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# Recent problems solved by 'AnlytcPhil'

Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289, >>Next

 Trigonometry-basics/748654: Show that the value of tan 3 α cot α cannot lie between 1/3and 3?1 solutions Answer 455637 by AnlytcPhil(1276)   on 2013-05-14 19:55:27 (Show Source): You can put this solution on YOUR website!```tan3α·cotα; It's easy if you're taking calculus, and can use L'Hopital's rule, but not if you aren't. Tell me in the thank-you note form if you are taking calculus or just taking trigonometry. Then I'll help you with it. Is this familiar? Numerator and denominator both approach 0, so we can use L'hopital's rule: Using reciplocal identity = 3 That's the start of the way to do it by calculus. I'll finish it if you tell me whether you're taking calculus or just trig. Edwin```
 Circles/748607: the circle passes through the point (0,0), (5,0) and (3,3)1 solutions Answer 455566 by AnlytcPhil(1276)   on 2013-05-14 14:50:26 (Show Source): You can put this solution on YOUR website!the circle passes through the point (0,0), (5,0) and (3,3) ```You can do this either by using the general form or the standard form. I'll first do it with the general form: x² + y² + Dx + Ey + F = 0 Substitute (x,y) = (0,0) 0² + 0² + D(0) + E(0) + F = 0 F = 0 So we have x² + y² + Dx + Ey + 0 = 0 x² + y² + Dx + Ey = 0 Substitute (x,y) = (5,0) 5² + 0² + D(5) + E(0) = 0 25 + 5D = 0 25 + 5D = 0 5D = -25 D = -5 So we have: x² + y² + Dx + Ey = 0 x² + y² - 5x + Ey = 0 Substitute (x,y) = (3,3) x² + y² - 5x + Ey = 0 (3)² + (3)² - 5(3) + E(3) = 0 9 + 9 - 15 + 3E = 0 3 + 3E = 0 3E = -3 E = -1 x² + y² - 5x + Ey = 0 x² + y² - 5x - y = 0 ------------------------------------------ You can also use the standard form equation for a circle: (x - h)² + (y - k)² = r² Substitute (x,y) = (0,0) (0 - h)² + (0 - k)² = r² h² + k² = r² Substitute (x,y) = (5,0) (5 - h)² + (0 - k)² = r² 25 - 10h + h² + k² = r² Substitute (x,y) = (3,3) (3 - h)² + (3 - k)² = r² 9 - 6h + h² + 9 - 6k + k² = r² 18 - 6h - 6k + h² + k² = r² So we have this system of equations: h² + k² = r² 25 - 10h + h² + k² = r² 18 - 6h - 6k + h² + k² = r² If we subtract the first equation term by term from each of the other two equations we have: 25 - 10h = 0 18 - 6h - 6k = 0 Solving the first for h 25 - 10h = 0 -10h = -25 h = h = Substituting in 18 - 6h - 6k = 0 18 - 6 - 6k = 0 18 - 15 - 6k = 0 3 - 6k = 0 -6k = -3 k = k = Substituting in h² + k² = r² So the equation (x - h)² + (y - k)² = r² becomes (x - )² + (y - )² = It was harder but mainly because the center and radius were fractions. Here is the graph. Notice that the circle passes through the given points and has center (,) or (2.5,.5) The radius is ≈ 2.55 Edwin```
 Matrices-and-determiminant/748403: how do i solve the following matrix by hand? 2x+y=-4 -2y+4z=0 3x-2z=-111 solutions Answer 455479 by AnlytcPhil(1276)   on 2013-05-13 22:02:05 (Show Source): You can put this solution on YOUR website!```eq. 1 2x+ y = -4 eq. 2 -2y+4z = 0 eq. 3 3x -2z = -11 The first equation has no z-term. So eliminate z from the other two terms. Multiply eq. 3 by 2 and add term by term: -2y+4z = 0 6x -4z = -22 -------------- 6x-2y = -22 Divide that through by 2 eq. 4 3x- y = -11 Now we have a system of two equations and two unknowns: eq. 1 2x+y = -4 eq. 4 3x-y = -11 We add them term by term and y will be eliminated: eq. 1 2x+y = -4 eq. 4 3x-y = -11 ------------------ 5x = -15 x = -3 Substitute x = -3 in eq. 4, eq. 4 3x-y = -11 3(-3)-y = -11 -9-y = -11 -y = -2 y = 2 Substitute x = -3 in eq. 3 eq. 3 3x-2z = -11 3(-3)-2z = -11 -9-2z = -11 -2z = -2 z = 1 Solution (x,y,z) = (-3,2,1) Edwin```
 logarithm/748387: Use a graphing calculator to find approximate solutions of the equation. xe^(3x)-1= 3 I have a graphing calculator and know how to graph, but have no idea how to go about it with using the logarithms.1 solutions Answer 455471 by AnlytcPhil(1276)   on 2013-05-13 21:21:17 (Show Source): You can put this solution on YOUR website!```If you have a TI-83 or 84, Press ON Press CLEAR Press Y= With cursor just after \Y1= Press the key that has X,T,θ,n printed on it Press 2ND Press LN Press 3 Press X,T,θ,n Press ) Press - Press 1 Press ENTER You should see \Y1=Xe3X-1 if you have a newer calculator and you'll need to press the right arrow after the exponent or \Y1=Xe^(3X)-1 if you have the older model With cursor just after \Y2= Press 3 You should see \Y1=Xe3X-1 \Y2=3 Press ZOOM Press 6 You should see a graph like this: Press 2ND Press TRACE See CALCULATE screen Press 5 Press ENTER Press ENTER Press ENTER Tou should read at the bottom of the screen under the graph Intersection X=.62093895 Y=3 Solution: x=.62093895 Edwin```
 Probability-and-statistics/748226: If the probability of not winning is 0.526, what are the odds for winning?1 solutions Answer 455394 by AnlytcPhil(1276)   on 2013-05-13 15:29:23 (Show Source): You can put this solution on YOUR website!what are the odds for winning? the probability of not winning is 0.526 ```That's the probability of losing. So the probability of winning is 1 - 0.526 = 0.474 0.474 : 0.526 Multiply both parts of the ratio by 1000 474 : 526 Divide both parts of the ratio by 2 237 : 263 Edwin```
 Trigonometry-basics/739523: sec(x)=tan(x)csc(x)1 solutions Answer 451159 by AnlytcPhil(1276)   on 2013-04-19 09:18:24 (Show Source): You can put this solution on YOUR website!sec(x)=tan(x)csc(x) ```You will need to know This quotient identity: #1. and these reciprocal identities #2. #3. We will work ONLY with the right side: sec(x) = tan(x)csc(x) Use identity #1 to substitute for tan(x). Use identity #2 to substitute for csc(x). sec(x) = Cancel the sin(x)'s sec(x) = That is equivalent to: sec(x) = Now use identity #3 to substitute sec(x) for sec(x) = sec(x) Edwin```
 Coordinate-system/739518: HOW DO YOU FIND ORDERED PAIRS IF YOU DO NOW KNOW THE EQUATION 1 solutions Answer 451158 by AnlytcPhil(1276)   on 2013-04-19 09:01:12 (Show Source): You can put this solution on YOUR website!```If you will, in the thank-you note space, state the problem and instructions exactly word-for-word as it is stated in your book, I will answer it. Phil```
 Probability-and-statistics/739508: : A bag contains 7 red marbles labeled {1,2,3,4,5,6,7} and 5 green marbles labeled {1,2,3,4,5}. Four marbles are pulled out at once (i.e. with no particular order). What is the probability that (10 marks) (a) all four marbles are red?1 solutions Answer 451157 by AnlytcPhil(1276)   on 2013-04-19 08:54:38 (Show Source): You can put this solution on YOUR website!``` The set of marbles consists of: {1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5} Edwin```
 Rate-of-work-word-problems/737587: Working alone, it takes Kristin 11 hours to harvest a field. Kayla can harvest the same field in 16 hours. Find how long it would take them if they worked together. 1 solutions Answer 450452 by AnlytcPhil(1276)   on 2013-04-14 19:14:44 (Show Source): You can put this solution on YOUR website!>>...Working alone, it takes Kristin 11 hours to harvest a field...<< ```So Kristin's havesting rate is 1 field per 11 hours or or ``` >>...Kayla can harvest the same field in 16 hours...<< ```So Kayla's havesting rate is 1 field per 16 hours or or ``` >>...Find how long it would take them if they worked together...<< ```Let the answer be x hours So their combined havesting rate is 1 field per x hours or or The equation comes from + = Multiply through by LCD 176x 16x + 11x = 176 27x = 176 x = x = hours Edwin```
 Trigonometry-basics/737560: Power reduce sin^3⁡x as a sum (or difference) of first degree trig terms1 solutions Answer 450445 by AnlytcPhil(1276)   on 2013-04-14 18:49:09 (Show Source): You can put this solution on YOUR website!```The trick is to start with sin(3x) and simplify and it will come out with an equation that you can solve for sin³(x). sin(3x) = sin(2x+x) = sin(2x)cos(x)+cos(2x)sin(x) = 2sin(x)cos(x)cos(x)+cos(2x)sin(x) = 2sin(x)cos²(x)+[1-2sin²(x)]sin(x) = 2sin(x)[1-sin²(x)]+[1-2sin²(x)]sin(x) = 2sin(x) - 2sin³(x) + sin(x) - 2sin³(x) = 3sin(x) - 4sin³(x) So sin(3x) = 3sin(x) - 4sin³(x) 4sin³(x) = 3sin(x) - sin(3x) sin³(x) = sin(x) - sin(3x) Edwin```
 Matrices-and-determiminant/737510: Use Gaussian elimination. 2w-3x+4y+z=7 w-x+3y-5z=10 3w+x-2y-2z=61 solutions Answer 450430 by AnlytcPhil(1276)   on 2013-04-14 17:10:21 (Show Source): You can put this solution on YOUR website!``` This will likely have infinitely many solutions, since there are more letters, variables, than equations. 2w - 3x + 4y + z = 7 w - x + 3y - 5z = 10 3w + x - 2y - 2z = 6 Swap rows 1 and 2, to get a 1 in the upper left corner: abbreviated R1<->R2 To get a 0 where the 2 is, multiply row 1 by -2 and add it to row 2: abbreviated -2R1+R2->R2 To get a 0 where the bottom left 3 is, multiply row 1 by -3 and add it to row 3: abbreviated -2R1+R2->R2 To get a 1 where the -1 is on Row 2, multiply row 2 by -1 abbreviated -R2->R2 To get a 0 where the 4 is, multiply row 2 by -4 and add it to row 3: abbreviated -4R2+R3->R3 To get a 1 where the -19 is on Row 3, multiply row 3 by -1/19 Then we change the matrix back to equations: w - x + 3y - 5z = 10 x + 2y - 11z = 13 y - 3z = 4 Solve each equation for the first letter w = 10 + x - 3y + 5z x = 13 - 2y + 11z y = 4 + 3z Now we do back substitution: Sunstitute the expression for y in the middle equation: x = 13 - 2(4 + 3z) + 11z x = 13 - 8 - 6z + 11z x = 5 + 5z Substitute the expressions for x and y in the top equation: w = 10 + x - 3y + 5z w = 10 + (5 + 5z) - 3(4 + 3z) + 5z w = 10 + 5 + 5z - 12 - 9z + 5z w = 3 + z So the solution is (w,x,y,z) = (3+z, 5+5z, 4+3z, z) Some teachers will tell you to use a different letter than z for z, such as "a", or "k". If they use "a" the solution would be: (w,x,y,z) = (3+a, 5+5a, 4+3a, a) Edwin```
 Numbers_Word_Problems/724884: The units digit of a two-digit number is 2 more than the tens digit. If the number is divided by the sum of it's digits, the partial quotient is 4 and the remainder is 3. Find the number. I have figured out the number is 35, but again I am lost in finding the algebraic way to write it out. 1 solutions Answer 443931 by AnlytcPhil(1276)   on 2013-03-11 23:04:13 (Show Source): You can put this solution on YOUR website!```t = the tens digit u = the units digit 10t+u = the number t+u = sum of digits ``` The units digit of a two-digit number is 2 more than the tens digit. ```So u = t+2 ``` If the number is divided by the sum of its digits, the partial quotient is 4 and the remainder is 3. ```We divide as below. Multiply the partial quotient 4 by the divisor, t+u, getting 4t+4u. Then subtract that from the dividend, 10t+u, and get remainder 6t-3u: 4 t+u)10t+ u 4t+4u 6t-3u = Remainder And we are given that the remainder is 3. So 6t-3u = 3 So we have to solve this system of equations: u = t+2 6t-3u = 3 Solve that by substitution and get t = 3 and u = 5. So the number is 35. Edwin```
 sets-and-operations/724863: a) Determine the sets A, B where A − B = {1, 3, 7, 11}, B – A= {2, 6, 8}, and A ∩ B= {4, 9}. b) Determine the sets C, D where C − D = {1, 2, 4}, D − C = {7, 8}, and C ∪ D= {1, 2, 4, 5, 7, 8, 9} 1 solutions Answer 443923 by AnlytcPhil(1276)   on 2013-03-11 22:28:54 (Show Source): You can put this solution on YOUR website! Determine the sets A, B where A - B = {1, 3, 7, 11} ```From that we have: 1 ∈ A, 1 ∉ B 3 ∈ A, 3 ∉ B 7 ∈ A, 7 ∉ B 11 ∈ A, 11 ∉ B B - A= {2, 6, 8}, From that we have: 2 ∈ B, 2 ∉ A 6 ∈ B, 6 ∉ A 8 ∈ B, 8 ∉ A A ∩ B = {4, 9} From that we have: 4 ∈ A, 4 ∈ B, 9 ∈ A, 9 ∈ B So we end up with A = {1,3,7,11,4,9}, B = {2,6,8,4,9} ---------------------------------------- ``` C - D = {1, 2, 4}, ```From that we have: 1 ∈ C, 1 ∉ D 2 ∈ C, 2 ∉ D 4 ∈ C, 4 ∉ D D − C = {7, 8}, From that we have: 7 ∈ D, 7 ∉ C 8 ∈ D, 8 ∉ C C ∪ D = {1, 2, 4, 5, 7, 8, 9} Since 5 and 9 are not in either C - D or in D - C, they must be elements of both sets, since they were removed in both cases of set subtraction. Therefore C = {1,2,4,5,9}, D = {7,8,5,9} Edwin```
 Equations/724873: One way to remember something is to explain it to another person. Suppose that you are studying this lesson with a friend who thinks that she should let x = 0 to find the x-intercept and let y = 0 to find the y-intercept. How would you explain to her how to remember the correct way to find intercepts of a line?1 solutions Answer 443919 by AnlytcPhil(1276)   on 2013-03-11 21:32:18 (Show Source): You can put this solution on YOUR website!```All points on the x-axis are like (2,0), (-4,0), (7,0), etc. They all have their y-coordinates as 0. x-intercepts are all on the x-axis, so to find them, we let y=0. All points on the y-axis are like (0,3), (0,-5), (0,6), etc. They all have their x-coordinates as 0. y-intercepts are all on the y-axis, so to find them, we let x=0. Edwin```
 Quadratic-relations-and-conic-sections/724585: What is the center and radius of the circle with the given equation? (x – 1)^2 + (y + 1)^2 = 4 center (–1, 1); radius 4 center (1, –1); radius 4 center (–1, 1); radius 2 center (1, –1); radius 21 solutions Answer 443810 by AnlytcPhil(1276)   on 2013-03-11 14:13:44 (Show Source): You can put this solution on YOUR website!(x – 1)² + (y + 1)² = 4 Compare it to (x – h)² + (y - k)² = r² x - 1 = x - h h = 1 y + 1 = y - k k = -1 So the center = (h,k) = (1,-1) r² = 4 r = 2 So the radius is 2. Edwimn
 Geometry_proofs/724533: i need the proof of law of cosines and sines1 solutions Answer 443807 by AnlytcPhil(1276)   on 2013-03-11 13:57:22 (Show Source): You can put this solution on YOUR website!```There are three cases for the law of cosines and 2 cases for the law of sines: Case 1: a² = b²+c²-2·b·c·cos(A) when ∠A and ∠B are both acute angles. Draw altitude CD ⊥ AB. Label CD h for the height of ߡABC. Label the left part of c, AD as x and the right part of c DB as c-x. ߡADC and ߡBDC are right triangles, so by the Pythagorean theorem, h² = b²-x² and also h² = a²-(c-x)² Therefore equate their right sides: b²-x² = a²-(c-x)² b²-x² = a²-(c²-2cx+x²) b²-x² = a²-c²+2cx-x² Add x² to both sides: b² = a²-c²+2cx From ߡADC, = cos(A) or x = b·cos(A) Substituting that for x: b² = a²-c²+2c·b·cos(A) Isolate a² on the right side: b² + c² - 2c·b·cos(A) = a² That is equivalent to a² = b²+c²-2·c·b·cos(A) or since c·b = b·c, a² = b²+c²-2·b·c·cos(A) ----------------------- The law of sines for this case = sin(A) and = sin(B) h = b·sin(A) and h =a·sin(B) So b·sin(A) = a·sin(B) Divide both sides by sin(A)sin(B) = That's not complete yet, for we haven't shown that those equal to but it will be when we finish the next case. -------------------------------------------- a² = b²+c²-2·b·c·cos(A) when ∠A is acute and ∠B is obtuse. Extend AB and draw CD ⊥ AB. Label CD h for the height of ߡABD. Label the left part of c, AD as x and the right part of c DB as c-x. ߡADC and ߡBDC are right triangles, so by the Pythagorean theorem, h² = b²-(c+x)² and also h² = a²-x² Therefore equate their right sides: b²-(c+x)² = a²-x² b²-(c²+2cx+x²) = a²-x² b²-c²-2cx-x² = a²-x² Add x² to both sides: b²-c²-2cx = a² From ߡADC, = cos(A) c+x = b·cos(A) x = b·cos(A)-c Substituting in b²-c²-2cx = a² b²-c²-2c(b·cos(A)-c) = a² b²-c²-2·c·b·cos(A)+2c² = a² b²+c²-2·c·b·cos(A) = a² That is equivalent to a² = b²+c²-2·c·b·cos(A) or since c·b = b·c, a² = b²+c²-2·b·c·cos(A) --------------------------- The law of sines for this case = sin(A) and = sin(∠CBD) h = b·sin(A) and h =a·sin(∠CBD) So b·sin(A) = a·sin(∠CBD) Divide both sides by sin(A)sin(∠CBD) = Now we use the fact that the sine of an angle is equal to the sine of its supplement. Therefore if we erase the extended part we can label ∠CBD as ∠B. = The law of sines is now complete for the first case was for two angles being acute, and this case is for when one angle is obtuse. Since the other two angles are necessarily acute, the first case takes care of them and we have the complete law of sines: = = ------------------------------------ To prove the law of cosines when A is an obtuse angle, we have to accept the definition from xy-plane trigonometry that cos(180°-A) = -cos(A). We'll just take the above triangle and swap angles A and B and sides a and b: Extend BA and draw CD ⊥ BD. Label CD h for the height of ߡABC. Label the extended segment, AD, as x. ߡADC and ߡBDC are right triangles, so by the Pythagorean theorem, h² = b²-x² and h² = a²-(c+x)² Therefore equate their right sides: b²-x² = a²-(c+x)² b²-x² = a²-(c²+2cx+x²) b²-x² = a²-c²-2cx-x² Add x² to both sides: b² = a²-c²-2cx From ߡADC, = cos(CAD), x = b·cos(CAD) and since ∠CAD and ∠BAC are supplementary, cos(∠CAD) = -cos(∠BAC) and x = -b·cos(∠BAC) b² = a²-c²-2cx becomes: b² = a²-c²-2c[-b·cos(∠BAC)] b² = a²-c²+2c·b·cos(∠BAC) Isolate a² on the right b²+c²-2c·b·cos(∠BAC) = a² which is equivalent to a² = b²+c²-2c·b·cos(∠BAC) or since c·b = b·c, a² = b²+c²-2·b·c·cos(∠BAC) and if we erase the extended segment x, we can write ∠BAC as ∠A and have a² = b²+c²-2·b·c·cos(A) --------------------------- Edwin```
 Polygons/724520: A Regular decagon A,B,C,D,E,F,G,H,I,J is in the x-y plane.Measure of angle ADH in degrees is ? 1 solutions Answer 443779 by AnlytcPhil(1276)   on 2013-03-11 06:27:20 (Show Source): You can put this solution on YOUR website! ```Central angle AOH is made up of the 3 red central angles. Each of those three central angles measures 360°÷10 or 36°. So the 3 of them have measure 3×36° or 108°, so angle AOH has measure 108°. Therefore minor arc AH which angle ADH subtends has measure 108°. Angle ADH is an inscribed angle and an inscribed angle gas the measure of one-half of the measure of its subtended arc. Angle ADH subtends the same arc which central angle AOH subtends. Thus the measure of angle ADH is one-half of 108° and is therefore 54°. Edwin```
 Word_Problems_With_Coins/724111: I have equal numbers of 25paise and 50paise coins ina bag. find the number of each type of coin;if the total amount of money with me is Rs150?1 solutions Answer 443593 by AnlytcPhil(1276)   on 2013-03-10 05:01:20 (Show Source): You can put this solution on YOUR website!I have equal numbers of 25paise and 50paise coins ina bag. find the number of each type of coin;if the total amount of money with me is Rs150? ```Let x = the number of 25 paise coins, so also x = the number of 50 paise coins. 25x + 50x = 15000 75x = 15000 x = 200 200 25paise coins and 200 50paise coins. Edwin```
 Sequences-and-series/723780: Could you please teach me how to evaluate the sum of the following series? 1 + 2 – 3 + 4 – 5 + 6 – 7 + … + 20001 solutions Answer 443392 by AnlytcPhil(1276)   on 2013-03-09 03:34:51 (Show Source): You can put this solution on YOUR website!```1 + 2 – 3 + 4 – 5 + 6 – 7 + … + 2000 That has 2000 terms. Let's write in the next to the last three terms: 1 + 2 – 3 + 4 – 5 + 6 – 7 + … + 1998 - 1999 + 2000 We put parentheses in like this: 1 + (2 – 3) + (4 – 5) + (6 – 7) + … + (1998 - 1999) + 2000 Only the 1 on the left and the 2000 on the right are not in parentheses. Those are the only two numbers that are not in parentheses so the other 1998 of the 2000 numbers are in parentheses. Since there are two numbers in each set of parentheses there are one-half of 1998 or 999 sets of parentheses. Every one of those 999 sets of parentheses has -1 in it. So we have 999 negative one's which makes -999 plus the 1 on the left and the 2000 on the right. That makes the sum -999+1+2000 = 1002 Edwin```
 Probability-and-statistics/723569: Event A and Event B are mutually exclusive. P(A) =.27 and P(B) = .31 . What is P(A or B) ?1 solutions Answer 443225 by AnlytcPhil(1276)   on 2013-03-08 08:17:13 (Show Source): You can put this solution on YOUR website!Event A and Event B are mutually exclusive. P(A) =.27 and P(B) = .31 . What is P(A or B) ```Since they are mutually exclusive, they cannot both happen, so P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) P(A or B) = .27 + .31 - 0 P(A or B) = .58 Edwin```
 Rate-of-work-word-problems/723565: It takes Jeff 90 mins. to mow the yard. It takes Joe 105 mins to mow the yard. What is the approximate time it will take if they work together to mow the yard? this is the problem, how do I work it out? thanks1 solutions Answer 443221 by AnlytcPhil(1276)   on 2013-03-08 08:02:27 (Show Source):
 Miscellaneous_Word_Problems/721481: Now, there are a different number of ordinary bicycles, tandem bikes, and tricycles in the shop. there are 135 seats,118 front handlebars, and 269 wheels. how many regular bikes, tandem bikes, and tricycles are there?1 solutions Answer 442405 by AnlytcPhil(1276)   on 2013-03-03 22:23:50 (Show Source): You can put this solution on YOUR website!```We could also do it using algebra: Let x = the number of ordinary bicycles, Let y = the number of tandem bikes, and Let z = the number of tricycles there are 135 seats, There are x seats on the ordinary bicycles, 2y seats on the tandem bikes, and z seats on the tricycles. So that gives us the equation: x + 2y + z = 135 118 front handlebars, There are x front handlebars on the ordinary bicycles, y front handlebars on the tandem bikes, and z front handlebars on the tricycles. So that gives us the equation: x + y + z = 118 and 269 wheels. There are 2x wheels on the ordinary bicycles, 2y wheels on the tandem bikes, and 3z seats on the tricycles. So that gives us the equation: 2x + 2y + 3z = 269 So the system of equations is x + 2y + z = 135 x + y + z = 118 2x + 2y + 3z = 269 Solve that and get x=68 regular bikes, y=17 tandem bikes, and z=33 tricycles. Edwin```
 Polynomials-and-rational-expressions/714337: I would like to know how to factor 64x^3-27y^3. Also, i need to be provided with the steps to getting to my solution as well! Thank you1 solutions Answer 438777 by AnlytcPhil(1276)   on 2013-02-13 20:45:52 (Show Source): You can put this solution on YOUR website!```We memorize the way to factor the sum or the difference of cubes: A³ ± B³ = (A ± B)(A² ∓ AB + B²) That is, the sum of two cubes factors like this: A³ + B³ = (A + B)(A² - AB + B²) And the difference of two cubes factors like this: A³ - B³ = (A - B)(A² + AB + B²) Be sure to memorize these. Your problem is 64x³ - 27y³ since 64 = 4³ and 27 = 3³, we can write that as (4x)³ - (3y)³ So we substitute A = 4x and B = 3y in A³ - B³ = (A - B)[A² + AB + B²] (4x)³ - (3y)³ = (4x - 3y)[(4x)² + (4x)(3y) + (3y)²] Then we simplify the bracket expression: = (4x - 3y)[16x² + 12xy + 9y²] = (4x - 3y)(16x² + 12xy + 9y²) Edwin```
 Graphs/714246: How can you graph absolute value equations? |x|+|y|=2 1 solutions Answer 438771 by AnlytcPhil(1276)   on 2013-02-13 20:33:47 (Show Source): You can put this solution on YOUR website!```|x| + |y| = 2 We find the x-intercepts by setting y=0 |x| + |0| = 2 |x| = 2 x = ±2 So the x-intercepts are (2,0) and (-2,0) We find the y-intercepts by setting x=0 |0| + |y| = 2 |y| = 2 y = ±2 So the y-intercepts are (0,2) and (0,-2) We plot those: We find some more points. Let x=1 and solve for y |1| + |y| = 2 1 + |y| = 2 |y| = 1 y = ±1 so we have the points (1,1) and (1,-1) Let x=-1 and solve for y |-1| + |y| = 2 1 + |y| = 2 |y| = 1 y = ±1 so we have the points (-1,1) and (-1,-1) We plot those four points: And we sketch in the graph: Edwin```
 pythagorean/707488: how do you find one leg when only the hypotenuse is known? 1 solutions Answer 435721 by AnlytcPhil(1276)   on 2013-01-28 21:34:31 (Show Source): You can put this solution on YOUR website!You can't unless an angle is also known and you can use trigonometry.
 Graphs/707202: Write the standrad ang general equation of a circle through (2,1) and (3,5) and having its center on the line 8x + 5y = 8. 1 solutions Answer 435604 by AnlytcPhil(1276)   on 2013-01-28 16:12:27 (Show Source): You can put this solution on YOUR website! ```Here's another way to do the problem: Suppose (x,y) is the center of the circle. Then the distance from (x,y) to (2,1) must equal the distance from (x,y) to (3,5), so we use the diatance formula and set them equal: = sqrt( (x-3)^2 + (y-5)^2)}}} (x-2)² + (y-1)² = (x-3)² + (y-5)² Rearrange to get difference of squares on each side: (x-2)² - (x-3)² = (y-5)² - (y-1)² [(x-2)-(x-3)][(x-2)+(x-3)] = [(y-5)-(y-1)][(y-5)+(y-1)] [x-2-x+3]{x-2+x-3] = [y-5-y+1][y-5+y-1] [1][2x-5] = [-4][2y-6] 2x-5 = -8y + 24 2x + 8y = 29 So we have the system of equations: Solve that and get the center, and do the rest like the other solution. Edwin```
 Average/704261: If the average of three different positive integers is 5, then the least possible product of these numbers is: Please help1 solutions Answer 433999 by AnlytcPhil(1276)   on 2013-01-20 21:07:02 (Show Source): You can put this solution on YOUR website!```To have an average of 5, the sum of the three must be 15. Here is a fact about positive numbers we can use: If two positive numbers have a certain sum, their product is largest when they are closest together and smallest when they are farthest apart. However, here we have three integers with that sum, not just two. So that complicates matters a bit. But, we can consider only two numbers and use that fact above. Here's how: Suppose the three integers are a, b, and c, where a < b < c. We can consider just two numbers, (a+b) and c. We want to choose (a+b) and c as far apart as possible. To do that we choose (a+b) as small as possible, which is 3, when a=1 and b=2, and that choice will make c as much larger than 3 as possible. In order to have sum 15, c must be 12, and 12 is as far away from 3 as we can get. So the smallest possible product is 1*2*12 = 24. Edwin```
 test/704229: In multiplying two positive integers a and b, Joey inadvertently reversed the digits of the two-digit number a. His erroneous product was 161. What is the value of (ab)^2?1 solutions Answer 433990 by AnlytcPhil(1276)   on 2013-01-20 19:47:51 (Show Source): You can put this solution on YOUR website!In multiplying two positive integers a and b, Joey inadvertently reversed the digits of the two-digit number a. His erroneous product was 161. What is the value of (ab)^2? ------------------------------ ```By dividing 161 by odd prime numbers, trying to find one that will give a whole number, when we come to dividing 161 by 7 we get the integer 23. So 161 = 7×23 So the two-digit number "a" with the digits reversed was 23, Therefore a = 32, and b = 7. So (ab)² = (32×7)² = 224² = 50176 Edwin```
 Square-cubic-other-roots/704154: I have to simplify I have no idea how to do this please help! 1 solutions Answer 433986 by AnlytcPhil(1276)   on 2013-01-20 19:17:39 (Show Source): You can put this solution on YOUR website!``` The procedure is: 1. Put parentheses around the denominator: 2. Create the conjugate of the denominator and put it in parentheses 3. Put that conjugate over itself (which has value 1) 4. Multiply by that. (multiplying it won't affect the value since it equals 1) 1. Put denominator in parentheses: 2. To create the conjugate of , just change the sign of the second term, making the conjugate . Put it in parentheses . 3. Put that conjugate over itself, like this 4. Multiply by that: FOIL out the bottom: The two middle terms cancel out on the bottom: On the bottom, is just 5 and is just 3 Divide top and bottom by 2 Edwin```
 Trigonometry-basics/704213: establish each identity 44) 1- 1/2sin (2 theta) = sin^3 theta + cos^3 theta over sin theta + cos theta1 solutions Answer 433973 by AnlytcPhil(1276)   on 2013-01-20 17:21:12 (Show Source): You can put this solution on YOUR website!``` Factor the numerator on the right side as the sum of two cubes, using Use the identity on the 1st and 3rd terms: Use the identity multiplied through by or Edwin```
 Polynomials-and-rational-expressions/704183: the zeroes of a polynomial function are 1,1,-5/2 and -1, develope the function and then sketch the graph.1 solutions Answer 433969 by AnlytcPhil(1276)   on 2013-01-20 16:50:23 (Show Source): You can put this solution on YOUR website!``` First we'll develop a polynomial equation with solutions 1,1,-5/2, and -1. x = 1, x = 1, x = -5/2, x = -1 x-1 = 0 x-1 = 0 x+5/2 = 0 x+1 = 0 (x-1)(x-1)(x+5/2)(x+1) = 0 To avoid fractions multiply through by 2 (x-1)(x-1)(2)(x+5/2)(x+1) = 0 (x-1)(x-1)(2x+5)(x+1) = 0 (x²-2x+1)(2x²+7x+1) = 0 2x4+3x³-7x²-3x+5 = 0 That's a quartic equation with those roots. Therefore a quartic function with those zeros is: p(x) = 2x4+3x³-7x²-3x+5 To graph it 1. the coefficient of the leading term 2x4 is positive and therefor the graph goes upward on the extreme right side. 2. the degree is 4, which is an even number, so the graph also goes up on the extreme left. The first zero on the left is -5/2 or -2.5, so the curve comes downard from the left and since -2.5 has odd multiplicity, 1, it cuts through the x-axis at -2.5 going downward to the right. 2. The next zero is -1, so the graph must turn around and go upward. Since -1 has odd multiplicity, 1, it cuts through the x-axis at -1 going upward to the right. 3. The next and final zero is 1, so the graph must turn around and go downward. Since 1 has even multiplicity, 2, the graph bounces off the x-axis at 1 going back upward to the right. You might plot these points: (-2.6,4.1), (-2.5,0), (-2,-9), (-1.3,-3.8), (-1,0), (0,5), (.5,2.25), (1,0),(1.5,5) Edwin```