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 Linear-equations/220636: i'm told to use an input/ output table for this problem, but because i dont have an X varriable i dont know what to do, i need a little help. 6y=18 1 solutions Answer 165591 by Theo(3461)   on 2009-10-02 09:03:48 (Show Source): You can put this solution on YOUR website!your equation is 6y = 18 solving for y, you get y = 18/6 = 3 this is true for all values of x. a graph of this equation is shown below: your input / output table would look like this: x / y -5 3 -4 3 -3 3 -2 3 -1 3 0 3 1 3 2 3 3 3 4 3 5 3 -----
 Trigonometry-basics/220645: Find the exact value of tan 15 degrees1 solutions Answer 165590 by Theo(3461)   on 2009-10-02 08:59:49 (Show Source): You can put this solution on YOUR website!tan (15) = .267949192 by the calculator. tan (15) = 0.267949192431123 by excel spreadsheet function. ----- I don't know if it just stops there or if excel doesn't have the capability to extend the fraction any further. ----- For it to be exact, it would have to be the result of the ratio of two integers. ---- I would be at a total loss to determine what integers would result in that fraction. ----- my guess is that it's probably irrational and so the fraction would go on endlessly without any repeating patterns to indicate that even if it's an endless fraction, it was the result of the ratio of two integers. ----- but that's just a guess. I really don't know. -----
 Equations/220640: A physician monitored the effects of aspirin in patients and found that if originally 200mg of aspirin is administered for a headache, sometime later, say 20minutes, only 50% of the aspirin will still be in the bloodstream. after 40 minutes, only 50 mg will still be in the bloodstream and after 60 minutes, only 25mg. a) determine a formula to establish the amount of aspirin in the bloodstream after t minutes. b) how long will it take for the bloodstream to be free of aspirin?1 solutions Answer 165583 by Theo(3461)   on 2009-10-02 08:12:14 (Show Source): You can put this solution on YOUR website!amount of aspirin in the bloodstream is 1/2 of the original amount every 20 minutes. ----- formula is: y = 200*(.5)^(.05x) ----- graph of this equation looks like this: ----- this equations looks like it models it pretty well. y is the amount of aspirin left in the bloodstream. x is the number of minutes. the horizontal lines are at 200,100,50,25 to show you the amount of aspirin left in the bloodstream after 0,20,40,60 minutes. ----- It does not appear that the bloodstream will ever be completely free of aspirin based on this formula. .5^(.05x) will never be equal to 0. It will become very small, but never zero. the calculator might show 0, but that's because the calculator can only go so far in the number of decimal places it can show, even with scientific notation. ----- My calculator shows a result of 0 when 200 is multiplied by .5^333 It shows 2.285974783 * 10^-98 when 200 is multiplied by .5^332 Since .05*x = 333 when the calculator give a result of 0, this means that x = 20*333 = 6660 minutes That's 4.625 days for the body to be completely free of aspirin enough so that my calculator can't determine how much is still left. ----- best thing to do with this is determine a point at which the level of aspirin in the blood is less than a certain amount. ----- that can be calculated. ----- assume that amount is 1 milligram. ----- formula would be 1 = 200*(.5)^(.05x) divide both sides of this formula by 200 to get: 1/200 = (.5)^(.05x) take the log of both sides of this equation to get: log(1/200) = (.05x)*log(.5) divide both sides of this equation by log(.5) to get: log(1/200)/log(.5) = (.05x) solve for (.05x) to get: (.05x = 7.64385619 solve for x to get: x = 152.8771238 minutes ----- this means that it will take 152.8771238 minutes for the aspirin in the bloodstream to be less than 1 milligrams. ----- the following graph should show that about as clear as can get with the tools available. I added the horizontal line at y = 1 to show you the cutoff point better. It's at somewhere around x = 153 which is just about right.
 Average/220634: From 11 positive integer scores on a 10-point quiz, the mean is 8, the median is 8, and the mode is 7. Find the maximum number of perfect scores possible on this test.1 solutions Answer 165572 by Theo(3461)   on 2009-10-02 06:24:10 (Show Source): You can put this solution on YOUR website!perfect score is 10 11 scores with: mean of 8 mode of 7 median of 8 ----- total of 11 scores with a median of 8 means that a maximum of 5 scores are below 8 and 5 scores are above 8. this will occur when there is only 1 number 8 in the distribution. we will assume only one number 8 in the distribution because that would allow for more number 10's if possible. we need at least one though, because we have 11 numbers and that means that one of the numbers has to be the median in order to get 5 on each side plus the 1 in the middle equal to 11 total. ----- mode of 7 means that 7 is the score that has occurred with the most frequency. ----- integer scores means that there are no fractions to worry about. ----- we need: 1 number 8 a minimum of 2 number 7's 5 numbers above 8 5 numbers below 8 ----- let's start off with 5 number 10's. that would be the maximum number of 10's without any other constraints except the fact that the median is 8 meaning the number of units above 8 had to be 5 at most. ----- 5 number 10's means that we would have to have 6 number 7's because 7 is the mode which is the number that occurs most frequently. we can't have 6 number 7's because the total of numbers below 8 has to be 5 and 6 is greater than 5. ----- so 5 number 10's is no good. ----- can we have 4 number 10's? ----- this means we must have at least 5 number 7's and number 7 would then be the most frequent number. if we have 4 number 10's, we would need 1 more number above 8 to make 5 numbers above 8. we'll pick the lowest number which is 9. so far we have: 4 number 10's equals a total of 40 1 number 9 equals a total of 9 1 number 8 equals a total of 8 5 number 7's equals a total of 35 ----- we have a total of 11 numbers with a grand total of 92 ----- we have exceeded the grand total of 88 so 4 number 10's is not possible given all the constraints. ----- can we have 3 number 10's? ----- 3 number 10's means we have to have a minimum of 4 number 7's to keep number 7 the most frequently used number in the set. we also need 2 more numbers above 8 which means we have to choose 9 because 9 is the smallest number above 8. so far we have: 3 number 10's equals a total of 30 2 number 9's equals a total of 18 1 number 8 equals a total of 8 4 number 7's equals a total of 28 ----- this give us a total of 10 numbers with a grand total of 84 ----- we need 1 more number that equals 4. adding 1 number 4 give us a total of 11 numbers with a grand total of 88. ----- the maximum number of 10's we can have is 3. since 10 is a perfect score, then: the maximum number of perfect scores we can have is 3. -----
 Numbers_Word_Problems/220635: Seven judges graded Joshua.s science project, and the mean of those seven scores was 8.0. Scores could be any integer value 1-10. When the highest and lowest scores were removed, the mean of the five remaining scores was 7.8. What was the least possible score that could have been removed? 1 solutions Answer 165568 by Theo(3461)   on 2009-10-02 05:22:26 (Show Source): You can put this solution on YOUR website!n = 7 m = 8 range = 1 to 10 high = h low = l n = 5 m = 7.8 least possible score to remove ----- T1 = 7*8 = 56 T2 = 5*7.8 = 39 R = 56-39 = 17 ----- H + L = 17 L = 17 - H ----- highest score is 10 L = 17 - 10 = 7 lowest score had to be 7 ----- the average times the number of items equals the total score ----- 7 * 8 = 56 total score ----- 5 * 7.8 = 39 total score ----- difference between 7 and 5 is 2 difference between 56 and 39 is 17 2 papers had to have a total score of 17 lowest score plus highest score had to add up to 17. lowest score could not be lower than 7 because that would make the highest score greater than 10 which is not possible. -----
 logarithm/220602: can you please solve for x. 2ln(x-5)=3.651 solutions Answer 165564 by Theo(3461)   on 2009-10-02 04:42:26 (Show Source): You can put this solution on YOUR website!2*ln(x-5) = 3.65 ----- your equation is: 2*ln(x-5) = 3.65 divide both sides of this equation by 2 to get: ln(x-5) = 1.825 by the basic laws of logarithms, this can happen if and only if: e^1.825 = (x-5) e^1.825 = 6.202795019 (use your calculator to get this) this means that: 6.202795019 = x-5 add 5 to both sides of this equation to get: x = 11.202795019 ----- the basic law of logarithms states: if and only if ----- with natural logarithms, the base is your x was actually (x-5) your y was 3.65 / 2 = 1.825 -----
 Percentage-and-ratio-word-problems/220420: Hi - I got the answer to this by working it out but I cannot get to an equation. Can you help me? A class has a boy-girl ratio of 5:3. Three more girls join the class, changing the ratio of 10:7. How many students are now in the class? By trying and just plugging in numbers I got 30 and 18 plus 3 more, so now the total is 51 - but I'm not sure how to set it up as a problem! Thank you!1 solutions Answer 165486 by Theo(3461)   on 2009-10-01 19:54:54 (Show Source): You can put this solution on YOUR website!A class has a boy-girl ratio of 5:3. Three more girls join the class, changing the ratio of 10:7. How many students are now in the class? By trying and just plugging in numbers I got 30 and 18 plus 3 more, so now the total is 51 - but I'm not sure how to set it up as a problem! ----- let b = number of boys. let g = number of girls. ----- b/g = 5/3 this means that b = 5g/3 ----- add 3 girls and the ratio becomes 10/7 this means that: b/(g+3) = 10/7 this means that: b = (10g + 30)/7 ----- since both these equations equal to b, then they must equal to each other. you get: 5g/3 = (10g + 30)/7 multiply both sides by 7 and you get: 35g/3 = 10g + 30 multiply both sides by 3 to get: 35g = 30g + 90 subtract 30g from both sides to get: 5g = 90 divide both sides by 5 to get: g = 18 ----- original ratio led to bg = 5g/3 5*18/3 = 5*6 = 30 original class size is 30 boys and 18 girls. ----- add 3 girls and the class size is 30 boys and 21 girls. boy to girl ratio is now 30/21 which equals 10/7 -----
 Geometry_Word_Problems/220403: the number of diagonals that can be drawn from one vertex in a convex polygon that has n verticles 1 solutions Answer 165469 by Theo(3461)   on 2009-10-01 19:16:16 (Show Source): You can put this solution on YOUR website!Triangle has none Rectangle has 1 Pentagon has 2 hexagon has 3 ----- Looks like it's n - 3 where n is the number of vertices. ----- Stands to reason because one vertex has two adjacent vertices that it can't connect to. The rest it can. ----- Octagon would be 5 ----- Click on the following hyperlink: Number of Diagonals of a Polygon ----- Look at the polygon from one of the vertices only. Click on more and less to change the number of sides of the polygon. You'll see that the number of diagonals from one vertex to the other vertices is n-3. ----- This website goes even further to tell you the total number of diagonals from all vertices. ----- The formula is n * (n-3) / 2 ----- Example: hexagon is 6 sides 6 * 3 = 18/2 = 9 total diagonals. There's 3 from each vertex to all the other vertices. There's 6 vertices in total. That's where the n * (n-3) comes from. Each diagonal touches 2 vertices, so the number of diagonals has to be cut in half. That's where the / 2 comes from. -----
 Equations/220294: I need an equation that has a domain of [-4, 5/3) U (5/3, infinity) if you could at least walk me through how to find it, that would be great.1 solutions Answer 165437 by Theo(3461)   on 2009-10-01 16:52:12 (Show Source): You can put this solution on YOUR website!5/3 can't be part of the domain. If you divide by (3x-5) then when x = (5/3) you will be dividing by 0 because 3*(5/3) - 5 = 5-5 = 0. That makes x = (5/3) not part of the domain. ----- Your equation will be something divided by (x-5/3). ----- Now we need to restrict your domain to values of x >= -4 ----- We can do that with square root of something that makes the something inside the square root negative when x < -4. ----- Equation of should do it. ----- When x < -4, expression within the square root sign is negative. This forces the domain to be >= -4. ----- When x = 5/3, denominator is 0 causing value of the equation to be undefined. This forces the domain to exclude x = 5/3. ----- Graph of this equation is shown below: ----- Table of Values from x = -10 to x = 10 are shown below: ----- x_______________y -10____________#NUM! -9____________#NUM! -8____________#NUM! -7____________#NUM! -6____________#NUM! -5____________#NUM! -4____________0 -3____________-0.071428571 -2____________-0.128564869 -1____________-0.216506351 0____________-0.4 1____________-1.118033989 (5/3)________#DIV/0! 2____________2.449489743 3____________0.661437828 4____________0.404061018 5____________0.3 6____________0.243252128 7____________0.207289049 8____________0.182321138 9____________0.163888694 10____________0.149666295
 Finance/220275: The monthly payment on a 2-year loan is \$500. The interest rate is 6.65% compunded monthly. What is the amount of the loan?1 solutions Answer 165421 by Theo(3461)   on 2009-10-01 15:10:28 (Show Source): You can put this solution on YOUR website!Formula for present value of a series of payments is: PRESENT VALUE OF A PAYMENT PV = present value PMT = payment per time period i = interest rate per time period n = number of time periods ----- 6.55% per year compounded monthly is: 6.55/12 = .554166667% per month which equals a rate of: .00554166667 per month. Number of month is 2 years * 12 months = 24 months payment is \$500 per month. ----- PV = what we want to find. PMT = \$500 per month i = .00554166667 per month n = 24 months Formula becomes: which equals: \$11,224.28901 ----- The loan was \$11,224.28901 The lender received \$12,778.06 by the end of the loan (Future Value of a Series of Payments). Principal was \$11,224.29 Interest was \$1,553.77
 Money_Word_Problems/220267: how many years would it take to double \$100 if it earned interest at a rate of 8% per year?1 solutions Answer 165410 by Theo(3461)   on 2009-10-01 14:43:16 (Show Source): You can put this solution on YOUR website! This becomes: This becomes: This becomes: x = 9.006468342 ----- to prove this is true, substitute in original equation to get: This becomes: 200 = 200 confirming the answer of x = 9.006468342 is good. ----- It would take 9.006468342 years to double your money at 8% compounded yearly. -----
 Mixture_Word_Problems/220184: how much pure acid should be mixed with 3 gallons of a 20% solution in order to get a 60% acid solution?1 solutions Answer 165362 by Theo(3461)   on 2009-10-01 11:31:51 (Show Source): You can put this solution on YOUR website!Let S = amount of new solution Then .6*S = the amount of acid in the new solution. ----- Amount of Acid in existing 3 gallon solution is equal to .2 * 3 = .6 gallons of acid. ----- Let X be the amount of acid We need to add to the existing solution. ----- The new solution is equal to the old solution plus X because we are adding pure acid which is represented by X. ----- This means that S = X + 3 ----- The amount of acid in the existing solution is .6 gallons We will be adding X gallons of acid. The Total amount of acid in the new Solution will be X + .6 That will equal to 60% of the new solution which equals .6*S ----- This means that (X+.6) = .6*S ----- Since we know that S = (X+3), we can susbtitute in our equation to get: (X+.6) = .6*(X+3) ----- This reduces one equation in two unknown into one equation in one unknown which we can then solve. ----- We remove parentheses to get: X + .6 = .6*X + .6*3 We Subtract .6 from both sides of the equation and we simplify to get: X = .6*X + .6*3 - .6 which becomes: X = .6*X + 1.2 We subtract .6*X from both sides of this equation to get: X - .6*X = 1.2 which simplifies to: .4*X = 1.2 We divide both sides of this equation by .4 to get: X = 1.2/.4 = 3 ----- Answer is you have to add 3 gallons of acid to 3 gallons of 20% acid to get a solution that is 60% acid. ----- .6 gallons of acid in the original mixture + 3 gallons of acid = 3.6 gallons in the new mixture. ----- 3 gallons of the original mixture + 3 gallons of of acid = 6 gallons of the new mixture. ----- 3.6 / 6 = .6 * 100% = 60% of acid in the new mixture.
 Quadratic_Equations/220173: Identify the coefficients that would be used in the quadratic formula for this equation. 5x^2 = 4x a. a = 5, b = 4, c = 1 b. a = -5, b = 4, c = 0 c. a = 5, b = 4, c = 0 d. a = 5, b = -4, c = 01 solutions Answer 165356 by Theo(3461)   on 2009-10-01 10:43:33 (Show Source): You can put this solution on YOUR website!The standard form of the quadratic equation is: ax^2 + bx + c = 0 ----- Your equation is 5x^2 = 4x Substract 4x from both sides of this equation to get: 5x^2 - 4x = 0 Your a coefficient is 5 Your b coefficient is -4 Your c coefficient is 0 That would be selection d. -----
 Rational-functions/220158: Give the domain f(x)= x^4+4/x^2+6x-16 Answer shows (-∞,-8)u(-8,2)u(2,∞) I am coming up with (-∞,-8)u(2,∞) Which is correct and why? Thank you1 solutions Answer 165333 by Theo(3461)   on 2009-10-01 09:02:35 (Show Source): You can put this solution on YOUR website! ----- They are right. Here's why. Your domain is all real values of x EXCEPT x = -8 and x = 2 ----- (-,8) covers the interval from minus infinity < x < -8. (2,) covers the interval from 2 < x < infinity. ----- you are missing the interval from -8 < x < 2 represented by: (-8,2) ----- number line would show that as follows: ------------------------(-8)---------------------(2)----------------
 Polynomials-and-rational-expressions/220141: What is a common factor? Where do you use the common factor in an expression consisting of various terms?1 solutions Answer 165332 by Theo(3461)   on 2009-10-01 08:45:52 (Show Source): You can put this solution on YOUR website!A common factor is a factor that applies to multiple terms. Suppose you have number 1 and 3 The only common factor between these two number is 1. 1//1 = 1 3/1 = 3 ----- As a point in fact, though, we do not consider the number 1 as a common factor because all number can be divided by 1. ----- We look for common factors greater than 1. ----- Take the numbers 2 and 4 A common factor is 2 because: 2/2 = 1 4/2 = 2 Both number can be divided by 2 and yield an integer result. ----- Integer result is another requirement, since all numbers can be divided by another number smaller than them and yield an answer if the answer does not have to be an integer. ----- Common factors for 7 and 5 are????? There are none becauswe there is no number than can be divided evenly into 7 and 5 at the same time. Matter of fact, there is no number smaller than these number and greater than 1 that can be divided into each of these numbers by itself because these numbers are prime numbers. ----- Take 9 and 3 3 goes evenly into 9 and 3 goes evenly into 3 so 3 is a common factor of 3 and 9. ----- Common factors are used to manipulate terms and expressions so they can be solved easier. ----- We can use them to simplify division. Take the case of: 5*10*7*3*9 / 5*7*3 ----- We have a common factor of 5 in the numerator that will cancel out a 5 in the denominator. Similarly with 7 and 3. Out equation reduces to: 10*9 which equals 90. ----- Using the common factor helped reduce the equation so it could be solved easier. ----- We can use it to simplify multiplication. Take the case of 5*7 + 3*7 + 6*7 We have a common factor of 7 that can change this equation to: 7 * (5+3+6) which equals 7 * 14 which equals 98 ----- Using the common factor helped reduce the equation so it coulde be solved easier. ----- Look at x^ + 2x + 1 / (x+1) The numerator in this equation can be factored to equals (x+1)^2 This changes our equation to: (x+1)^2 / (x+1) which becomes ( (x+1) * (x+1) ) / (x+1) Take out the common factor of (x+1) from the numberator and denominator and you are left with: x+1 ----- It's true this equation could also have been solved by dividing (x+1) into x^2 + 2x + 1 as follows: equation of x^2+2x+1 divided by (x+1) = x with a remainder of x+1 remainder of x+1 divided by (x+1) = 1 with no more remainder. Solution is x+1 ----- We got the same answer but we have to work harder to get it. Recognizing the common factor made solution of the problem easier. -----
 Linear-equations/220119: How do I solve the following problem? A boat manufacturer makes fishing boats that are sold for a profit of \$500. each, and canoes that are sold for a profit of \$400 each. Each fishing boat requires 100 assembly hours and 25 finishing hours, while each canoe requires 75 assembly hours and 50 finishing hours. How many fishing boats and how many canoes should be made to maximize the manufacturer's profit? Let x= the number of fishing boats and y=the number of canoes. thanks for your help.1 solutions Answer 165329 by Theo(3461)   on 2009-10-01 08:25:45 (Show Source): You can put this solution on YOUR website!x = number of fishing boats y = number of canoes ----- profit per fishing boat is \$500 profit per canoe is \$400 profit equation would be: p = 500*x + 400*y ----- your constraints appear to be: total assembly hours = 100*x + 75*y total finishing hours = 25*x + 50*y ----- To keep these straight, it helps to create a table that would look something like: ----- craft_________________assembly time required__________finishing time required fishing boat_________________100 hours______________________25 hours canoe______________________75 hours______________________50 hours ----- you appear to be missing some constraints like total assembly hours <= ... and total finishing hours <= ... ----- I will insert some hours and solve for that. You can then replace whatever I have with whatever you have and solve the problem for yourself. ----- Without those constraints, the problem is already solved because if he has unlimited time to assemble and finish then he should sell all fishing boats because he gets more profit per fishing boat. ----- I'll put in some constraints and we'll see how it works out. ----- Total hours available for assembly are <= 1000 Total hours available for finishing are <= 500 ----- Our constraints become: 100x + 75y <= 1000 25x + 50y <= 500 ----- Summary: Profit equation is 500*x + 400*y Constraint equation number 1 is 100x + 75y <= 1000 Constraint equation number 2 is 25x + 50y <= 500 ----- We graph the constraint equations. ----- We do that by solving for y in both equations. ----- We solve for equality, not for inequality. This allows us to graph the equations. Inequality is handled by looking at the area above or below the graph of the equations. ----- Constraint equation 1 becomes y = (1000-100x)/75 Constraint equation 2 becomes y = (500-25x)/50 ----- graph of these equations is shown below: ----- Picture of this graph with the area of possible solutions can be viewed by clicking on the following hyperlink. ----- Graph of Equations and Area of Possible Solutions ----- The area of possible solution is bounded by: The x-axis because the value of y, which is number of canoes, has to be >= 0. The y-axis because the value of x, which is number of fishing boats, has to be >= 0. The line y = (1000-100x)/75 because the number of hours to assemble a fishing boat (x) and a canoe (y) has to be <= 1000. the line y = (500-25x)/50 because the number of hours to finish a fishing boat (x) and a canoe (y) has to be <= 500. ----- Remember that x equals the number of fishing boats, and y equals the number of canoes. ----- The equation for total assembly time is 100*x + 75*y <= 1000 We converted this to: y <= (1000-100*x)/75 in order to graph it, and then we graphed: y = (1000-100*x)/75 ----- When y equals the value of this equation, y will be ON the line. When y is smaller than the value of this equation, y will be BELOW the line. ----- Our area of possible solutions for this equation are thye points ON the line and the points BELOW the line. ----- Same for the other equation. ----- The possible solutions for the x-axis are the value of y ON the x-axis and the values of y ABOVE the x-axis. The possible solutions for the y-axis are the values of x ON the y-axis and the values of x TO THE RIGHT OF the y-axis. ----- The rules of linear programming states that the maximum/minimum values will be at the intersection of the boundaries of the constraints. ----- This means we have to look at the points: (0,10) (0,0) (10,0) (4,8) ----- We apply these points to our revenue equation of 500*x + 400*y Point (0,10) yields 500*0 + 400*10 = \$4,000.00 Point (0,0) yields 500*0 + 400*0 = \$0.00 Point (10,0) yields 500*10 + 400*0 = \$5,000.00 Point (4,8) yields 500*4 + 400*8 = \$2,000.00 + \$3,200 = \$5,200 ----- Our maximum profit can be attained when we build and sell 4 fishing boats and 8 canoes. ----- Why couldn't we get more profit when we build all fishing boats and no canoes? It would seem that a greater profit on each fishing boat should result in building all fishing boats. ----- Here's what happened. ----- When we built 10 fishing boats, we used up all the assembly time available so we had no assembly time left over for the canoes. That means we had to stop, even though we had some finishing time left for the canoes. ----- When we built 4 fishing boats, we still have assembly time left over for 8 canoes. 4*100 = 400 subtracted from 1000 = 600 divided by 75 hours for each canoe equals 8 canoes able to be assembled with the time left over. This resulted in a total of 12 craft being assembled and finished and sold (presumption is you are able to sell all that you build). Even though the profit for each canoe was less than the profit for each fishing boat, more craft total yielded more total profit. 4*500 + 8*400 = 5200 while 10*500 = 5000. -----