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i need the answer to the problem one over three plus seven over ten

i am having trouble with the problem  one over three plus seven over ten

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I need help quick. What is the quotient of three... - OpenStudy


I need help quick. What is the quotient of three and one over five divided by seven over ten in simplest form? three and two over seven ... Not the answer you are ...
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Add fractions, subtract fractions -- Problems - TheMathPage


To see the answer, pass your mouse over the ... ("Reload"). Do the problem yourself first! 1 1 ... Three candidates were running for office. One of them got ...
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Mathematical Puzzles: What is () + () + () = 30 using 1,3,5,7 ...


... I have added the previous answer to the present one. ... Mathematical Puzzles: What is + + = 30 using ... 30_9[/math], is equivalent to twenty seven in base ten ...
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Quick MathSpeak Tutorial


Quick MathSpeak Tutorial. ... StartFrac six-halves Over 3 EndFrac equals three-thirds equals 1. ... 5 plus StartFrac x minus one-half Over x plus one-half EndFrac .
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Think of a number, double it, add six, divide it in half ...


... so I must have my original number plus three Subtract the number you started ... have done so in other answers. No one has ... What number can you add to ten, ...
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Answers to Riddles - Israeli English Teachers Network


Answers to Riddles. ... When can you add two to eleven and get one as the correct answer? Answer: ... one two three four five six seven: Feb. 28
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Adding and Subtracting Fractions with Pictures Pt 1 of 2 ...


Feb 03, 2014 · Adding and Subtracting Fractions with Pictures Pt ... to our answer. We need to add three more ... at one last addition problem. Five-fourths plus ...

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i need the answer to the problem one over three plus seven over ten

1/3 + 7/10 10/30 + 21/30 31/30 1   1/30
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seven sets of seven plus sixty two sets of seven

Treat “sets of seven” like one object. Call it a box if you like. Now read the question again using boxes: “7 boxes + 62 boxes”. Of course, the answer is 69 boxes. So the answer to the problem is 69 sets of seven. You don't need to know what sets of seven means.
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solve the system: 5x-10y+4z=-73; -x+2y-3z=19; 4x-3y=5z=-42

solve the system: 5x-10y+4z=-73; -x+2y-3z=19; 4x-3y=5z=-42 1)  5x - 10y + 4z = -73 2)  -x +  2y - 3z =  19 3)  4x -  3y + 5z = -42 Equation three had "=5z" so I changed the equal sign to a plus sign. It happens a lot; the shift key doesn't register, so what was supposed to be "+" comes out "=". We're going to eliminate z so we can work on finding x and y. We need the coefficients of z to be equal (disregarding the sign). Equation one; multiply by 3: 3 * (5x - 10y + 4z) = -73 * 3 4)  15x - 30y + 12z = -219 Equation two; multiply by 4: 4 * (-x +  2y - 3z) =  19 * 4 5)  -4x + 8y - 12z = 76 Add equation five to equation four, eliminating z.   15x - 30y + 12z = -219 +(-4x +  8y - 12z =   76) -----------------------------------   11x - 22y         = -143 6)  11x - 22y = -143 Follow the same procedure with equations two and three. Equation two: multiply by 5 this time: 5 * (-x +  2y - 3z) =  19 * 5 7)  -5x + 10y - 15z = 95 Equation three: multiply by 3: 3 * (4x -  3y + 5z) = -42 * 3 8)  12x - 9y + 15z = -126 Add equation eight to equation seven, eliminating z.    -5x + 10y - 15z =    95 +(12x -  9y + 15z = -126) ------------------------------------     7x +    y         =  -31 9)  7x + y = -31 Multiply equation nine by 22; then we will add equation six. 22 * (7x + y) = -31 * 22  154x + 22y = -682 +(11x - 22y = -143) ---------------------------  165x         = -825 x = -5  <<<<<<<<<<<<<<<<<<<<<<<<<< Substitue that into equations six and nine to find y. They will serve to verify the calculations if you do it with two equations. 11x - 22y = -143 11(-5) - 22y = -143 -55 - 22y = -143 -22y = -143 + 55 -22y = -88 y = 4  <<<<<<<<<<<<<<<<<<<<<<<<<< 7x + y = -31 7(-5) + y = -31 -35 + y = -31 y = -31 + 35 y = 4     same value for y Now, we can go back to the original equations, plug in both x and y, and solve for z. To check and verify, we will use all three of the original equations. One: 5x - 10y + 4z = -73 5(-5) - 10(4) + 4z = -73 -25 - 40 + 4z = -73 -65 + 4z = -73 4z = -73 + 65 4z = -8 z = -2  <<<<<<<<<<<<<<<<<<<<<<<<<< Two: -x +  2y - 3z =  19 -(-5) +  2(4) - 3z =  19 5 + 8 - 3z = 19 13 - 3z = 19 -3z = 19 - 13 -3z = 6 z = -2     same answer for z Three: 4x -  3y + 5z = -42 4(-5) -  3(4) + 5z = -42 -20 - 12 + 5z = -42 -32 + 5z = -42 5z = -42 + 32 5z = -10 z = -2    again, verified x = -5, y = 4, z = -2
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A math word problem using the Quadratic Equation formula (URGENT)

A. The Golden Ratio I have a line divided into two unequal sections so that the ratio of the shorter length to the longer length is the same as the ratio of the longer length to the whole length of the line. What is the ratio? Solution Call the lengths of the two sections A and B. A/B=B/(A+B). Cross-multiply: A^2+AB=B^2, so A^2+AB-B^2=0 and, using the quadratic formula, A=(-B+sqrt(B^2+4B^2)/2=(-B+Bsqrt(5))/2=B(-1+sqrt(5))/2. Therefore the fraction A/B=(sqrt(5)-1)/2=0.618 or -1.618 approx. The positive root applies because A and B are considered positive lengths. An alternative solution is to let r=A/B, which is the ratio, then r=1/(r+1); r^2+r-1=0 and r=(sqrt(5)-1)/2.  This Golden Ratio keeps popping up in different contexts and seems to be aesthetically pleasing. B. The A sizes A rectangular piece of paper is folded in half so that the ratio of length to width remains the same after folding as before folding. What is the ratio? Solution If A and B are the length and width, the area is AB, and after folding in half the area becomes 1/2(AB). In the process of folding, only one side (choose the length A) is halved so the new dimensions of the rectangle are A/2 and B, but the ratio of the sides is still A/B, and the length and width are B and A/2 (the length has become the width and the width has become the length). So A/B=length/width=B/(A/2)=2B/A. If A/B=2B/A then A^2=2B^2. The ratio of the sides is A/B which we'll call r. So r^2=2 and r=sqrt(2)=1.414 approx. So the length is 1.414 times the width. The paper we use today is largely based on this ratio. A4 is the standard size used for printers. A3 is larger and tends to be used for drawings and posters. A5 is smaller and tends to be used for booklets and leaflets. All use the square root of 2 as the ratio of the sides. C. Identity Prove that the difference between the cubes of two consecutive integers is one more than three times their product. Solution The difference between the cubes of consecutive integers (x+1)^3-x^3 where x is an integer. The product of consecutive integers is x(x+1). The difference between the cubes is x^3+3x^2+3x+1-x^3=3x^2+3x+1=3x(x+1)+1 which is three times the product of the integers plus 1. These questions use in different ways a quadratic expression. Although this reply is later than you wished for, what goes round comes round, and you may have the opportunity of using the ideas.
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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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math triangle sums equals 20 on all sides using the numbers 1 - 9 only once

Note that if we take the numbers 1 to 19 in 9 pairs, the sums of the paired numbers come to 20: (1,19), (2,18), (3,17), (4,16), (5,15), (6,14), (7,13), (8,12), (9,11). Using the numbers 1 to 9 only, we can break down 4 of these pairs by splitting the larger number of each pair: (3,8,9), (4,7,9), (5,6,9), (5,7,8). These are the only unique groups of three, but they do not include 1 and 2. But if we include 1 and 2 we know the remaining 2 numbers to make up a group of 3 must add up to 19, which can't be achieved using a pair of numbers 1 to 9. Therefore, in some cases we need 4 numbers to make up the sum. We can only use  the numbers 1 to 9 and we need to use them all and we only have 3 sides of the triangle, so if we have 4 numbers on one side, one of the other sides will have only 2 numbers, because we're only left with 5 when we've taken out 4 for one side. What to do? We have to use the vertices so that the numbers at each vertex are used twice. Let's assume that all three vertices contain such numbers, then we will have 4 numbers a side. Instead of looking at groups of three numbers, we have to look for groups of 4: (1,2,8,9), (1,3,7,9), (1,4,6,9), (2,3,6,9), (1,4,7,8), (2,3,7,8), (1,5,6,8), (2,4,5,9), (2,4,6,8), (3,4,5,8), (3,4,6,7). Out of these we need to find where different groups contain a common number. These common numbers will form the vertices. Let's try putting the numbers 1, 2 and 3 at the vertices. This gives us a choice of 5 groups for each of the numbers 1, 2 and 3 (we can arrange the order so that the numbers occupy the first element in each set). Here are the sets: (1,2,8,9), (1,3,7,9), (1,4,6,9), (1,4,7,8), (1,5,6,8) (2,1,8,9), (2,3,6,9), (2,3,7,8), (2,4,5,9), (2,4,6,8) (3,1,7,9), (3,2,6,9), (3,2,7,8), (3,4,5,8), (3,4,6,7) Now we have to rearrange the numbers so that there are two vertices within each set, so that means we pick groups containing 1 and 2, 2 and 3, and 3 and 1. Yes, there are some, but unfortunately some contain duplicates of the two numbers that are not at the vertices. Look at the following, for example: (1,8,9,2), (2,6,9,3), (3,7,9,1) 9 is duplicated and we have no 4 or 5. In fact, we can"t have 1 as a vertex with 2 and 3 as the other vertices, because there's always a duplicate to spoil the arrangement. The same thing happens with 2, 3 and 4, and 3, 4 and 5. Symmetry is a common occurrence in mathematics and there's nothing more pleasing to a mathematician than an "elegant" solution. So I'm going to assume that symmetry applies to this problem and go for vertices in the middle of the range 1 to 9. The middle is occupied by numbers 4, 5 and 6. Let's see what happens if we use them as vertices: (4,2,9,5), (5,1,8,6), (6,3,7,4) This time it works! There are no duplications in the central pair of numbers so all the numbers from 1 to 9 are there. The other combination of 4 and 5 in the same group, (4,3,8,5), leads to a duplicated 8 in (5,1,8,6). So one answer is:  The vertices are A, B and C=4, 5 and 6 respectively, with 2 and 9 along AB, 1 and 8 along BC, and 3 and 7 along AC.  
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why in februry there 28 or 29 days and other month 30 or 31?

Please dont put these kinds of questions here anymore but here is the answer i copied it off of an amazon ask.com web sit which you could have just googled for./ ********************************************************************************** To meticulous persons such as ourselves, having the calendar run out in December and not pick up again until March probably seems like a pretty casual approach to timekeeping. However, we must realize that 3,000 years ago, not a helluva lot happened between December and March. The Romans at the time were an agricultural people, and the main purpose of the calendar was to govern the cycle of planting and harvesting. February has always had 28 days, going back to the 8th century BC, when a Roman king by the name of Numa Pompilius established the basic Roman calendar. Before Numa was on the job the calendar covered only ten months, March through December. December, as you may know, roughly translates from Latin as "tenth." July was originally called Quintilis, "fifth," Sextilis was sixth, September was seventh, and so on. Numa, however, was a real go-getter-type guy, and when he got to be in charge of things, he decided it was going to look pretty stupid if the Romans gave the world a calendar that somehow overlooked one-sixth of the year. So he decided that a year would have 355 days--still a bit off the mark, admittedly, but definitely a step in the right direction. Three hundred fifty five days was the approximate length of 12 lunar cycles, with lots of leap days thrown in to keep the calendar lined up with the seasons. Numa also added two new months, January and February, to the end of the year. Since the Romans thought even numbers were unlucky, he made seven of the months 29 days long, and four months 31 days long. ********************************************************************************** But Numa needed one short, even-numbered month to make the number of days work out to 355. February got elected. It was the last month of the year (January didn't become the first month until centuries later), it was in the middle of winter, and presumably, if there had to be an unlucky month, better to make it a short one. ********************************************************************************** Many years later, Julius Caesar reorganized the calendar yet again, giving it 365 days. Some say he made February 29 days long, 30 in leap year, and that Augustus Caesar later pilfered a day; others say Julius just kept it at 28. None of this changes the underlying truth: February is so short mainly because it was the month nobody liked much.
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Please help

You need to see through the problem and apply whatever is necessary to reduce the number of variables till eventually there's only one to find. Remember a simple fact: if you have two variables you always need two independent equations to find them; for three variables, three equations; four variables, four equations. You use the multiplication property if it helps you to eliminate a variable between two equations. Take some examples: x+y=10, x-y=3; simply adding these two equations will eliminate y and help you find x. 2x=13 so x=6.5 and y=3.5; 2x+y=10, x-2y=10; we could double one equation or the other so as to match the coefficients of one or other of the variables; but since it's easier to add two equations rather than to subtract them, where we have a minus in one equation and a plus in the other, we would prefer to use the multiplier for the relevant variable. So we double the first equation and add to the second: 4x+2y=20 PLUS x-2y=10: 5x=30, making x=6 and y=-2. The last pair of equations could have been written: 2x+y=x-2y=10, but it's still two equations. There is no one way to solve equations, and you can save yourself a lot of stress by not assuming you have to remember a rigid technique or formula as “The Way to do it”. You'll find mathematics is more fun when you intelligently try different methods and use your natural creativity to guide you. And here's another interesting thing. Those questions about finding a missing number in a series can be tackled in many cases as solving simultaneous equations. You only need n equations to find n variables, and a series can be seen as a set of terms generated by a function y=f(x) for different values of x (the position in the series), giving different values of y (the terms in the series). f(x) is a polynomial of the type ax^n+bx^(n-1)+cx^(n-2)+... If there are four given terms n=3 and the variables are a, b, c and d; if there are 3 given terms, n=2 and the variables are a, b and c. There is always a solution, we just have to work through and find it!
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what is the answer to 18-4t/0.5=20 ?

Perhaps it was the fraction that put you off, or was it the decimal in the denominator. I guess you want to find the value of t. Let's get rid of the fraction by multiplying both sides of the equation by 0.5. It's not clear in the question whether the problem is 18-(4t/0.5)=20 or (18-4t)/0.5=20. The brackets tell us what has to be worked out first. I'm going to provide solutions whichever one of these is the problem. First, 18-(4t/0.5)=20 Multiplying both sides by 0.5 we get 18*0.5-4t=20*0.5 giving us 9-4t=10. The decimals have been swallowed up in the multiplication. 0.5 is the same as 1/2, but if you didn't know that, multiply by 5 and move the decimal point back one place. So, for example, 5*18=90. The decimal point invisibly sits after 90, so moving it back a place makes it 9.0 which is just 9. So 9-4t=10. We need to get the numbers on one side of the equation and the unknown on the other side. At this stage we don't want to separate 4 from 4t. OK, let's keep the unknown on the left, so we get -4t=10-9, giving us -4t=1. When we move things from one side of an equation to the other plus becomes minus, and minus becomes plus, multiply becomes divide, and divide becomes multiply. So this is the same as 4t=-1, and t=-1/4. Second, (18-4t)/0.5=20 Multiply both sides by 0.5 and we get 18-4t=20*0.5, in other words 18-4t=10. Let's separate the unknown from the numbers, but this time we'll take the unknown over to the right (it doesn't matter which side really). So we get 18-10=4t. Therefore 8=4t, the same as 4t=8, so t=2. That looks like the simpler of the two answers, so my guess is that it's what the question really was, don't you think? It's important that questions are written correctly because, as you can see, we came up with two quite different answers.
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Use the following clues to find the ages of the three children.

Welcome to Math Homework Answers - Math Homework Answers is a math help site where students, teachers, and math lovers can ask and answer math homework questions to help create a useful fun math community. Ask or answer math questions in basic math, pre-algebra, algebra, trigonometry, geometry, pre-calculus, calculus, unit conversion or any other math related topic. Register and play the math game against others to see how many points you can earn. All categories Pre-Algebra Answers (518) Algebra 1 Answers (1,109) Algebra 2 Answers (450) Geometry Answers (245) Trigonometry Answers (71) Calculus Answers (238) Statistics Answers (71) Word Problem Answers (372) Site News (11) Use the following clues to find the ages of the three children. Use the following clues to find the ages of the three children. The product of their ages is 96. The middle child is two years older then the youngest.   A = Eldest B = Middle child C = Youngest   AxBxC = 96 <= eq1 B = 2 + C <= eq2   You need to supply one more clue to solve the problem    
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