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# Recent problems solved by 'Maths68'

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 Human-and-algebraic-language/671072: What are the numerical expression of... 5 decreased by the product of a number and 3? The sum of -9 times a number and 6 is 72? Thank-you for your consideration if you have time to answer my question.1 solutions Answer 417262 by Maths68(1474)   on 2012-10-24 00:18:55 (Show Source): You can put this solution on YOUR website!Let number = x (1) 5 decreased by the product of a number and 3 = 3x-5 (2) The sum of -9 times a number and 6 is 72 = 6-9x=72
 Travel_Word_Problems/671087: a car travels 50 mph for 3 hours.how far has it traveled 1 solutions Answer 417261 by Maths68(1474)   on 2012-10-23 23:59:06 (Show Source): You can put this solution on YOUR website!Distance = Speed * time Distance = 50 * 3 Distance = 150 miles
 Rational-functions/671058: Solve for x i am stuck on this one problem. i have it half done and stuck on figuring out the rest. the formula is: Sqrt(6x-8) = x1 solutions Answer 417260 by Maths68(1474)   on 2012-10-23 23:55:43 (Show Source): You can put this solution on YOUR website! or or
 Linear_Equations_And_Systems_Word_Problems/671083: Using X as a variable, translate twice a number increased by 7 equals 19 solve for X1 solutions Answer 417259 by Maths68(1474)   on 2012-10-23 23:43:26 (Show Source): You can put this solution on YOUR website!Let number = x Twice a number = 2x Increased by 7 = 2x+7 2x+7=19 2x=19-7 2x=12 2x/2=12/2 x=6 Number is = x = 6
 Equations/637959: Half the sum of two numbers is 20 and three times their difference is 18. Find the numbers1 solutions Answer 401956 by Maths68(1474)   on 2012-08-16 12:55:49 (Show Source): You can put this solution on YOUR website!Let Smaller Number = x Larger Number = y Half the sum of two numbers is 20 (x+y)/2=20 x+y=2*20 x+y=40...............(1) Three times their difference is 18 3(y-x)=18 3(y-x)/3=18/3 y-x=6..............(2) Add (1) and (2) x+y=40...............(1) y-x=6..............(2) --------------------- 2y=46 2y/2=46/2 y=23 Put the value of y in (1) x+y=40...............(1) x+23=40 x=40-23 x=17 Smaller Number = x = 17 Larger Number = y = 23
 Age_Word_Problems/637635: How old are Edz and Noe now, if Edz is twice as old as Noe and the sum of their ages is 66? ITS MY QUIZ. PLEASE ANSWER :)1 solutions Answer 401824 by Maths68(1474)   on 2012-08-15 12:50:41 (Show Source): You can put this solution on YOUR website!Let Edz present age = e Noe present age = n Edz is twice as old as Noe e=2n..............(1) The sum of their ages is 66 e+n=66............(2) Put the value of e in (2) 2n+n=66 3n=66 3n/3=66/3 n=22 Put the value of n in (1) e=2n..............(1) e=2(22) e=44 Edz present age = e = 44 Noe present age = n = 22
 Travel_Word_Problems/637387: The sum of the speeds of two trains is 723.7. If the speed of the first train is 10.3 faster than the second train, find the speeds of each. HELP!! this is killing me that i cant figure this out1 solutions Answer 401633 by Maths68(1474)   on 2012-08-14 13:05:15 (Show Source): You can put this solution on YOUR website!Let Speed of slower train = x Speed of faster train = x+10.3 Speed of slower train + Speed of faster train = Sum of speeds x+x+10.3=723.7 2x+10.3=723.7 2x=723.7-10.3 2x=713.4 2x/2=713.4/2 x=356.7 Speed of slower train = x = 356.7 Speed of faster train = x+10.3 = 356.7+10.3=367
 Travel_Word_Problems/637089: Keiko runs 7 miles in 80 mins. At the same rate, how many miles would she run in 32mins?1 solutions Answer 401418 by Maths68(1474)   on 2012-08-13 13:59:56 (Show Source): You can put this solution on YOUR website!In 80 mins...............7 miles In 80/80 mins............7/80 miles In 1 mins............7/80 miles In (32)*1 mins.........(7/80)*(32)miles In 32 mins.........(7/80)(32)miles In 32 mins.........(7/5)(2)miles In 32 mins.........14/5miles In 32 mins.........2.8miles Or In 1 minute keiko runs = 7/80 = 0.0875 miles In 32 minutes keiko run = 0.875*32 = 2.8 miles
Miscellaneous_Word_Problems/637069: In the following sum the same two numbers are needed. When they are added or mutliplied together, they form the answers for (a) and (b). I need to work out what theose numbers are

(a) ............+........... = 2.7
(b) ............x........... = 1.8

Can anone help me with the following I just cnat seem to get it thanks

I have already submitted this question but my email address was wrong
1 solutions

Answer 401416 by Maths68(1474)   on 2012-08-13 13:50:30 (Show Source):
You can put this solution on YOUR website!
Let x and y be the two numbers
x+y=2.7
x+y=27/10
10(x+y)=27
10x+10y=27.............(1)

xy=1.8
xy=18/10
x=18/10y...............(2)

Put the value of x from (2) to (1)
10(18/10y)+10y=27
18/y+10y=27
(18+10Y^2)/y=27
18+10Y^2=27y
10y^2-27y+18=0
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=9 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.5, 1.2. Here's your graph:

So
y=1.5 or 1.2

Put the values of y in (1)
10x+10y=27.............(1)
Put y=1.5
10x+10(1.5)=27
10x+15=27
10x=27-15
10x=12
x=12/10
x=1.2

Put the values of y in (1)
10x+10y=27.............(1)
Put y=1.2
10x+10(1.2)=27
10x+12=27
10x=27-12
10x=15
x=15/10
x=1.5
So numbers are 1.2 and 1.5

Check
=====
xy=1.8
(1.2)(1.5)=1.8
1.8=1.8

x+y=2.7
1.2+1.5=2.7
2.7=2.7

 Age_Word_Problems/636780: Two years hence, John's mother will be twice his age.Five years hence his father will be thrice his present age.What is the difference between the ages of his parents,if the sum of all the three ages is 105?1 solutions Answer 401243 by Maths68(1474)   on 2012-08-12 13:47:10 (Show Source): You can put this solution on YOUR website!Let Present age of mother = m years old Present age of father = f years old Present age of John = j years old Two years hence mother will be = m+2 years old John will be = j+2 years old Then mother will be twice his age m+2=2(j+2) m+2=2j+4 m=2j+4-2 m=2j+2.................(1) Five years hence father will be = f+5 years old John will be = j+5 years old Then father will be thrice his present age f+5=3(j+5) f+5=3j+15 f=3j+15-5 f=3j+10..................(2) the sum of all the three ages is 105 m+j+f=105 Put the value of m and f from (1) and (2) to above equation 2j+2+j+3j+10=105 6j+12=105 6j=105-12 6j=93 6j/6=93/6 j=31/2 Put the value of j in (1) m=2j+2.................(1) m=2(31/2)+2 m=31+2 m=33 Put the value of j in (2) f=3j+10..................(2) f=3(31/2)+10 f=93/2+10 f=(93+20)/2 f=113/2 f=56.5 Difference b/w parents ages = f-m=56.5-33= 23.5 years.
 Money_Word_Problems/636786: Lauren paid for movie tickets which cost \$8.50 for adults and \$4.75 for students. She ordered six times as many student tickets as adult tickets. The total cost is \$111. How many student tickets did she purchase?1 solutions Answer 401236 by Maths68(1474)   on 2012-08-12 13:19:39 (Show Source): You can put this solution on YOUR website!Lauren Purchased adult tickets = x Lauren Purchased student tickets = 6x Cost of a Adult ticket = 8.50 Cost of a student ticket = 4.75 Total Cost of all tickets = 111 (Lauren Purchased adult tickets)(Cost of a Adult ticket)+(Lauren Purchased student tickets)(Cost of a student ticket) = Total Cost of all tickets (x)(8.50)+(6x)(4.75)=111 8.50x+28.50=111 37x=111 37x/37=111/37 x=3 Lauren Purchased adults tickets = x = 3 Lauren Purchased students tickets = 6x = 6*3 = 18 Check ====== Total Cost of Adults tickets = 8.50*3=25.50 Total Cost of students tickets= 4.75*18=85.50 Total Cost = 25.5+85.5=111
 Linear-equations/636427: Find the slope‐intercept equation of the line with the following properties: Parallel to the line y= 6; containing the point (5,7)1 solutions Answer 401046 by Maths68(1474)   on 2012-08-11 02:34:00 (Show Source): You can put this solution on YOUR website!Equation of the line slope-intercept form y=mx+b Given Equation of the Given line y=6 Compare with the equation of line slope-intercept form Slope of the given = m = 0 Since lines are parallel their slope will be same Slope of the required line = m = 0 Point (5, 7) We have a point and slope we can get easily the require line equation. Use Equation of the line point-slope form m = y2-y1/x2-x1 Put the values in above equation, we have m=(y-2)/(x-5) 0(x-5)=(y-7) 0=y-7 y=7 Above equation is the equation of the required line Graph ====== y=6 y=7
 Age_Word_Problems/636450: Paul is 11 years younger than Wendy. In 10 years, Wendy will be twice as old as Paul. How old is Paul now?1 solutions Answer 401045 by Maths68(1474)   on 2012-08-11 02:21:41 (Show Source): You can put this solution on YOUR website!Let Present age of Paul = p years old Present age of Wendy = w years old Paul is 11 years younger than Wendy. w=p+11............(1) In 10 years, Paul will be = p+10 years old Wendy will be = w+10 years old Then Wendy will be twice as old as Paul w+10=2(p+10) w+10=2p+20 w=2p+20-10 w=2p+10................(2) Put the value of w from (1) to (2) p+11=2p+10 p-2p=10-11 -p=-1 p=1 Put the value of p in (1) w=p+11............(1) w=1+11 w=12 Present age of Paul = p = 1 year old Present age of Wendy = w = 12 years old
 Mixture_Word_Problems/636474: How much of a 5% cream and a 0.5% cream would be required to compound 75 g of a 2% cream? 1 solutions Answer 401037 by Maths68(1474)   on 2012-08-11 01:03:00 (Show Source): You can put this solution on YOUR website!Cream A ======= Amount = x Cream = 5%=0.05 Cream B ======= Amount = 75-x Cream = 0.5%=0.005 Mixture Cream ============= Amount = 75g Cream = 2%=0.02 (Amount)(cream A)+(Amount)(Cream B)= (Amount)(Mixture Cream) 0.05x+(75-x)0.005=(75)(0.02) 0.05x+(75-x)0.005=(0.02)(75) 0.05x+(75-x)0.005=1.5 0.05x+0.375-0.005x=1.5 0.045x+0.375=1.5 0.045x=1.5-0.375 0.045x=1.125 0.045x/0.045=1.125/0.045 x=25 Cream A ======= Amount = x = 25g Cream = 5%=0.05 Cream B ======= Amount = 75-x=75-25=50g Cream = 0.5%=0.005 Mixture Cream ============= Amount = 75g Cream = 2%=0.02 25g of a 5% cream and 50g of a 0.5% cream would be required to compound 75 g of a 2% cream.
 Mixture_Word_Problems/636461: How much 20% antifreeze solution must be mixed with a 80% antifreeze solution to produce 30 gallons of a 60% disinfectant solution?1 solutions Answer 401035 by Maths68(1474)   on 2012-08-11 00:43:58 (Show Source): You can put this solution on YOUR website!Solution A =========== Amount = x gallons Concentration = 20%=0.2 Solution B =========== Amount = 30-x gallons Concentration = 80%=0.8 Mixture ======= Amount = 30 gallons Concentration = 60% = 0.6 (Amount of Solution A)(Concentration of Solution A)+(Amount of Solution B)(Concentration of Solution B)=(Amount of Mixture)(Concentration of Mixture) (x)(0.2)+(30-x)(0.8)=(30)(0.6) 0.2x+24-0.8x=18 0.2x-0.8x=18-24 -0.6x=-6 -0.6x/-0.6=-6/-0.6 x=10 10 gallon of 20% antifreeze solution must be mixed with 20 gallons of 80% antifreeze solution to produce 30 gallons of a 60% disinfectant solution.
 Mixture_Word_Problems/636254: A solution of 61% bleach is to be mixed with a solution of 27% bleach to make 68 gallons of a 44% solution. How many gallons of each should be used? 1 solutions Answer 400964 by Maths68(1474)   on 2012-08-10 14:13:18 (Show Source): You can put this solution on YOUR website!Solution A =========== Amount = x gallons Concentration = 61%=0.61 Solution B =========== Amount = 68-x Concentration = 27%=0.27 Mixture ======= Amount = 68 gallons Concentration = 44% = 0.44 (Amount of Solution A)(Concentration of Solution A)+(Amount of Solution B)(Concentration of Solution B)=(Amount of Mixture)(Concentration of Mixture) (x)(0.61)+(68-x)(0.27)=(68)(0.44) 0.61x+18.36-0.27x=29.92 0.34x+18.36=29.92 0.34x=29.92-18.36 0.34x=11.56 0.34x/0.34=11.56/0.34 x=34 Solution A =========== Amount = x = 34 gallons Concentration = 61%=0.61 Solution B =========== Amount = 68-x = = 68-34 = 34 gallons Concentration = 27%=0.27 Mixture ======= Amount = 68 gallons Concentration = 44% = 0.44 34 gallons of each should be used.
 Rectangles/636311: A rectangle 15cmx30cm is increased on all sides by the same amount. It's area is increased by 144cm^2. By how much is each dimension increased?1 solutions Answer 400960 by Maths68(1474)   on 2012-08-10 13:59:00 (Show Source): You can put this solution on YOUR website!Let Increase on all sides = x Area of rectangle = length * width A=L*W A=15*30 A=450 After Increase A=450+144 A=594 Length = 30+x Width = 15+x 594=(15+x)(30+x) 594=450+15x+30x+4x^2 594=450+45x+x^2 594-450=45x+x^2 144=45x+x^2 x^2+45x-144=0 x^2+48x-3x-144=0 x(x+48)-3(x+48)=0 (x-3)(x+48)=0 x-3=0 or x+48=0 x=3 or x=-48 (unacceptable) x=3 Each dimension increased by 3cm Length = 30+x = 30+3 = 33cm Width = 15+x = 15+3 = 18cm
 Polynomials-and-rational-expressions/636330: Essay; show all work. Find the product: (1/7x^2-2/5)(1/2x-1/2)1 solutions Answer 400949 by Maths68(1474)   on 2012-08-10 13:33:57 (Show Source): You can put this solution on YOUR website!= = = = =
 Polynomials-and-rational-expressions/636331: Short answer. Find the product: -5x (x3 – 3x2 + 4x + 6)1 solutions Answer 400947 by Maths68(1474)   on 2012-08-10 13:22:51 (Show Source): You can put this solution on YOUR website!= =
 Polynomials-and-rational-expressions/636352: Add. (4v^4-4v^3+4v^2+16v-10)+(v^5+8v^3+7v^2-4v+6)+(-7v^4+v^2-8v-7) The answer is_________?1 solutions Answer 400946 by Maths68(1474)   on 2012-08-10 13:19:53 (Show Source): You can put this solution on YOUR website!= = = = =
 Age_Word_Problems/636356: the sum of julies age and her fathers age is 58. if julies father is 10 years younger than triple Julie's age, find their age.1 solutions Answer 400940 by Maths68(1474)   on 2012-08-10 13:08:32 (Show Source): You can put this solution on YOUR website!Let Present age of Julies = j years old Present age of Father = f years old the sum of julies age and her fathers age is 58 j+f=58 j=58-f.............(1) julies father is 10 years younger than triple Julie's age 3j=f-10............(2) Put the value of j from (1) to (2) 3j=f-10 3(58-f)=f-10 174-3f=f-10 -3f-f=-10-174 -4f=-164 -4f/-4=-164/-4 f=41 Put the value of f in (1) j=58-f.............(1) j=58-41 j=17 Present age of Julies = j = 17 years old Present age of Father = f = 41 years old
 Quadratic_Equations/636086: 3/4x + 8 = 5/8x - 161 solutions Answer 400722 by Maths68(1474)   on 2012-08-09 13:17:51 (Show Source): You can put this solution on YOUR website! if you question is as follows:
 Age_Word_Problems/635687: the square of katie's age six year ago is equal to her age in six years time. find her present age1 solutions Answer 400668 by Maths68(1474)   on 2012-08-09 03:52:07 (Show Source): You can put this solution on YOUR website!Let Present age of Katie = k years old Six years ago Katie was = k-6 years old In Six years Katie will be = k+6 years old The square of katie's age six year ago is equal to her age in six years time (k-6)^2=k+6 k^2-12k+36=k+6 k^2-13k+30=0 Solve for K k^2-10k-3k+30=0 k(k-10)-3(k-10)=0 (k-3)(k-10)=0 k-3=0 or k-10=0 k=3 (unacceptable) or k=10 k=10 Present age of Katie = k = 10 years old
 Age_Word_Problems/635970: When I was 4 - my brother was half my age, now I am 50, how old is he?1 solutions Answer 400667 by Maths68(1474)   on 2012-08-09 03:31:25 (Show Source): You can put this solution on YOUR website!You were = 4 years Your brother was = 2 years (half of your age) You are = (4+46) =50 years old now Your brother is = (2+46) = 48 years old now
 Age_Word_Problems/635971: When I was 4 - my brother was half my age, now I am 50, how old is he?1 solutions Answer 400666 by Maths68(1474)   on 2012-08-09 03:30:14 (Show Source): You can put this solution on YOUR website!You were = 4 years Your brother was = 2 years (half of your age) You are = (4+46) =50 years old now Your brother is = (2+46) = 48 years old now
 Percentage-and-ratio-word-problems/635955: A part-time instructor is receiving \$950 per credit taught. If the instructor receives a 4% increase, how much will the new per credit compensation be1 solutions Answer 400665 by Maths68(1474)   on 2012-08-09 02:42:08 (Show Source): You can put this solution on YOUR website!New Per credit compensation = Old compensation + increase% New Per credit compensation = 950+(950)(4/100) New Per credit compensation = 950+(950)(1/25) New Per credit compensation = 950+38 New Per credit compensation = 988
 Numeric_Fractions/635965: (-1 1/2) (-1 1/4)1 solutions Answer 400664 by Maths68(1474)   on 2012-08-09 02:30:38 (Show Source): You can put this solution on YOUR website!= = = =
 Expressions-with-variables/635945: 8 y - 1 = x 3 x = 2 y I need to find x and y. I'm stuck on it, thanks in advance.1 solutions Answer 400646 by Maths68(1474)   on 2012-08-09 00:36:13 (Show Source): You can put this solution on YOUR website!8y-1=x..............(1) 3x=2y...............(2) Substitute the value of x from (1) to (2) 3(8y-1)=2y 24y-3=2y 24y-2y=3 22y=3 22y/22=3/22 y=3/22 Put the value of y in (2) 3x=2(3/22) 3x=3/11 3x/3=(3/11)/3 x=1/11 x,y=1/11,3/22 Check ===== Put the values of x and y in (1) 8y-1=x..............(1) 8(3/22)-1=1/11 24/22-1=1/11 (24-22)/22=1/11 2/22=1/11 Put the values of x and y in (1) 3x=2y...............(2) 3(1/11)=2(3/22) 3/11=6/22
 Miscellaneous_Word_Problems/635919: tickets to the local circus cost 3 dollars for children and 5 dollars for adult. There were 3 times as many children tickets sold as adult tickets. All together the circus made 700 dollars. How many children and how many adult bought tickets to the circus?1 solutions Answer 400641 by Maths68(1474)   on 2012-08-08 23:50:36 (Show Source): You can put this solution on YOUR website!Adults tickets sold = x Children tickets sold= 3x Total Amount = \$700 (Cost of a children ticket)(Children tickets sold)+(Cost of an adult ticket)( Adult tickets sold)= Total Amount 3(3x)+5(x)=700 9x+5x=700 14x=700 14x/14=700/14 x=50 Adults tickets sold = x = 50 Children tickets sold= 3x = 3*50 = 150 Check ===== 3*150+5*50=700 450+250=700 700=700
 Age_Word_Problems/635677: a father is now 3 times as old as his son. twelve years ago, he was six times as old as his son. find the present age of each1 solutions Answer 400481 by Maths68(1474)   on 2012-08-08 03:32:10 (Show Source): You can put this solution on YOUR website!Let Present age of of Father = f years old Present age of Son = s years old Father is 3 times as old as his son f=3s...............(1) 12 years ago father was = f-12 years old Son was = s-12 years old Then f-12=6(s-12) f-12=6s-72 f=6s-72+12 f=6s-60..............(2) Compare (1) and (2) 3s=6s-60 3s-6s=-60 -3s=-60 -3s/-3=-60/-3 s=20 Put the value of s in (1) f=3s f=3(20) f=60 Present age of of Father = f = 60 years old Present age of Son = s = 20 years old
 Word_Problems_With_Coins/635658: the same number of each coin have a total of 6.00\$.There are nickels ,dimes and quarters. how many of each are there1 solutions Answer 400478 by Maths68(1474)   on 2012-08-08 02:41:47 (Show Source): You can put this solution on YOUR website!0.05x+0.1x+0.25x=6 0.4x=6 x=6/0.4 x=15 15 of coins of each nickels, dimes,quarters are there.