 11 the snow was 2 in deep and at 11:24 it was gone, therefore He shoveled at a rate of 2 in .4 hr or at a rate of 2/.4 = 5 inches/hr : Let x = snowfall in inches/hr: : Fallen snow - shoveled snow = 0 9x - .9(5) = 0 9x - 4.5 = 0 9x = +4.5 x = 4.5/9 x = .5 inches per hr : Did this make some sense to you? Any questions?
 Travel_Word_Problems/104939: Brittani and Lea, two cyclists, start simultaneously. Brittani at point A and Lea at Point B. They travel toward each other at constant sppeds. They pass each other in 30 minutes, and Brittani, headed for point B, arrives there in another 20 minutes. How many minutes after brittani arrives at point B will Lea arrive at Point A?1 solutions Answer 76450 by ankor@dixie-net.com(15652)   on 2007-10-18 22:07:53 (Show Source): You can put this solution on YOUR website!Brittani and Lea, two cyclists, start simultaneously. Brittani at point A and Lea at Point B. They travel toward each other at constant speeds. They pass each other in 30 minutes, and Brittani, headed for point B, arrives there in another 20 minutes. How many minutes after Brittani arrives at point B will Lea arrive at Point A? : We know that it took B 20 min to go the distance L required 30 min to go. : We need to know how long it will take L the cover the distance that B covered in 30 min : Let d1 = dist from B to where they met. Speed = Dist/time (in min) B's speed = d1/20 L's speed = d1/30 : Find out the distance (d2) from A to where they met: Dist = speed * time d2 = * 30 d2 = d1 d2 = 1.5*d1 : Find Time required for L to travel 1.5*d1: Time = dist/speed L's time = 1.5d1/(d1/30) L's time = 1.5d1 * 30/d1 L's time = 1.5 * 30; d1's cancel L's time = 45 min : It asked how long after B gets to pt B will L get to point A It would be 45 - 20 = 25 minutes. : Did this help, hope I made it understandable.
 Numbers_Word_Problems/104695: find a six-digit number in which the first is two less that the fifth, the second digit is one more that the fourth, and the fifth digit is four less than the last. The sum of the third and last digits equals the second and the sum of all the digits is 30.1 solutions Answer 76439 by ankor@dixie-net.com(15652)   on 2007-10-18 19:53:13 (Show Source): You can put this solution on YOUR website!: find a six-digit number in which the first is two less that the fifth, the second digit is one more that the fourth, and the fifth digit is four less than the last. The sum of the third and last digits equals the second and the sum of all the digits is 30. : Let the digits by a, b, c, d, e, f In that order. : Write an equation for each statement and all it's combinations: Designated * were used in the equation : " the first is two less that the fifth," a = e-2 e = a+2 : "second digit is one more that the fourth," b = d+1 d = b-1 : "fifth digit is four less than the last." e = f-4 * f = e+4 From the 1st equation a = e-2 a = (f-4) -2 a = f-6 * : "The sum of the third and last digits equals the second" c + f = b b = c+f * f = b-c c = b-f From d = (b-1) and b = c+f, we can say d = c+f-1 * : "the sum of all the digits is 30." a + b + c + d + e + f = 30 : Substitute all we can in terms of f from the above equations: --a-------b-----c------d---- --e-----f (f-6) + (c+f) + c + (c+f-1)+ (f-4) + f = 30 Combine like terms f + f + f + f + f + c + c + c -6 - 1 -4 = 30 3c + 5f - 11 = 30 3c + 5f = 30 + 11 3c + 5f = 41 A manageable equation; we know that f has to be odd, it cannot be 9 Try f = 7 and solve for c 3c + 5(7) = 41 3c = 41 - 35 3c = 6 c = 2 : From c = 2, and f = 7, find the rest of the numbers: b = c + f b = 2 + 7 b = 9 : From d = b-1 d = 9 - 1 d = 8 : From e = f-4: e = 7 - 4 e = 3 : From a = e-2 a = 3 - 2 a = 1 : Check and see if a b c d e f add up to 30 1 + 9 + 2 + 8 + 3 + 7 = 30 : Our number: 192837
 Miscellaneous_Word_Problems/104879: Hi please help me solve this word problem. The function A=A(base of 0)e^-0.01152x models the amount of pounds of a particular radioactive material stored in a concrete vault, where x is the number of years since the material wsa put into the vault. If 800 pounds of material are initially put into the vault, how many pounds will be left after 180 years? Thank you1 solutions Answer 76393 by ankor@dixie-net.com(15652)   on 2007-10-18 12:58:19 (Show Source): You can put this solution on YOUR website!The function A=A(base of 0)e^-0.01152x models the amount of pounds of a particular radioactive material stored in a concrete vault, where x is the number of years since the material wsa put into the vault. If 800 pounds of material are initially put into the vault, how many pounds will be left after 180 years? : Get to know the formula: A = Ao*(e^-.01152x) Where: Ao = original amt (800 lb in this problem) A = resulting amt x = years of decay (180 yrs in this problem) : A = 800(e^(-.01152*180)) A = 800(e^-2.0736) On good calc find e^-2.0736, should get: .12573, so you have: A = 800 * .12573 A = 100.586 lb remaining after 180 yrs
 Travel_Word_Problems/104935: This question is from textbook Question : A twin-engine plane can fly 800 mi in the same time that it takes a single-engine plane to fly 600 mi. The rate of the twin engine plane is 50 miles faster than that of the single engine plane. Find the rate of the twin engine plane. This is what I have tried: distance/time=rate. Twin engine 800 50+r Single engine 600 50-r So 800/50+r & 600/50-r (50+r)(50-r)800/50+r=(50+r)(50-r)600/50-r (50-r)800=(50+r)600 40000-800=30000+600 -1400=-10000 r=? I have been trying to find wesites that could help me with this problem all week. The week for this problem is over tomorrow, October 18,2007, at 11:59pm. Please help me ASAP! I have many other questions that need answering and I am looking forward to someone helping me out. 1 solutions Answer 76389 by ankor@dixie-net.com(15652)   on 2007-10-18 12:36:36 (Show Source): You can put this solution on YOUR website!A twin-engine plane can fly 800 mi in the same time that it takes a single-engine plane to fly 600 mi. The rate of the twin engine plane is 50 miles faster than that of the single engine plane. Find the rate of the twin engine plane. : You got things a little mixed up here, here's how I would do it: : Let r = speed of the twin engine plane then (r-50) = speed of the single engine plane : The time of the two planes is given as the same so write a time equation: Time = Distance/speed : Twin time = single time = Twin: Cross multiply and you have: 800(r-50) = 600r : 800r - 40000 = 600r; multiplied what's inside the brackets : 800r - 600r = +40000 200r = 40000 r = r = 200 mph is the speed of the twin : : WE can check our solutions by finding if the times are the same Single plane speed is 200 - 50 = 150 mph : Twin: 800/200 = 4 hrs Single: 600/150 = 4 hrs, confirms our solution: : Did that help? I think you made it more complicated than it is:
 Age_Word_Problems/104848: amaqn born on the first half of the 19th century will be x years old in x to the power 2AD.find his year of birth.1 solutions Answer 76371 by ankor@dixie-net.com(15652)   on 2007-10-18 09:57:12 (Show Source): You can put this solution on YOUR website!amaqn born on the first half of the 19th century will be x years old in x to the power 2AD.find his year of birth. : Who dreams up these goofy names? : Find his approximate age with calc: = 43.588... : If his age is 43: 43^2 = 1849 Therefore: x = 43 years old x^2 = 1849, the year he was 43 1849 - 43 = 1806, born in the 1st half of the 19th century
 Miscellaneous_Word_Problems/104703: Train A leaves a station traveling at 80 kilometers per hour. 8 hours later, train B leaves the same staion traveling in the same direction at 90 kilometers per hour. How long does it take train B to catch up to train A?1 solutions Answer 76288 by ankor@dixie-net.com(15652)   on 2007-10-17 20:13:32 (Show Source): You can put this solution on YOUR website!Train A leaves a station traveling at 80 kilometers per hour. 8 hours later, train B leaves the same station traveling in the same direction at 90 kilometers per hour. How long does it take train B to catch up to train A? : One thing to notice about these "catch-up" problems, is that when one catches up with the other, they will have traveled the same distance, we don't what the distance is, but we do know that they are the same. We can make a distance equation from this simple fact. : Let t = time required by Train B to catch up then (t+8) = time when Train A is caught : Distance = speed * time: : 90t = 80(t+8) 90t = 80t + 640 90t - 80t = 640 10t = 640 t = 640/10 t = 64 hr for B to catch up with A : : Now, we can find out the distances, and by them being equal, we know that our solutions are correct: Train A time: 64+8 = 72 : A: 80 * 72 = 5760 mi B: 90 * 64 = 5760 also : Did this make sense to you? Any questions?
 Linear_Equations_And_Systems_Word_Problems/104736: I'm in grade 10 math and have finished solving word problems using linear systems. We were given a sheet where I now have to try and make up a linear solution using the given variables, but I'm having trouble trying to work this out and my parents are unsure also. The question is below and if you could help me it would be greatly appreciated. Question: Make up a word problem that you can solve using a linear system. The solution is (5,35). 1 solutions Answer 76285 by ankor@dixie-net.com(15652)   on 2007-10-17 19:56:42 (Show Source): You can put this solution on YOUR website!Question: Make up a word problem that you can solve using a linear system. The solution is (5,35). : Since you have a choice, choose a convenient slope, say +2, then use the point/slope equation which is y - y1 = m(x - x1) : In our problem x1 = 5; y1 = 35; m = +2 (which we chose) : y - 35 = 2(x - 5) y - 35 = 2x - 10 y = 2x - 10 + 35 y = 2x + 25; this our equation : Now think of the word problem that fits this equation. : Example: You rent a U-haul truck. It cost \$25 per day + \$2 a mile : y = total daily cost, and x = number of miles driven per day : Therefore, if you drive it 5 mi (x) then cost will be \$35 (y), right?
 Surface-area/104629: This question is from textbook saxon algebra 2 Two fences in a field meet at 120 degrees. A cow is tethered at their intersection with a 15-foot rope, as shown in the figure. Over how many square feet may the cow graze?1 solutions Answer 76283 by ankor@dixie-net.com(15652)   on 2007-10-17 19:38:57 (Show Source): You can put this solution on YOUR website!Two fences in a field meet at 120 degrees. A cow is tethered at their intersection with a 15-foot rope, as shown in the figure. Over how many square feet may the cow graze? : First, find the area the cow could graze if there were no fence, just a circular area with 15' radius : A = pi*15^2 A = pi*225 A = 706.85 sq ft : The portion of the circle limited by the fence would be: = * 706.85 = 235.62 sq ft
 Miscellaneous_Word_Problems/104786: A wire 36 inches long is cut into two pieces and then bent into two square frames. The two frames have sides differing by five inches. What is the sum of the two areas?1 solutions Answer 76280 by ankor@dixie-net.com(15652)   on 2007-10-17 19:27:20 (Show Source): You can put this solution on YOUR website!A wire 36 inches long is cut into two pieces and then bent into two square frames. The two frames have sides differing by five inches. What is the sum of the two areas? : Let x = length of the side of the smaller square: Perimeter = 4x and (x+5) = length of the side of the larger square Perimeter = 4(x+5) : The total perimeter of the two squares = 36 in, solve for x: 4x + 4(x+5) = 36 4x + 4x + 20 = 36 8x = 36 - 20 8x = 16 x = 16/8 x = 2" length of the smaller square and 2 + 5 = 7" length of the larger square : Sum of the Areas: 2^2 + 7^2 = 4 + 49 = 53 sq/inches