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11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299, >>NextMiscellaneous_Word_Problems/685499: a candy manufacturer has two frades of candy selling at Php100/kilo and Php120/kilo respectively. how many kilos of each must be used to make a mixture of 50 kilos, to sell at Php105/ kilo?
please help me .
1 solutions
Answer 424463 by mananth(12270)   on 2012-11-26 20:29:05 (Show Source):
You can put this solution on YOUR website! Php ---------------- quantity
Candy type I 100 ---------------- x Kg
candy type II 120 ------ 50 - x Kg
Mixture 105 ---------------- 50
600
100 x + 120 ( 50 - x ) = 105 * 50
10000 x + 12000 ( 50 - x ) = 525000
10000 x + 600000 - 12000 x = 525000
10000 x - 12000 x = 525000 - -600000
-2000 x = -75000
/ -2000
x = 37.5 Kg Php100.00 Candy type I
12.5 Kg Php 120.00 candy type II
m.ananth@hotmail.ca
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Percentage-and-ratio-word-problems/685475: An investor invested a total of $2000 in 2 mutual funds. One fund earned an 8% profit and the other earned a 4% profit. If his total profit was $132, how much was invested in each fund?
1 solutions
Answer 424451 by mananth(12270) on 2012-11-26 20:04:51 (Show Source):
You can put this solution on YOUR website!Part I 8.00% per annum ------------- Amount invested =x
Part II 4.00% per annum ------------ Amount invested = y
2000
Interest----- 132
Part I 8.00% per annum ---x
Part II 4.00% per annum ---y
Total investment
x + 1 y= 2000 -------------1
Interest on both investments
8.00% x + 4.00% y= 132
Multiply by 100
8 x + 4 y= 13200.00 --------2
Multiply (1) by -8
we get
-8 x -8 y= -16000.00
Add this to (2)
0 x -4 y= -2800
divide by -4
y = 700
Part I 8.00% $ 1300
Part II 4.00% $ 700
CHECK
1300 --------- 8.00% ------- 104.00
700 ------------- 4.00% ------- 28.00
Total -------------------- 132.00
m.ananth@hotmail.ca
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Linear-systems/685486: please help me solve this problem:
x+3y=11
4x+y=6
using (3,-6)
may you determine if the given ordered pair is a solution to the linear equations and explain why. please and thank you
1 solutions
Answer 424445 by mananth(12270) on 2012-11-26 19:58:59 (Show Source):
You can put this solution on YOUR website!1 x + 3 y = 11 .............1
4 x + 1 y = 6 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by -3
1 x 3 y = 11
-12 x -3 y = -18
Add the two equations
-11 x = -7
/ -11
x = 1
plug value of x in (1)
1 x + 3 y = 11
1 + 3 y = 11
3 y = 11 -1
3 y = 10
y = 3
x= 1
y= 3
(1,3) is the solution.
If (3,-6) is a solution then it must satisfiy both equations.
plug (3,-6) in one of the equations
x+3y=11
3+(-6)3=-15.
This is not equal to 11 hence (3,-6) is not the solution of this system
m.ananth@hotmail.ca
|
Linear-equations/685042: My question is this in the slope-intercept formula I know what is y-intercept and which is slope. Take the formula y=-x+4, now I know the coordinate for y is (0,4), but now how do you figure the slope when you only have a -x. If it was like a -2x I could at least change that to -2/1, but what can you do with something like a -x.
1 solutions
Answer 424314 by mananth(12270) on 2012-11-26 07:19:47 (Show Source):
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Travel_Word_Problems/685046: a biker travels a rate of 12 mphwhen there is no wind. Against the wind, he bikes 8 miles in the same time that it takes to bike 14 miles with the wind. what is the speed of the wind?
1 solutions
Answer 424313 by mananth(12270) on 2012-11-26 07:18:00 (Show Source):
You can put this solution on YOUR website!biker speed = 12 mph
windspeed = x mph
speed against wind 12 - x
speed with wind 12 + x
Distance against wind 8 miles
Distance with wind 14 miles
t=d/r time against = time with wind -
8/(12-x )=14/(12+x)8
8( 12 + x )= 14 (12 -x )
96 + 8 x = 168 - -14 x
14 x 8 x 168 + -96
22 x = 72
/ 22 check
x = 3.27 mph wind speed
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Numbers_Word_Problems/685055: The numerator and denominator of a fraction are in a ratio 2:3. If 8 is added to both numbers, the new fraction becomes 10:13. Find the original fraction.
1 solutions
Answer 424312 by mananth(12270) on 2012-11-26 07:09:48 (Show Source):
You can put this solution on YOUR website!Let the common factopr of the ratio be x
numberator will be 2x
denominator will be 3x
cross multiply
13(2x+8)=10(3x+8)
26x+104 = 30x+80
30x-24x=104-80
6x=24
x= 4
2x=8
3x=12
8/12 is the fraction
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Polynomials-and-rational-expressions/684867: I need help with find the equation of the line passing through the given point and having the given slope. (1,-8) and m is undefined . I know the answer is x=1 but I want to know how to solve the problem. Thank you
1 solutions
Answer 424230 by mananth(12270) on 2012-11-25 20:08:55 (Show Source):
You can put this solution on YOUR website!The slope of a line is defined as the change in the y values divided by the change in x values between two points on the line
(rise over run)
In a vertical line change in x values is =0.
m= Change in y values /0 = undefined
The point passes through (1,-8)
so the equation is x=1
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test/684858: Solve by graphing
5x+2y=8
X-y=3
1 solutions
Answer 424229 by mananth(12270) on 2012-11-25 19:59:27 (Show Source):
You can put this solution on YOUR website!5 x + 2 y = 8
y= -5/ 2x +4
Assume values of x and plug in the equation to get values of y.
Plot the points
x 0 4 2
y= 4 -6 -1
(x,y) (0,4),(4,-6),( 2-1)
--------
1 x + -1 y = 3
y= 1 x + -3
Assume values of x and plug in the equation to get values of y.
Plot the points
x 0 4 2
y= -3 1 -1
(x,y) (0,-3),(4,1),( 2,-1)
The solution is (2,-1)
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Quadratic_Equations/684844: University theater sold 515 tickets for a play. Tickets cost $20 per adult and $13 per senior citizen. If total receipts were $7822, how many senior tickets were sold?
1 solutions
Answer 424225 by mananth(12270) on 2012-11-25 19:42:40 (Show Source):
You can put this solution on YOUR website!x= number of adult tickets
y= number of senior citizen tickets
Total tickets sold
1 x + 1 y = 515 .............1
20 x + 13 y = 7822 .............2
Eliminate y
multiply (1)by -13
Multiply (2) by 1
-13 x -13 y = -6695
20 x + 13 y = 7822
Add the two equations
7 x = 1127
/ 7
x = 161
plug value of x in (1)
1 x + 1 y = 515
161 + y = 515
y = 515 -161
y = 354
y = 354
x= 161 number of adult tickets
y= 354 number of senior citizen tickets
m.ananth@hotmail.ca
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Quadratic_Equations/684401: A telephone pole was bent over at the point 4/9 of the distance from its base to the top. The top of the pole reaches a point on the ground 9 meters from the base of the pole (thus forming a triangle with the ground). What was the original height of the pole?
1 solutions
Answer 424092 by mananth(12270) on 2012-11-24 19:40:12 (Show Source):
You can put this solution on YOUR website!let the length of pole be x
4x/9 is erect and 5x/9 is the bent portion which is the hypotenuse of the triangle formed.
distance from ground = 9 m
Using Pythagoras theorem
multiply equation by 81
/9
x^2= 81*81/9
x^2=81*9
x= +/- 9*3
x= 27 ignore negative value.
Length of pole = 27m
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Linear-systems/684251: the school theater sold 49 tickets to their play today. tickets for adults cost $15 and tickets for children cost $8 each. they collected a total of $469 in ticket sales today. how many of each kind of ticket did they sell?
1 solutions
Answer 424001 by mananth(12270) on 2012-11-24 07:26:15 (Show Source):
You can put this solution on YOUR website!x= number of adult tickets
y= number of child tickets
Total tickets sold
1 x + 1 y = 49 .............1
15 x + 8 y = 469 .............2
Eliminate y
multiply (1)by -8
Multiply (2) by 1
-8 x -8 y = -392
15 x + 8 y = 469
Add the two equations
7 x = 77
/ 7
x = 11
plug value of x in (1)
1 x + 1 y = 49
11 + y = 49
y = 49 -11
y = 38
y = 38
x= 11 number of adult tickets
y= 38 number of child tickets
m.ananth@hotmail.ca
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Sequences-and-series/684177: if 5th term of an AP is equal to 19 times the 19th term show that the 24th term of an AP is zero
1 solutions
Answer 424000 by mananth(12270) on 2012-11-24 07:18:49 (Show Source):
You can put this solution on YOUR website!if 5th term of an AP is equal to 19 times the 19th term show that the 24th term of an AP is zero
I think the problem should be
if 5 times the 5th term of an AP is equal to 19 times the 19th term show that the 24th term of an AP is zero
T5 = a+4d
t19= a+18d
5(a+4d)=19(a+18d)
5a +20d=19a+342d
14a+322d=0
14(a+23d)=0
a+23d=0
..
.
t24= a+23d
But a+23d=0
therefore
T24=0
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Age_Word_Problems/684176: Dear Tutor, please help me on this problem. Thank you.
Paul is now 3 times as old as his son Jason. 5 years ago, Paul was 4 years younger than 5 times Jason's age. How old is Jason now?
Let Jason be N years old now
1 solutions
Answer 423964 by mananth(12270) on 2012-11-23 20:06:06 (Show Source):
You can put this solution on YOUR website!Jason's age = N
Paul's age =3N
5 years ago
Jason's age = N-5
Paul's age = 3N-5
According to the condition
3N-5=4(N-5)-4
3N-5=4N-20-4
4N-3N=-5+20+4
N=19
Jason's age = 19 years
Paul's age = 3*19 = 57 years
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Average/684038: 3x+y=95,
x+y=29
1 solutions
Answer 423880 by mananth(12270) on 2012-11-23 08:00:35 (Show Source):
You can put this solution on YOUR website!3 x + 1 y = 95 .............1
1 x + 1 y = 29 .............2
Eliminate y
multiply (1)by -1
Multiply (2) by 1
-3 x -1 y = -95
1 x + 1 y = 29
Add the two equations
-2 x = -66
/ -2
x = 33
plug value of x in (1)
3 x + 1 y = 95
99 + y = 95
y = 95
y = -4
y = -4
x= 33
y= -4
m.ananth@hotmail.ca
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test/683891: A boat sails 30 miles to the east from a point P, then it changes direction and sails to the south. If this boat is sailing at a constant speed of 10 miles/hr, at what rate is its distance from the point P increasing
a) 2 hours after it leaves the point P
b) 7 hours after it leaves the point P
1 solutions
Answer 423859 by mananth(12270) on 2012-11-22 20:37:20 (Show Source):
You can put this solution on YOUR website!The route taken by the boat is a right angle
so the actual distance from starting point is any point of the hypotenuse
d(east)= 30 mi
d(south)=10 mi/h
After2 hours it would be 20 miles down south
distance from origin = sqrt(30^2+20^2) ( the legs of the right triangle)
After 2 hours it is sqrt(1300) from the origin
its rate with respect to origin = d/t
sqrt(1300)/2
10sqrt(13)/2
5sqrt(13) mi/h is the rate with respect to origin P after 5 hours.
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Travel_Word_Problems/683929: two cars start from the same place and at the same time but travel in opposite directions.after 6 hours, they are 540 metres apart. if one car travel 10 metres per hour slower than the other, what are their speed per hour?
1 solutions
Answer 423855 by mananth(12270) on 2012-11-22 20:17:14 (Show Source):
You can put this solution on YOUR website!speed of car A=x
speed of II car = x+10
They are moving away. so add up their speed
x+x+10=2x+10 m/h
distance = 540 m
time = 6 hours
speed = distance / time
2x+10 = 540/6
6(2x+10)=540
12x+60=540
12x=480
x=40 m/h
speed of car A = 40 m/h
speed of car II = 50 m/h
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Expressions-with-variables/683988: I need help with this problem what is 3x + 2 = 11 and the next one is 4:8 is 8: is what
1 solutions
Answer 423853 by mananth(12270) on 2012-11-22 20:09:29 (Show Source):
You can put this solution on YOUR website!3x+2 = 11
we have to find the value of x
so make x the subject
add -2 to both sides
3x+2-2=11-2
3x=9
divide by 3
3x/3 = 9/3
x=3
-
the denominators are 8 & x
so product is 8x
multiply equation by 8x
we get
4x= 64
/4
x=16
So
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Linear-systems/683913: Would you please help me solve this equation:
Solve the system analytically:
x+y+z=-4
2x+6y+2z=8
-x+7y-3z=40
Thanks
1 solutions
Answer 423850 by mananth(12270) on 2012-11-22 20:00:09 (Show Source):
You can put this solution on YOUR website!1 x + 1 y + 1 z = -4 --------------1
2 x + 6 y 2 z = 8 -------------- 2
-1 x + 7 y + -3 z 40 -------------- 3
consider equation 1 &2 Eliminate y
Multiply 1 by -6 -5
Multiply 2 by 1 4
we get
-6 x + -6 y + -6 z = 24
2 x + 6 y + 2 z = 8
Add the two
-4 x + 0 y + -4 z = 32 ------------- 4
consider equation 2 & 3 Eliminate y
Multiply 2 by -7
Multiply 3 by 6
we get
-14 x + -42 y + -14 z = -56
-6 x + 42 y + -18 z = 240
Add the two
-20 x + 0 y + -32 z = 184 -------------5 5
Consider (4) & (5) Eliminate x
Multiply 4 by -5
Multiply (5) by 1
we get
20 x + 20 z = -160
-20 x + -32 z = 184
Add the two
0 x + -12 z = 24
/ -69840
z = -2
Plug the value of z in (5)
-20 x + -32 z = 184
-20 x = 120
x = -6 OR
plug value of x & z in 1
-6 + 1 y + -2 = -4
1 y = -4 + 6 + 2
y = 4
x=-6,y=4,z=-2
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Age_Word_Problems/683983: if a boys age and his fathers age amount together to 24 years. fourth part of the product of their ages exceeds the boys age by 9 years.find how old they are.
1 solutions
Answer 423847 by mananth(12270) on 2012-11-22 19:52:58 (Show Source):
You can put this solution on YOUR website!boy's age =x
father's age = y
x+y=24
(xy)/4 -x=9
multiply equation by 4
xy-4x=36
x(y-4)=36
x=36/(y-4)
Now x+y = 24
substitute x in the equation
36/(y-4) +y = 24
multiply equation by (y-4)
36+y(y-4)=24(y-4)
36+y^2-4y=24y-96
y^2-4y-24y+36+96=0
y^2-28y+132=0
y^2-22y-6y+132=0
y(y-22)-6(y-22)=0
(y-22)(y-6)=0
y=22 OR 6
Father's age = 22
Son's age = 2 (because x+y=24)
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Linear-equations/683956: find the equation of the line passes through the given point and has the given slope. Express your answer in slope incept form. (8,0), m = -3
1 solutions
Answer 423846 by mananth(12270) on 2012-11-22 19:43:52 (Show Source):
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Travel_Word_Problems/683982: harry takes 2 hours less to travel 50 km than cara takes to travel 35 km. cara travels at a speed that is 5km/hr slower than that of harry. find the time taken by harry and cara to complete their respective journeys and the speed at which they traveled.
1 solutions
Answer 423844 by mananth(12270) on 2012-11-22 19:35:35 (Show Source):
You can put this solution on YOUR website!Harry speed =x km/h
Harry distance = 50 km
Cara speed = x-5
Cara distance = 35 km
Time = d/r
Time taken by cara - time taken by Harry = 2
35/(x-5) - 50/x= 2
LCD = x(x-5)
multiply equation by x(x-5)
35x-50(x-5)=2x(x-5)
35x-50x+250=2x^2-10x
2x^2-10x+50x-35x-250=0
2x^2+5x-250=0
2x^2-20x+25x-250=0
2x(x-10)+25(x-10)=0
(x-10)(2x+25)=0
x=10 OR -12.5
speed cannot be negative
so x= 10 Harry's speed
Cara's speed = 10-5 = 5 km/h
time = d/r
50/10 = 5 hours time taken by Harry
35/(10-5)= 7 hours Time taken by Cara
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Points-lines-and-rays/683785: help please! a triangle has its vertices at (6,0), (5,3), and (-1,-4) find the equation of the line through the vertices parallel to the opposite side.
1 solutions
Answer 423717 by mananth(12270) on 2012-11-21 20:12:17 (Show Source):
You can put this solution on YOUR website!(6,0), (5,3), and (-1,-4)
Let A (6,0), B(5,3), and C(-1,-4)be the vertices of the triangle
Let P,Q be the mid points of AB & AC respectively
PQ is parallel to BC
Determine co ordinates of P & Q by mid point formula
A (6,0), B(5,3),
Let P(x3,y3)
p=((11/2),(3/2))
Similarly find co ordinates of Q the mid point of AC
Q=(x4,y4)
A (6,0), C(-1,-4)
Q=((5/2),(-2))
The equation of PQ
use formula 
p=((11/2),(3/2)), Q=((5/2),(-2))
equation of PQ -> 
multiply equation by 4
14x-12y=59 is the equation of PQ
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Polynomials-and-rational-expressions/683782: Hello there,my question is, how do I find all integers k so that the trinomial can be factored? The problem is: 2m squared minus km minus 11. My online course does not, to my knowledge, have the "formula" to find all possible integer solutions for a trinomial. If You will please walk me through this process, I would appreciate Your help very much! Thank You!Also, Happy Thanksgiving!
1 solutions
Answer 423709 by mananth(12270) on 2012-11-21 19:36:33 (Show Source):
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Geometry_Word_Problems/683578: one side of right traingle is 12 cm long. The hyphotenuse is 3 cm more than twice as long as the other side. What are the dimensions of the rectangle?
1 solutions
Answer 423625 by mananth(12270) on 2012-11-21 07:24:50 (Show Source):
You can put this solution on YOUR website!one side = 12 cm
other side =x
hypotenuse = 2x+3
By Hypotenuse theorem
(2x+3)^2=x^2+12^2
4x^2+12x+9=x^2+144
3x^2+12x-135=0
/3
x^2+4x-45=0
x^2+9x-5x-45=0
x(x+9)-5(x+9)=0
(x+9)(x-5)=0
therefore x= -9 OR 5
ignore negative value
x=5 cm the other side
Hypotenuse = 2x+3.
Plug x=6
we get hypotenuse = 13
The dimensions are , 5,12,13
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Money_Word_Problems/683584: A person invested $20,000, part at an interest rate of 6% annually and the remainder at 7% annually. The total interest at the end of 1 year was equivalent to an annual 6.75% rate on the entire $20,000. How much was invested at each rate?
1 solutions
Answer 423624 by mananth(12270) on 2012-11-21 06:55:14 (Show Source):
You can put this solution on YOUR website!6.75% on 20,000 = 1350 total interest earned.
Part I 6.00% per annum ------------- Amount invested =x
Part II 7.00% per annum ------------ Amount invested = y
20000
Interest----- 1350
Part I 6.00% per annum ---x
Part II 7.00% per annum ---y
Total investment
x + 1 y= 20000 -------------1
Interest on both investments
6.00% x + 7.00% y= 1350
Multiply by 100
6 x + 7 y= 135000.00 --------2
Multiply (1) by -6
we get
-6 x -6 y= -120000.00
Add this to (2)
0 x 1 y= 15000
divide by 1
y = 15000
Part I 6.00% $ 5000
Part II 7.00% $ 15000
CHECK
5000 --------- 6.00% ------- 300.00
15000 ------------- 7.00% ------- 1050.00
Total -------------------- 1350.00
m.ananth@hotmail.ca
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Money_Word_Problems/683587: A person invested $20,000, part at an interest rate of 6% annually and the remainder at 7% annually. The total interest at the end of 1 year was equivalent to an annual 6.75% rate on the entire $20,000. How much was invested at each rate?
1 solutions
Answer 423623 by mananth(12270) on 2012-11-21 06:52:41 (Show Source):
You can put this solution on YOUR website!6.75% on 20,000 = 1350 total interest earned.
Part I 6.00% per annum ------------- Amount invested =x
Part II 7.00% per annum ------------ Amount invested = y
20000
Interest----- 1350
Part I 6.00% per annum ---x
Part II 7.00% per annum ---y
Total investment
x + 1 y= 20000 -------------1
Interest on both investments
6.00% x + 7.00% y= 1350
Multiply by 100
6 x + 7 y= 135000.00 --------2
Multiply (1) by -6
we get
-6 x -6 y= -120000.00
Add this to (2)
0 x 1 y= 15000
divide by 1
y = 15000
Part I 6.00% $ 5000
Part II 7.00% $ 15000
CHECK
5000 --------- 6.00% ------- 300.00
15000 ------------- 7.00% ------- 1050.00
Total -------------------- 1350.00
m.ananth@hotmail.ca
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