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45 divided by 108

what is 45 divided by 108????

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108 divided by 45 |Math problem 108 divided by 45 long division


108 divided by 45 . ... 108/45 divided by 2 » (108/45) ÷ 2 » 2.4 ÷ 2 = 1.2 . 108/45 long division below 108/45 conversion. Answer to math problems Solution Steps.
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What can divide into both 45 and 108 - Answers.com


What can divide into both 45 and 108? SAVE CANCEL. already exists. Would you like to merge this question ... 108 divided by 6 is 18 because 18 times 6 is 108.
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What does 108 divided by 12 equal - Answers.com


What does 108 divided by 12 equal? SAVE CANCEL. already exists. Would you ... What is 12 divided by 108? It is .111111111 - etc. Edit. Share to: Answer Bot.
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Long Division - Math Is Fun


Long Division. Below is the process written out in full. You will ... Divide this number by the divisor. The whole number result is placed at the top.
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Divide Two Numbers- WebMath


Divide Two Numbers - powered by WebMath ... This page will show you a complete "long division" solution for the division of two numbers.
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Division (mathematics) - Wikipedia


Division (mathematics) This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced ...
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4 Easy Ways to Do Long Division (with Pictures) - wikiHow


How to Do Long Division. A part of basic arithmetic, long division is a method of solving and finding the remainder for division problems that involve numbers with at ...
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what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices

what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices how to solve this math problem using matrices?   2  -1  -2  │  1   4   1  -1  │  5   1   1   4  │ -5 Multiply line 2 by 2.   2  -1  -2  │  1   8   2  -2  │ 10   1   1   4  │ -5 Subtract line 2 from line 1, replace line 1.  -6  -3   0  │ -9   8   2  -2  │ 10   1   1   4  │ -5 Multiply line 2 by 2.  -6  -3   0   │ -9  16   4  -4  │ 20    1   1   4  │ -5 Add line 3 to line 2, replace line 2.  -6   -3   0  │ -9  17   5   0  │ 15    1   1   4  │ -5 Multiply line 3 by 6.  -6  -3   0  │  -9  17   5   0  │  15   6   6  24  │ -30 Add line 1 to line 3, replace line 3.  -6  -3   0  │  -9  17   5   0  │  15   0   3  24  │ -39 Multiply line 1 by 5; multiply line 2 by 3. -30 -15   0  │ -45  51  15   0  │  45    0   3  24  │ -39 Add line 2 to line 1, replace line 1.  21    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Divide line 1 by 21.   1     0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Multiply line 1 by 51.  51    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Subtract line 1 from line 2, replace line 2.  51   0   0  │   0   0  15   0  │  45   0   3  24  │ -39 Divide line 2 by 5.  51   0   0  │   0    0   3   0  │   9   0   3  24  │ -39 Subtract line 2 from line 3, replace line 3.  51   0   0  │   0    0   3   0  │   9   0   0  24  │ -48 Three-in-one: divide line 1 by 51, divide line 2 by 3, divide line 3 by 24.   1   0   0  │   0   0   1   0  │   3   0   0   1  │  -2 This shows that x = 0, y = 3, z = -2  
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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d square y divided by dt square+2 dy divided by dx_3y=sint, given that y=dy divided by dx=0 when t=0

d(dy/dt)/dt+2dy/dx-3y=sin(t); y=dy/dx=0 when t=0. y=f(t); x=g(t); dy/dt=f'; dx/dt=g'; d(dy/dt)/dt=f"; dy/dx * dx/dt=dy/dt, g'dy/dx=f'; dy/dx=f'/g'. f"+2f'/g'-3f=sin(t) or g'f"+2f'-3fg'=g'sin(t). f(0)=0; f'(0)/g'(0)=0, so f'(0)=0 (because dy/dx=0=f'(0)/g'(0)). Also, g'(0) cannot be zero, otherwise dy/dx would be undefinable at t=0. But for f'/g', we would have an expression on the left that could be solved as an auxiliary, or characteristic, equation and we could apply the homogeneous solution fH=Ae^-3t + Be^t where fH denotes the homogeneous component of the solution, and A and B are constants and the exponents are simply the solutions of m^2+2m-3=(m+3)(m-1)=0 giving us roots -3 and 1. If g'=1 then clearly g' is non-zero, which is a requirement. But if g'=1 x=g(t)=t+C where C is a constant and fH=Ae^-3t + Be^t. xH=gH=t, where H denotes the homogeneous component. Could this be a solution? This is the homogeneous component of the solution fH, where complete f=fH+fP, where fP denotes the particular solution. Now we need to resolve the right-hand side. This is the particular solution fP and we can assume that it is of the form fP=Csin(t)+Dcost(t). Applying the derivatives we have: Ccos(t)-Dsin(t) as 1st, -Csin(t)-Dcos(t) as 2nd. These must satisfy the original DE. So, -Csin(t)-Dcos(t)+(2/g')(Ccos(t)-Dsin(t))-3(Csin(t)+Dcos(t))=sin(t). Comparing coefficients: sin(t): -C-2D/g'-3C=1; cos(t): -D+2C/g'-3D=0. 4C+2D/g'+1=0 and 2C/g'=4D, so C=2Dg'. 8Dg'+2D/g'+1=0, 8Dg'^2+2D+g'=0. 2D(4g'^2+1)=-g'. This gives us fP=-sin(t)/5-g'cos(t)/(8g'^2+2). Combining the auxiliary and particular components, y=f(t)=fH+fP=Ae^-3t + Be^t-sin(t)/5-g'cos(t)/(8g'^2+2). If we can replace g' with 1 then D=-1/10, C=-1/5, y=Ae^-3t + Be^t-sin(t)/5-cos(t)/10 and x=t. The initial conditions have f(0)=0=f'(0), so 0=A+B-1/10 and -3A+B-1/5=0. From these, -3A+1/10-A-1/5=0, -4A-1/10=0, A=-1/40, 0=-1/40+B-1/10, making B=1/8. Thus, y=(e^t)/8-(e^-3t)/40-sin(t)/5-cos(t)/10, x=t or y=(e^x)/8-(e^-3x)/40-sin(x)/5-cos(x)/10. CHECK: y'=(e^t)/8+3e^(-3t)/40-cos(t)/5+sin(t)/10; x'=g'=1, y'=f', dy/dx=dy/dt=f'; y"=(e^t)/8-9e^(-3t)/40+sin(t)/5+cos(t)/10.  y"+2dy/dx-3y becomes: (e^t)/8-9e^(-3t)/40+sin(t)/5+cos(t)/10 [this is y"] +(e^t)/4+3e^(-3t)/20-2cos(t)/5+sin(t)/5 [this is 2dy/dx=2y'] -3(e^t)/8+3(e^-3t)/40+3sin(t)/5+3cos(t)/10= [this is -3y] (e^t)(1/8+1/4-3/8) [=0] +e^(-3t)(-9/40+3/20+3/40) [=0] +sin(t)(1/5+1/5+3/5) [=sin(t)*1]  +cos(t)(1/10-2/5+3/10)= [=0] sin(t) [ADDENDUM: Why the auxiliary, or characteristic, equation works. Consider the equation y=Ae^ax+Be^bx, where a and b are constants. y'=aAe^ax+bBe^bx; y"=a^2Ae^ax+b^2Be^bx. Now consider a 2nd degree DE: y"+Py'+Qy=0, where P and Q are constants. We can replace the derivatives: a^2Ae^ax+b^2Be^bx+PaAe^ax+PbBe^bx+QAe^ax+QBe^bx=0, which can be written: Ae^ax(a^2+Pa+Q)+Be^bx(b^2+Pb+Q)=0. Since the exponential components cannot be <0, the quadratics in a and b must each be zero and they must be the roots of the same equation: e.g., z^2+Pz+Q=0 where the roots are z=a and b, with P and Q as the coefficients. In the question, P=2 and Q=-3 so z^2+2z-3=0 is the quadratic, so (z+3)(z-1)=0 and a=-3 and b=1, so fH=Ae^-3t+Be^t.]
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53 divided 898 ??

do you need a  decimal quotient? or fraction? if decimal:     53  /  898  =   0.0590200... 1st: add a decimal point to 53 then 0... put the point above in the quotient 2nd: divide as 530 by 898   , the quotient is 0 3rd: place another 0 in 530 it becomes 5300, which when divided by 898 = 5 4th: multiply 5 by 898 = 4490, subract this from 5300 = 810, then put another 0. 5th: 8100 divided by 898 = 9 , multiply 9 by 898 = 8082, subtract this from 8100.. 6th: continue putting 0 to the difference then divide again...
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how to find slope and y-intercept of y=3x-5

Any line equation which is in the form y=m*x + b is called slope-intercept form. What that gives you is the slope is the number which is multiplied by X (called the coefficient) while the slope intercept is the Y value when X=0. So if we take your first equation: y = 3x - 5 The slope (or m) = 3 The Y intercept = -5 Slope is defined as either "the change in y divided by the change in x" or "rise over run", so that 3, really can be considered as 3/1. Each change in X of 1 will change Y by 3. So slope = 3/1. Graphing any line can come from 2 methods. 1) create a table of values or 2) calculate a single point (x,y) and then apply the slope to find a second coordinate pair (x,y). For the equation y=3x - 5: y = 3x - 5 x y 0 -5 1 -2 Replacing the values for X into the original equation, will come out to the values for Y. So when X=0, y = 3*0 - 5, or simply -5.  When X = 1, Y = 3*1 -5 or simply -2. With these two coordinate pairs of points, you can plot a dot on your graph at each (0,-5) and (1, -2) then draw a straight line which goes through each point and continues straight in each direction, probably ending each end of this line with an arrow to show it continues. I do not have a way to include a picture here of a graph. The second way to graph a line is as follows. You need a starting point that will be on the line. Given the form of y=mx + b, you have a simple point which can be used at the y-intercept. The point is always in the form of (0, b), so in this case it is (0,-5). From that first point, you will apply your slope. The slope is 3 (or technically 3/1 which is a big help). From the initial point (0,-5) you will go UP 3 and RIGHT 1 and that will be the next point that is easy to find. Connect those two points and continue the line in each direction and that will be a graph of your line. Anytime your slope is positive, you will use it by going UP the top number (numerator of the slope) and going RIGHT the bottom number (denominator of the slope). But if your slope is negative (like your second problem is) you will use it by going DOWN the numerator and then RIGHT the denominator. The equation is y= -2/3x + 4 ( / = divided). I need to state the slope and y-intercept. I will not walk through the details on the second equation, but you should have enough information to get the answer from the above example.
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explain what it means if you divide a quantity between 2 people in this ratio 1:1,0:1 and 1:0

Dividing a quantity between 2 people in the ratio 1:1.  You get the same amount as me. Dividing a quantity between 2 people in the ratio 0:1.  You get none, I get all of it. Dividing a quantity between 2 people in the ratio 1:0.  You get all of it, I get none. Actually the last two (1:0 and 0:1) are really the same thing because the problem doesn't define ahead of time which person/quantity is the first one and which is the second one.
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does 5 divide by 0 and 0 divide by 5 have same the answer?

yu dont getta divide bi zero...that wood giv infinity yu will get all the guvt krats mad at yu
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how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0
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how to solve for x with fractions

Simplifying x3 + 3x2 + -4x = 0 Reorder the terms: -4x + 3x2 + x3 = 0 Solving -4x + 3x2 + x3 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), 'x'. x(-4 + 3x + x2) = 0 Factor a trinomial. x((-4 + -1x)(1 + -1x)) = 0 Subproblem 1 Set the factor 'x' equal to zero and attempt to solve: Simplifying x = 0 Solving x = 0 Move all terms containing x to the left, all other terms to the right. Simplifying x = 0 Subproblem 2 Set the factor '(-4 + -1x)' equal to zero and attempt to solve: Simplifying -4 + -1x = 0 Solving -4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '4' to each side of the equation. -4 + 4 + -1x = 0 + 4 Combine like terms: -4 + 4 = 0 0 + -1x = 0 + 4 -1x = 0 + 4 Combine like terms: 0 + 4 = 4 -1x = 4 Divide each side by '-1'. x = -4 Simplifying x = -4 Subproblem 3 Set the factor '(1 + -1x)' equal to zero and attempt to solve: Simplifying 1 + -1x = 0 Solving 1 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + -1x = 0 + -1 Combine like terms: 1 + -1 = 0 0 + -1x = 0 + -1 -1x = 0 + -1 Combine like terms: 0 + -1 = -1 -1x = -1 Divide each side by '-1'. x = 1 Simplifying x = 1 Solution x = {0, -4, 1}
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What is Mike's speed given the information below

In the first part of the problem, Andrew, traveling at a speed of v1, travels 4 miles, while Mike, traveling at speed v2, travels (d - 4) miles. They leave from their respective starting points at the same time, so the time it takes for them to meet and pass is the same for both. t = d / s 1. t1 = 4/v1 = (d - 4) / v2 Multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 2. (4 / v1) * v1 * v2 = ((d - 4) / v2) * v1 * v2 3. 4v2 = (d - 4) v1 Divide both sides by four to get the value of v2 4. v2 = ((d - 4)v1) / 4 In the second part of the problem, Andrew has reached Simburgh (d) and turned around, travelling another 2 miles, or (d + 2), while Mike has reached Kirkton and turned around, travelling another (d - 2) miles, for a total of d + (d - 2) = (2d - 2) miles. Again, their times are equal when they meet and pass. 5. t2 = (d + 2) / v1 = (2d - 2) / v2 As in the first part, multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 6. ((d + 2) / v1) * v1 * v2 = ((2d - 2) / v2) * v1 * v2 7. (d + 2)v2 = (2d - 2)v1 Divide both sides by (d + 2) go get the value of v2 8. v2 = ((2d - 2)v1) / (d + 2) We have two equations for v2, equation 4 and equation 8. The problem states that v2 remains the same throughout the journey. Therefore: ((d - 4)v1) / 4 = ((2d - 2)v1) / (d + 2) Once again, multiply both sides by both denominators. (((d - 4)v1) / 4) * 4 * (d + 2) = (((2d - 2)v1) / (d + 2)) * 4 * (d + 2) v1 * (d - 4) * (d + 2) = v1 * (2d - 2) * 4 Divide both sides by v1, eliminating speed from this equation. (d - 4) * (d + 2) = (2d - 2) * 4 d^2 - 4d + 2d - 8 = 8d - 8 d^2 - 2d - 8 = 8d - 8 Subtract 8d from both sides and add 8 to both sides. (d^2 - 2d - 8) - 8d + 8 = (8d - 8) + 8d + 8 = 0 d^2 - 10d = 0 Factor out a d on the left side. d * (d - 10) = 0 One of those factors is equal to 0 (to give a zero answer). d = 0 doesn't work; we already know the distance is more than 4 miles. d - 10 = 0 d = 10    <<<<<   That's the answer to the first question, how far is it? We'll substitute that into equation 4 to find v2 in relation to v1. v2 = ((d - 4)v1) / 4 = ((10 - 4)v1) / 4 v2 = 6v1 / 4 = (6/4)v1 v2  = 1.5 * v1   <<<<<< That's the answer to the second question No matter what speed you choose for Andrew (v1), Mike's speed is one-and-a-half times faster. Let's set Andrew's speed to 6mph and solve equation 1. t1 = 4/v1 = (d - 4) / v2 t1 = 4mi / 6mph = (10 - 4) / (1.5 * 6mph) 4/6 hr = 6/9 hr 2/3 hr = 2/3 hr With Andrew travelling at 4 mph, and Mike travelling at 6 mph, it took both of them 2/3 of an hour to reach a point 4 miles from Kirkton.
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