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Word story problem square number

How old am I? My age is a perfect square number of years greater than a perfect square number; it is also a perfect square number of years less than a perfect square. Next year, my age will be a prime number less than 50.

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Square Root Word Problems - TuLyn


Square Root Word Problems. Are you looking for word problems on square root?
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WORD PROBLEMS INVOLVING PERFECT SQUARES


WORD PROBLEMS INVOLVING PERFECT SQUARES. ... When I take one less than three times this number, and then square the result, I end up with the number $\,25\,$.
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Math - Square Root - Word Problem - Solving Math Problems


Math - Square Root - Word Problem. by Andrew (Marikina City, Philippines ... Replace the words “sum of a number and 12” with n + 12 √ of n + 12 is 5
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Square Root Word Problems - Math Worksheets Land


Exact Square Root Word Problems. We covered just about every scenario we have ever seen on national exams with these problems.
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Solving Word Problems Involving Square Roots - Braingenie


Improve your skills with free problems in 'Solving Word Problems Involving Square Roots and Area' and thousands of other ... Graphing and Ordering Real Numbers
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Square roots word problems - Raymond Junior


Subject: Image Created Date: 8/20/2011 9:28:40 AM
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Word problems - A complete course in algebra


Word problems that lead to a linear equation. The whole is equal to the sum of the parts. ... Every word problem has an unknown number. In this problem, ...
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"Number" Word Problems - Purplemath | Home


"Number" Word Problems ... In other words, letting the first number be "n" and the second number be "n + 2", I have: (n) (n + 2) = 24 n 2 + 2n = 24 n 2 + 2n ...
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Square Roots Word Problems Lesson - Math Worksheets Land


Square Roots Word Problems Lesson ... Square footage is determined by multiplying the length and width of the room. Length x Width = Square Footage
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Suggested Questions And Answer :


Word story problem square number

So the answer is a number less than 50.  The  perfect square numbers are.. 4, 9, 16, 25, 36 and 49 which are all less than 50.  The question states it is a number that is a perfect square greater than a perfect square and a perfect square less than a perfect square.   The only number that would work is 40 which is a perfect square of 4 greater than 36 another perfect square and a perfect square of 9 less than 49.  No other numbers fit the perfect square.  And next year  41 is a prime number less than 50.
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Word problems formulas

############# grate...yu dont hav a klue, so yu leev out essenshal data us gotta hav tu solv problem me bet guvt rule bout size av furnus is a number * size av hous size av hous=area av floor (square ft) number...based on past speerans with witnter wether theer
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how to translate mathematical sentence?

It can't be answered with that exact wording because "the sum of a number. . ." is missing the what else part, like "the sum of a number and something else." If the problem was "translate the square of the sum of a number and another number," it would be (x+y)^2
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magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
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Need help with word problem

Let the age of one surfer be 10a+b, then the age of the other is 10b+a. Add them together: 11a+11b, so 1/11 is a+b=sqrt(10b+a-10a-b)+1=sqrt(9b-9a)+1=sqrt(9(b-a))+1=3sqrt(b-a)+1. If b-a is to be a perfect square, the only ones are 0, 1, 4, 9, making a+b=1, 4, 7, 10. Now there's a problem: if we use the elimination method using two equations b+a=M and b-a=N, where M and N are positive integers, we add the equations together to eliminate a and we get 2b=M+N, so b=(M+N)/2. Therefore M and N must be both odd or both even. Unfortunately, if M is in the set 1, 4, 7, 10 then the corresponding N values are 0, 1, 4, 9. But when M is even N is odd or vice versa, and we would end up with fractional values for a and b. If the question is interpreted as a+b=sqrt(9(b-a)+1) (the 1 is added before taking the square root instead of after), then the only perfect square occurs when b-a=7, so we have sqrt(9*7+1)=8=a+b. Again, an odd and an even number. (There is actually another trivial solution: when a=b, a+b=1, and we still end up with fractions!) There is a solution if "plus 1" is omitted: 15 and 51 added together make 66; divide by 11 to get 6; subtract 15 from 51=36, the square root of which is 6.
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multiple equations and word problems?

x^2=(1/9), so x=(1/3) x^3=(1/27), so x=(1/3) x^2=(1/16), so x=(1/4)
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HOW DO I CALCULATE WEIGHT OF DUPLEX PAPER REEL WHICH WIDTH IS 21" ,DIAMETER OF REEL IS 102" AND GSM IS 230

I assume that 230 GSM  means grams per square metre? A reel of paper is effectively a spiral, for which there is a mathematical formula: r=R+At, where R is the inner radius of the reel of paper and t is an angle measured in radians. A radian is approximately 57 degrees and 360 degrees is 2(pi) radians. In other words, to convert degrees to radians we multiply degrees by the quantity (pi)/180. The mathematical number pi is about 3.142, the ratio of the circumference of a circle to its diameter. If we draw a line length R, the inner radius, this is where the paper starts from the middle. This is the starting point and angle t=0. When t=2(pi) we have one wrapping of paper and the length of paper is 2(pi)R. The next wrapping lies over the first wrapping, so the radius of the second wrapping is slightly more than R: it's R+p, so the length of the paper is 2(pi)R+2(pi)(R+p). The angle t after 2 wrappings is 4(pi). The weight of one wrapping is related to the surface area of the cylinder of paper=2(pi)Rw where w is the width of the paper=21". So we need to convert 21" into metres so that we can bring in the given density of 230GSM. 1"=2.54cm=0.0254m, so 21"=0.5334m. We convert the surface area into a weight by multiplying it by the density d=230. Now we have the equation weight=2(pi)Rwd grams. Because the paper is thin we can consider the reel as being made up of concentric cylinders of paper until we reach the outer radius of 102/2=51"=1.2954m. Now we have a series: surface area of 1st cylinder + SA of 2nd cylinder + SA of 3rd cylinder +... This is: 2(pi)Rw+2(pi)(R+p)w+2(pi)(R+2p)w+2(pi)(R+3p)w+... This can be written: 2(pi)w(R+R+p+R+2p+R+3p+...) which is an arithmetic series. How long is the series? To find out we need to divide the difference between the inner and outer radii by p to get the number of wrappings. So since the outer radius is 1.2954m N, the number of wrappings is (1.2954-R)/p. The series is 2(pi)w(NR+p(1+2+3+...))=2(pi)w(NR+p(N(N-1)/2) and the weight of this is 2(pi)Nwd(R+p(N-1)/2). To this has to be added the weight of the empty reel, which we can call W: gross weight=W+2(pi)Nwd(R+p(N-1)/2), and N=(1.2954-R)/p. CONCLUSION: the inner radius R and the paper thickness p in metric units need to be found to find the weight, plus, possibly, the weight of the empty reel. [In the equation r=R+At, A can be calculated: when t=2(pi), R=R+p, so R+p=R+2(pi)A and A=p/2(pi). r=R+pt/2(pi), and calculus can then be used to solve the problem in a different way. The arithmetic series, however, is probably sufficient to determine the weight to a good approximation.]
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Using a system of equations to solve a word problem?

assuming x times she raked the leaves she mowed lawns x+12 times 6x + 8(x+12) = 918 14 x = 918-96 x = 58 (if the total is 908, otherwise x is not a whole number)
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solve word problem using polynomials

Problem: solve word problem using polynomials Joe has a collection of nickles and dimes worth $2.15 If the number of dimes was doubled and the number of nickles was increased by 28, the value of the coins would be $4.65. How many nickles and dimes does he have? Let's say he has n nickles and d dimes. 0.05n + 0.10d = 2.15 The problem proposes to increase d to 2d and increase n to n + 28. 0.05(n + 28) + 0.10(2d) = 4.65 We have two equation to work with. Let's multiply both equations by 100 to eliminate the decimals. 100(0.05n + 0.10d) = 2.15 * 100 5n + 10d = 215 100(0.05(n + 28) + 0.10(2d)) = 4.65 * 100 5(n + 28) + 10(2d) = 465 We need to expand this second equation. 5(n + 28) + 10(2d) = 465 5n + 140 + 20d = 465 5n + 140 + 20d - 140 = 465 - 140 5n + 20d = 325 Here's what we have: 1) 5n + 10d = 215 2) 5n + 20d = 325 We can immediately eliminate the n by subtracting equatioin 1 from equation 2.    5n + 20d = 325 -(5n + 10d = 215) ------------------------           10d = 110 10d = 110 10d/10 = 110/10 d = 11 This tells us that Joe currently has 11 dimes. Let's use equation 1 directly above to solve for n. 5n + 10d = 215 5n + 10(11) = 215 5n + 110 = 215 5n + 110 - 110 = 215 - 110 5n = 105 5n/5 = 105/5 n = 21 This one tells us that Joe currently has 21 nickels. We need to check the two numbers against what the problem originally stated. 0.05n + 0.10d = 2.15 0.05(21) + 0.10(11) = 2.15 1.05 + 1.10 = 2.15 2.15 = 2.15 Verified! Answer: Joe has 21 nickels and 11 dimes.  
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In your own words, explanation what it means to find the absolute value of each number, use a number line example ti support your explanation

The absolute value of any number is its magnitude regardless of whether it is positive or negative. So absolute value is always positive. It's the same as taking the positive square root of the number squared. So the absolute value of 10 and -10 is 10, because both have a square of 100. If we take a straight line and mark its centre point as 0, then numbers to the left of centre are negative and those to right of centre are positive. Taking the absolute value is the same as measuring how far the number is from the centre point. -10 is 10 away from zero and so is +10.
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