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Recent problems solved by 'stanbon'

stanbon answered: 57151 problems
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 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 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56970..56999 , 57000..57029 , 57030..57059 , 57060..57089 , 57090..57119 , 57120..57149 , 57150..57179, >>Next

Square-cubic-other-roots/187477: simplify sqrt(6) (8 sqrt(5)-6 sqrt(6)) 1 solutions Answer 140519 by stanbon(57219)   on 2009-03-19 21:14:59 (Show Source): You can put this solution on YOUR website! simplify sqrt(6) (8 sqrt(5)-6 sqrt(6)) ---------- = 8sqrt(30) - 6sqrt(36) = 8sqrt(30) - 6*6 = 8sqrt(30) - 36 ===================== Cheers, Stan H

Linear_Equations_And_Systems_Word_Problems/187480: Can you please help me with this problem and how you got to the answer. A carpenter goes to a hardware store to buy a specific number of pounds of nails for his business. The first store he enters sells his favorite-sized nails at nine cents per pound. At this price, the carpenter realizes that he would be short $7.15 if he tried to buy the nails. The carpenter goes to a competing hardware store and finds the nails sell for six cents per pound. He realizes that after paying for the nails, he would have $2.45 left over. How many pounds of nails is he trying to buy? Thanks :) 1 solutions Answer 140518 by stanbon(57219)   on 2009-03-19 21:09:58 (Show Source): You can put this solution on YOUR website! A carpenter goes to a hardware store to buy a specific number of pounds of nails for his business. -- Let # of pounds be "x" -------------------- The first store he enters sells his favorite-sized nails at nine cents per pound. At this price, the carpenter realizes that he would be short $7.15 if he tried to buy the nails. Let "y" be the money he has. 9x = y + 715 ------------------------------------- The carpenter goes to a competing hardware store and finds the nails sell for six cents per pound. He realizes that after paying for the nails, he would have $2.45 left over. 6x = y -245 ----------------------------- How many pounds of nails is he trying to buy? ----------- Rearrange the equations: y= 9x - 715 y = 6x + 245 ---------------- Substitute to solve for "x": 9x - 715 = 6x + 245 3x= 960 x = 320 lbs (# of pounds he is trying to buy) ================================================== Cheers, Stan H.

Rectangles/187430: is a rhombus a parallelogram 1 solutions Answer 140499 by stanbon(57219)   on 2009-03-19 18:44:54 (Show Source): You can put this solution on YOUR website! Yes a rhombus is a parallelogram. Cheers, Stan H.

Trigonometry-basics/187426: #1) If tan of theta = (6/5) and cos of theta < 0, use the fundamental identities to evaluate the other five trigonometric functions of theta. #2) use the fundamental identities to simplify csc squared if beta (1-cos squared of beta) #3) factor and simplify. (sec "to the power of 4" x- tan "to the power of 4" x/ sec "squared" x + tan "squared" x) 1 solutions Answer 140497 by stanbon(57219)   on 2009-03-19 18:40:25 (Show Source): You can put this solution on YOUR website! #1) If tan of theta = (6/5) and cos of theta < 0, use the fundamental identities to evaluate the other five trigonometric functions of theta. --- If tan(theta) = 6/5 and cos(theta) < 0 then y = -6 and x = -5 ------------------------------- Solve for "r": r = sqrt(6^2+5^2) = sqrt(61) --- sin(theta) = y/r = -6/sqrt(61) cos(theta) = x/r = -5/sqrt(61) tan(theta) = y/x = 6/5 ----------------- The csc, sec, and cot are the inverse of sin, cos, tan ================================================================ #2) use the fundamental identities to simplify csc squared of beta (1-cos squared of beta) --- csc^2(1 - cos^2) = csc^2 - (cos^2/sin^2) = (1/sin^2) - (cos^2/sin^2) = (1 - cos^2)/sin^2 = sin^2/sin^2 = 1 ------------------------------------------------------ ================================================================= #3) factor and simplify. (sec "to the power of 4" x- tan "to the power of 4" x/ sec "squared" x + tan "squared" x) ----------------------------------------- [sec^4(x) - tan^4(x)] / [sec^2(x) + tan^2(x)] = [(sec^2+tan^2)(sec^2-tan^2)] / [sec^2 + tan^2] = sec^2(x) - tan^2(x) = 1 ==================================================== Cheers, Stan H.

Polynomials-and-rational-expressions/187410: Solve for c. a/b=1/c-2 1 solutions Answer 140494 by stanbon(57219)   on 2009-03-19 18:12:41 (Show Source): You can put this solution on YOUR website! Solve for c. a/b = 1/(c-2) --- Invert both sides to get the "c" into the numerator: (c-2)/1 = b/a c = 2 + (b/a) c = (2a + b)/a ====================== Cheers, Stan H.

Linear-equations/187416: please solve using subsitution x=-y+6 y=-1x-4 1 solutions Answer 140492 by stanbon(57219)   on 2009-03-19 18:05:27 (Show Source): You can put this solution on YOUR website! please solve using subsitution x=-y+6 y=-1x-4 ----------- substitute for x to solve for "y": y = -(-y+6) + 6 y = y -6 + 6 y = y -------------- That means the two equations are really one equation. The solution is y = -x-4 ===================================== Cheers, Stan H.

Square-cubic-other-roots/187417: This question is from textbook (5+3Squareroot sign 2) ( squareroot sign3+squareroot2) 1 solutions Answer 140491 by stanbon(57219)   on 2009-03-19 18:00:05 (Show Source): You can put this solution on YOUR website! (5+3Squareroot sign 2) ( squareroot sign3+squareroot2) --------------- Use FOIL: F: 5sqrt(3) O: 5sqrt(2) I: 3sqrt(6) L: 3*2 = 6 ======================= Combine to get: 5sqrt(3) + 5sqrt(2) + 3sqrt(6) + 6 ======================== Cheers, Stan H.

Average/187425: 12(4r-5r/6)+ 20= 19r- 15+45r/2 1 solutions Answer 140490 by stanbon(57219)   on 2009-03-19 17:53:29 (Show Source): You can put this solution on YOUR website! [12(4r-5r/6)+ 20] = [19r- 15+45r/2] = [12[(24r-5r)/6] + 20] = [19r - ???? You need some parentheses here to clarity terms, numerators, and denominators???. =========================== cheers, Stan H.

Exponents-negative-and-fractional/187424: This question is from textbook Algebra 1 2r^-5 y^3 over n^2 divided by r^2 y^5 over 2n I have no clue where to start with this. Thank you for your help 1 solutions Answer 140487 by stanbon(57219)   on 2009-03-19 17:43:54 (Show Source): You can put this solution on YOUR website! 2r^-5 y^3 over n^2 divided by r^2 y^5 over 2n --------------------------------------------- [(2y^3/r^5)/n^2] / [r^2y^5/2n] = [(2n^2y^3/r^5] * [2n/r^2y^5] Cancel factors that are common to any numerator and a denominator, such as y^3 = [(2n^2/r^5] * [2n/r^2y^2] = [4n^3] / [r^7y^2] ================================== Cheers, Stan H.

Expressions-with-variables/187412: Simplify (C^0)^2+(2B)^1+(2B^3)^-2 Thank you! 1 solutions Answer 140481 by stanbon(57219)   on 2009-03-19 17:05:25 (Show Source): You can put this solution on YOUR website! Simplify (C^0)^2+(2B)^1+(2B^3)^-2 --- (1)^2 + (2B) + [1/(4B^6)] --- 1 + (2B) + (1/4B^6) lcd = 4B^6 Convert each term to a fraction with the lcd as its denominator: = (4B^6/lcd) + (8B^7/lcd) + 1/lcd Combine the numerators over the lcd: = [4B^6 + 8B^7 + 1]/(4B^6) ============================== Chers, Stan H.

Percentage-and-ratio-word-problems/187418: I need more help on this problem. I got the answer from my professor and he said it was 1/8. I don't know how to get the fraction as the answer. Any help would be great! The scruffy dog company sold 5600 dog booties last year and 4800 dog booties this year. By what FRACTION did their sales decrease this year as compared to last year? Thanks! 1 solutions Answer 140480 by stanbon(57219)   on 2009-03-19 16:55:32 (Show Source): You can put this solution on YOUR website! The scruffy dog company sold 5600 dog booties last year and 4800 dog booties this year. By what FRACTION did their sales decrease this year as compared to last year? -------------------- Quantity Decrease: 5600-4800 = 800 Fractional Decrease: 800/5600 = 1/7 --------------------------- Cheers, Stan H.

Proportions/187422: This question is from textbook mathematics for machine technology a sheet of steel 8 1/4 feet long weight 325 pounds. A piece 2 1/2 feet long is sheared from the sheet. determine the weight of the 2 1/2- foot piece to the nearest whole pound. 1 solutions Answer 140479 by stanbon(57219)   on 2009-03-19 16:49:43 (Show Source): You can put this solution on YOUR website! a sheet of steel 8 1/4 feet long weight 325 pounds. A piece 2 1/2 feet long is sheared from the sheet. determine the weight of the 2 1/2- foot piece to the nearest whole pound. --------------------------------------- Let the weight of the 2 1/2- foot piece be "x". Proportion: x/2 1/2 = 325/ 8 1/4 -------------------------- x/(5/2) = 325/(33/4) -- x = (5/2)(325/(33/4)) x = 98.48 lbs. =================== Cheers, Stan H.

Systems-of-equations/187420: 2x + 2y=8 and 4x-6y=36 solution (6,-2) how did my text book come up with thesesolutions 1 solutions Answer 140478 by stanbon(57219)   on 2009-03-19 16:43:30 (Show Source): You can put this solution on YOUR website! 2x + 2y=8 4x - 6y=36 ----------- Divide thru the 2nd equation by 2 to get: 2x + 2y = 8 2x - 3y = 18 ----------------- subtract the 2nd from the 1st equation to solve for "y": 5y = -10 y = -2 -------------- Substitute into 2x + 2y = 8 to solve for "x": 2x + 2*-2 = 8 2x - 4 = 8 2x = 12 x = 6 -------------------------------------------- solution (6,-2) ============================================= Cheers, Stan H.

Systems-of-equations/187421: 2x+ 4y=16 x+3=5 solution:(14,-3) need a break down of how my text book got these solutions 1 solutions Answer 140477 by stanbon(57219)   on 2009-03-19 16:38:44 (Show Source): You can put this solution on YOUR website! 2x+ 4y=16 x+3y =5 ------------ Divide thru the 1st equation by 2 to get: x + 2y = 8 x + 3y = 5 --------------- Subtract the 1st from the 2nd to solve for "y": y = -3 -------------- Substitute into x + 3y = 5 to solve for "x": x + 3*-3 = 5 x - 9 = 5 x = 14 --------------- solution:(14,-3) ====================== Cheers, Stan H.

Probability-and-statistics/187334: Seven girls and six boys walk into a video store. There are five salespeople available to help them. Find the probability that the salespeople will first help the given numbers of girls and boys. P(4 girls or 4 boys) P(all girls or all boys) P(at least 4 girls or at least 4 boys) (Sorry, I don't have the ISBN but it's on page 661 of Adding Probablilites of the www.algebra2.com book http://www.glencoe.com/sec/math/algebra/algebra2/algebra2_05/ ) 1 solutions Answer 140473 by stanbon(57219)   on 2009-03-19 14:50:27 (Show Source): You can put this solution on YOUR website! Seven girls and six boys walk into a video store. There are five salespeople available to help them. Find the probability that the salespeople will first help the given numbers of girls and boys. ----------------------------------------------- P(4 girls or 4 boys) = P(4 girls) + P(4 boys) = (7C4 + 6C4)/13C4 = 50/715 =0.0699 ----------------------- P(all girls or all boys) = P(5 girls) + P(5 boys) = 27/1287 = 0.0210 ------------------------------------------------------------------------ P(at least 4 girls or at least 4 boys) = 0.0699 + 0.0210 = 0.0909 ===================================================== Cheers, Stan H.

Exponential-and-logarithmic-functions/187395: After 13 years, 2.1 pounds of radioactive material remains from a 7 pound sample. What is the half-life of this material? 1 solutions Answer 140469 by stanbon(57219)   on 2009-03-19 13:49:39 (Show Source): You can put this solution on YOUR website! After 13 years, 2.1 pounds of radioactive material remains from a 7 pound sample. What is the half-life of this material? ------------------------------------ A(t) = A(0)(1/2)^(13/h)t 2.1 = 7 (1/2)^(13/h)t 0.3 = (1/2)^(13/h) Take the log to get: (13/h)log(0.5) = log(0.3) (13/h) = [log(0.3)]/[log(0.5)] 13/h = 1.7370 h = 13/1.7370 h = 7.4843 years (this is the half-life of the material) ------------------------------------------------------------- Cheers, Stan H.

Probability-and-statistics/187399: What is the probability of getting 2 heads if a coin is tossed 3 times? 1 solutions Answer 140466 by stanbon(57219)   on 2009-03-19 13:35:30 (Show Source): You can put this solution on YOUR website! What is the probability of getting 2 heads if a coin is tossed 3 times? ----------------- It's a binomial problem with n=3, p = 1/2 , x = 2 P(x=2) = 3C2(1/2)^2(1/2)^1 = 3(1/2)^3 = 3/8 ------------------------------------------------ Cheers, Stan H.

Probability-and-statistics/187394: What is the probability that if you have 8 kids, that at least 2 of them will be males? (Assume 7 births and P(male)=P(female)=.5 and births are independent events.) 1 solutions Answer 140464 by stanbon(57219)   on 2009-03-19 13:31:12 (Show Source): You can put this solution on YOUR website! What is the probability that if you have 8 kids, that at least 2 of them will be males? (Assume 7 births and P(male)=P(female)=.5 and births are independent events.) ------------- It's a binomial problem with n = 8,p = 1/2 ; 2 <= x <= 8 ---------------------------------------------------------------- P(2 <= x <= 8) = 1 - binomcdf(8,0.5,1) = 0.9648 ================================================== Cheers, Stan H.

Systems-of-equations/187389: 2x-3y =18 and 5x +3y =3 solutions ( 3 ,-4) dont understand how my book got the solution 3 need a full break down 1 solutions Answer 140452 by stanbon(57219)   on 2009-03-19 10:29:47 (Show Source): You can put this solution on YOUR website! 2x-3y =18 5x +3y =3 --------- Add to solve for "x": 7x = 21 x = 3 --------- Substitute into 5x + 3y = 3 to solve for "y": 5*3 + 3y = 3 15 + 3y = 3 3y = -12 y = -4 ------------ Solution: (3,-4) ===================== Cheers Stan H.

Probability-and-statistics/187355: A couple plan to have exactly three children a) Construct a tree diagram and list the sample space b) Find the probability that the family has at least 2 girls. 1 solutions Answer 140450 by stanbon(57219)   on 2009-03-19 10:26:44 (Show Source): You can put this solution on YOUR website! answered today

Probability-and-statistics/187358: This question is from textbook Survey of Mathematics w/ Applications expanded edition I'm totally lost with the probability and statistics, I have 2 discussions questions and don't know where to start. 1.) Explain using the definition of probability why: A) the probability of an event that cannot occur is 0. B) The probability of an event that must occur is 1. 2.) How do I explain permutations and combinations and the difference between the two, and what are some examples to illustrate the 2. 1 solutions Answer 140443 by stanbon(57219)   on 2009-03-19 09:01:35 (Show Source): You can put this solution on YOUR website! 1.) Explain using the definition of probability why: A) the probability of an event that cannot occur is 0. P(success) = (# of ways success can happen)/(# of ways to succeed + # of ways to fail) If the event cannot occur the # of ways to succeed is 0. Then the numerator of the probability fraction is zero. Then the probability is zero. ---------------------------------------------------------------------- B) The probability of an event that must occur is 1. If it must occur the # of failures is zero. So P(success) = (# successes)/(# successes + 0) = (# successes)/(# successes)=1 ============================================================================== 2.) How do I explain permutations and combinations and the difference between the two, # of permutations of n objects is the # of ways to arrange n objects: n ways to choose position #1 n-1 ways to choose position #2 n-2 wya to choose position #3 etc. --------------------------------- # of combinations of n object is the number of distinct groups of size r that can be formed using the n objects. The number of ARRANGEMENTS of r of the n objects is n*n-1*n-2*...*(n-r+1) But each set of r of these objects appears r*r-1*r-2*...*1 times is that large set of arrangements. So the # of unique sets of r of the n objects is (# of arrangements of the object) divided by (# of arrangements of r of the objects) ------------------------------------------------------------------ and what are some examples to illustrate the 2. Say you have three objects: a,b,c # of arrangements of all three is 3*2*1 = 6 # of combinations of all three is 6/6 = 1 --------------------- # of arrangements of 2 of the objects is 3*2 = 6 # of combinations of 2 of the 3 objects is 6/(2*1) = 3 {a,b},{a,c} and {b,c} are the three combinations. ============================ Cheers, Stan H.

Linear-equations/187361: This question is from textbook ALGEBRA 1 I DONT UNDERSTAND THIS HOMEWORK ASKING ME TO DECIDE WEHTER IS SOLUTION OR INFINELY MANY SOLUTIONS Y=3X-1 Y=-2X+4 1 solutions Answer 140442 by stanbon(57219)   on 2009-03-19 08:41:01 (Show Source): You can put this solution on YOUR website! DECIDE WEHTER IS SOLUTION OR INFINELY MANY SOLUTIONS Y=3X-1 Y=-2X+4 ------------- Substitute to get: 3x-1 = -2x+4 5x = 5 x = 1 ---- Substitute to solve for "y": y = 3*1-1 = 2 --------------------------- Solution: (1,2) It does not have an infinite number of solutions. ====================================================== It would have an infinite number of solutions if the slope and the y-intercept were the same for the two equations. ========================================================= Cheers, Stan H.

Probability-and-statistics/187362: This question is from textbook Survey of Mathematics w/ Applications expanded edition How many different four-digit numbers can be formed from the digits 0 through 9 if the first digit must be even and cannot be zero? 1 solutions Answer 140441 by stanbon(57219)   on 2009-03-19 08:36:33 (Show Source): You can put this solution on YOUR website! See the Answered file. Cheers, Stan H.

Probability-and-statistics/187363: This question is from textbook Survey of Mathematics w/ Applications expanded edition A box of chocolates contains 20 identically shaped chocolates. Five of them are filled with jelly, three are filled with caramel, and twelve are filled with nuts. What is the probability that one chocolate chosen at random is filled with jelly, caramel, or nuts? 1 solutions Answer 140440 by stanbon(57219)   on 2009-03-19 08:35:29 (Show Source): You can put this solution on YOUR website! See the Answered file. Cheers, Stan H.

Probability-and-statistics/187364: This question is from textbook Survey of Mathematics w/ Applications expanded edition A couple plan to have exactly three children. (a) Construct a tree diagram and list the sample space. (b) Find the probability that the family has at least two girls 1 solutions Answer 140439 by stanbon(57219)   on 2009-03-19 08:34:22 (Show Source): You can put this solution on YOUR website! A couple plan to have exactly three children. (a) Construct a tree diagram and list the sample space. 1st column: b,g 2nd column: b,g;b,g 3rd column: b,g;b,g;b,g ------------------------ (b) Find the probability that the family has at least two girls ggb ; gbg ; bgg ; ggg P(at least 2 girls) 4/8 = 1/2 =================================== Cheers, Stan H.

Probability-and-statistics/187365: This question is from textbook Survey of Mathematics w/ Applications expanded edition A card is selected from a deck of 52 playing cards. Find the probability of selecting a) a diamond given the card is black b) a queen given the card is a picture card. 1 solutions Answer 140438 by stanbon(57219)   on 2009-03-19 08:30:11 (Show Source): You can put this solution on YOUR website! This was answered in the last few days. Cheers, Stan H.

Finance/187313: Ann has 1.70 in dimes and nickles. She has two more dimes than nickles. How many nickles does she have. 1 solutions Answer 140423 by stanbon(57219)   on 2009-03-18 22:45:07 (Show Source): You can put this solution on YOUR website! Ann has 1.70 in dimes and nickles. She has two more dimes than nickles. How many nickles does she have. --------------------------------- Value Equation: 10d + 5n = 170 cents Quantity Eq...: d = n + 2 ---------------------------------- Substitute to solve for "n": 10(n+2) + 5n = 170 10n + 20 + 5n = 170 15n = 150 n = 10 (# of nickels) --- Substitute into d = n+2 to solve for "d": d = 10+2 = 12 (# of dimes) =============================== Cheers, Stan H.

Rational-functions/187327: I have a math project for school and it involves finding the equations of the following companies' data. I know how to graph these and the equations for each part of each graph; however, I don't know how to find an equation that will take into account the change in cost per minute. Thanks for any help! Here is the data: Company A: $30 fee, 10 cents a minute for the first 300 minutes, then 8 cents a minute after that. Company B: No fee, 25 cents a minute for 200 minutes, 10 cents a minute for the next 500 minutes then 5 cents a minute. Company C: $50 flat fee for the first 500 minutes, then 10 cents per minute. 1 solutions Answer 140422 by stanbon(57219)   on 2009-03-18 22:19:17 (Show Source): You can put this solution on YOUR website! find an equation that will take into account the change in cost per minute. Thanks for any help! Here is the data: Company A: $30 fee, 10 cents a minute for the first 300 minutes, then 8 cents a minute after that. Cost = 3000 + 10x for 0300 ----------------------------------------------------------- Company B: No fee, 25 cents a minute for 200 minutes, 10 cents a minute for the next 500 minutes then 5 cents a minute. Cost = 25x for 0 < x <=200 or 10x for 200< x <=500 or 5x for x > 500 ------------------------------------------------------------ Company C: $50 flat fee for the first 500 minutes, then 10 cents per minute. Cost 50 for 0 < x <= 500 or 10x for x> 500 ====================================================== Cheers, Stan H.

Linear_Algebra/187324: Factoring completely: 5x^8 – 1280 Could the answer be: 5(x^4 + 16)(x^4 -16) ? __________________ 2x²(p+q³) – (p + q³) Could the answer be (2-x^2) (p + q^3) ? __________________ Simplify: (6+ 1/3)/(4- 2/9) Could the answer be 57 / 17 ? __________________ Solve the problem: The perimeter of a rectangle is 62m. If the width were doubled and the length were increased by 21m, the perimeter would be 120m. What are the length and width of the rectangle? __________________ Solve the equation: 3/y = y/(Sy-12) __________________ Subtract. Simplify if possible. (2x+5)/(x²+8x+15) – (x+2)/(x²+8x+15) Could the answer be 1 / x+5 ? ___________________ 1 solutions Answer 140421 by stanbon(57219)   on 2009-03-18 22:01:32 (Show Source): You can put this solution on YOUR website! Factoring completely: 5x^8 – 1280 Could the answer be: = 5(x^4 + 16)(x^4 -16) = 5(x^4+16)(x^2+4)(x+2)(x-2) __________________ 2x²(p+q³) – (p + q³) = (p+q^3)(2x^2-1) = (p+q^3)(sqrt(2)x -1)(sqrt(2)x + 1) ------------------------------------------ Simplify: (6+ 1/3)/(4- 2/9) = (19/3) / (34/9) = (19/3)*(9/34) = (19)*(3/34) = 57/34 --------------------------------------------- Solve the problem: The perimeter of a rectangle is 62m. If the width were doubled and the length were increased by 21m, the perimeter would be 120m. What are the length and width of the rectangle? Equations: L + W = 31 (L+21) + 2W = 60 -------------------------- Rearrange: L + W = 31 L +2W = 39 ---------------- Subtract 1st from 2nd to solve for "w": W = 8 meters -- Substitute to solve for "L L + 8 = 31 L = 23 meters __________________ Solve the equation: 3/y = y/(Sy-12) Note: Did you mean to put "S" in the equation? __________________ Subtract. Simplify if possible. (2x+5)/(x²+8x+15) – (x+2)/(x²+8x+15) --- The denominators are the same ; combine the numerators over the denominator: [2x+5 - (x+2)]/[(x+3)(x+5)] = [x+3]/[(x+3)(x+5)] = 1/(x+5) You are correct. ========================== Cheers, Stan H. ___________________

Equations/187326: I have a story problem!! Thank you. 223 yards of rope is to be cut into 3 parts. The second part is three more than three times the first part and the third part is 10 yards less than the first. What is the length of each part. I have this: Let x = length of the 1st part Let 3 + x * 3 = 2nd part Let x - 10 = 3rd part From there I cant figure out how to write in an equation. Gets all confusing from there. x + 3 + ( x * 3 ) + x - 10 = 223 3x * 3 - 10 = 223 3x * ( - 7) = 223 -7 3x = 216 --------- both by 3 x = 72 1 solutions Answer 140418 by stanbon(57219)   on 2009-03-18 21:34:21 (Show Source): You can put this solution on YOUR website! 223 yards of rope is to be cut into 3 parts. The second part is three more than three times the first part and the third part is 10 yards less than the first. What is the length of each part. --------------------------------------- Equations: a + b + c = 223 b = 3a + 3 c = a - 10 ---------------------------- Rearrange the equations: a + b + c = 223 3a - b + 0 = -3 a + 0 - c = 10 --------------------------- Use any method to solve to get: a = 46 b = 141 c = 36 ================ Cheers, Stan H.

Equations/187325: Christina makes a $9.25 purchase at a bookstore in Reno with a $20 bill. The store has no bills and gives her the change in quarters and 50-cent piece. There are 30 coins in all. How many of each kind are there? 1 solutions Answer 140417 by stanbon(57219)   on 2009-03-18 21:26:23 (Show Source): You can put this solution on YOUR website! Christina makes a $9.25 purchase at a bookstore in Reno with a $20 bill. The store has no bills and gives her the change in quarters and 50-cent piece. There are 30 coins in all. How many of each kind are there? -------------------------- Change = 20 - 9.25 = $10.75 ----------------------------- Equations: Quantity Equation: q + f = 30 Value Equation:..25q +50f= 1075 ------------------------------------- Modify to set up for elimination: 25q + 25f = 25*30 25q + 50f = 1075 ---------------------- Subtract 1st equation from the 2nd to solve for "f": 25f = 325 f = 13 (# of fifty-cent pieces) --- Substitute to solve for "q": q + 13 = 30 q = 17 (# of quarters) ================================== Cheers, Stan H.