Question about Wise Solutions Wise Installation System 9 Professional Edition for PC

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Take a no. = X

the no. X is 5 greter than another no.

hence another no. is X-5

sum of the squares = smaller no.

[(X*X)]+[(X-5)(X-5)] = X-5

solve this and u will get the desired answer.

THNKS FOR ASKING.ASK SOME MORE QUESTIONS.

Posted on Nov 11, 2008

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Posted on Jan 02, 2017

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I don't think you can do it! Could you check your numbers.

If we are using the numbers 1-9 once and only once in a 3x3 magic square, the sum of the rows + the sum of the columns should be 90, since 1+2+3+4+5+6+7+8+9 in the rows adds up to 45 and 1+2+3+4+5+6+7+8+9 in the columns adds up to 45.

When trying to solve, the magic number seem to be the sums 24 and 10. To get 24, the only three numbers that add to 24 are 7 + 8 +9. Similarly to get 10, the lowest numbers re 1 and 2 and the smallest big number to use is 7. I then ran out of number trying to get 21, 13 and 15 sums.

Good luck,

Paul

If we are using the numbers 1-9 once and only once in a 3x3 magic square, the sum of the rows + the sum of the columns should be 90, since 1+2+3+4+5+6+7+8+9 in the rows adds up to 45 and 1+2+3+4+5+6+7+8+9 in the columns adds up to 45.

When trying to solve, the magic number seem to be the sums 24 and 10. To get 24, the only three numbers that add to 24 are 7 + 8 +9. Similarly to get 10, the lowest numbers re 1 and 2 and the smallest big number to use is 7. I then ran out of number trying to get 21, 13 and 15 sums.

Good luck,

Paul

May 21, 2015 | Office Equipment & Supplies

When you say "to numbers" I assume you meant "two numbers."

There is no solution.

For the product of two numbers to be 280, either both numbers must be positive or both numbers must be negative. For the sum of two numbers to be -18, one or both numbers must be negative. Taken together, both numbers must be negative.

We can find two numbers that multiply to 280 and add to +18, then simply take the negative of both numbers.

The square root of 280 is about 16.7. The sum of any two numbers that multiply to 280 must be at least twice that, or at least 33.4. Since 33.4 is greater than 18, there is no solution.

If you want to do it the hard way, take any pair of negative numbers that multiply together to 280, add them together, and see how close you come to -18. For example, -1 and -280 add to -281. -2 and -140 add to -142. -4 and -70 add to -74. -5 and -56 add to -61. -7 and -40 add to -47. -8 and -35 add to -43. -10 and -28 add to -38. -14 and -20 add to -34, which is as close as you can come in the integers.

There is no solution.

For the product of two numbers to be 280, either both numbers must be positive or both numbers must be negative. For the sum of two numbers to be -18, one or both numbers must be negative. Taken together, both numbers must be negative.

We can find two numbers that multiply to 280 and add to +18, then simply take the negative of both numbers.

The square root of 280 is about 16.7. The sum of any two numbers that multiply to 280 must be at least twice that, or at least 33.4. Since 33.4 is greater than 18, there is no solution.

If you want to do it the hard way, take any pair of negative numbers that multiply together to 280, add them together, and see how close you come to -18. For example, -1 and -280 add to -281. -2 and -140 add to -142. -4 and -70 add to -74. -5 and -56 add to -61. -7 and -40 add to -47. -8 and -35 add to -43. -10 and -28 add to -38. -14 and -20 add to -34, which is as close as you can come in the integers.

Jun 25, 2014 | Office Equipment & Supplies

[3, 10, 16], [4, 5, 18], [5, 12, 14], and [10, 11, 12] all work. Of course, there are an infinite number of numbers that are the sum of three squares.

Aug 22, 2012 | Office Equipment & Supplies

Assume this:

A is the greater number

B is the lesser

You have 2 equations

A + B = 20

A = 3xB - 4

Substitute the solution for the second eqation into the first

(3xB - 4 ) + B = 20

solve for B

3xB + B = 20 + 4

4xB = 24

B = 6

solve for A with solution for B

go back to equation 2

A = 3xB - 4

A = 3x6 - 4

A = 14

So A = 14

and B = 6

Otherwise use trial and error.

Hope that helps.

A is the greater number

B is the lesser

You have 2 equations

A + B = 20

A = 3xB - 4

Substitute the solution for the second eqation into the first

(3xB - 4 ) + B = 20

solve for B

3xB + B = 20 + 4

4xB = 24

B = 6

solve for A with solution for B

go back to equation 2

A = 3xB - 4

A = 3x6 - 4

A = 14

So A = 14

and B = 6

Otherwise use trial and error.

Hope that helps.

Jun 29, 2011 | HP 12c Calculator

"5 and 25"

5+25 = 30

The square root of 25 is 5.

5+25 = 30

The square root of 25 is 5.

May 21, 2011 | Ubi Soft Catz 5 (Windows) (680901) for PC

If the base is one, and it is a right angle triangle, then the hypotenuse MUST be greater than one, because it is the sum of the squares of the other two sides, hence one squared, which is one, plus another number squared, which is some positive number, so the sum is greater than one.

So, if the hyp is one and the base is one, the hyp is not actually a hyp, because it is not a right angle triangle.

What a strange question to ask on a PC forum!

So, if the hyp is one and the base is one, the hyp is not actually a hyp, because it is not a right angle triangle.

What a strange question to ask on a PC forum!

Oct 22, 2010 | Acer Aspire One PC Notebook

There are no such pair of numbers. The problem has no solution.

Sep 21, 2010 | Garden

From Joe-Bob's handy-dandy reference:

(wikiedia...)

Normal magic squares exist for all orders*n* ≥ 1 except *n* = 2, although the case *n* = 1 is trivial—it consists of a single cell containing the number 1

The constant sum in every row, column and diagonal is called the magic constant or magic sum,*M*. The magic constant of a normal magic square depends only on *n* and has the value

For normal magic squares of order*n* = 3, 4, 5, …, the magic constants are:

15, 34, 65, 111, 175, 260, …

(wikiedia...)

Normal magic squares exist for all orders

The constant sum in every row, column and diagonal is called the magic constant or magic sum,

For normal magic squares of order

15, 34, 65, 111, 175, 260, …

Feb 23, 2009 | Computers & Internet

Is this somebody's home work assignment or what??? You got the wrong forum dude.

Feb 12, 2008 | Office Equipment & Supplies

You may want COUNTIF if you're specifying criteria. For instance, if my prices are found in b3 to b7, here's a formula that will find all those that are less than 6 ($6.00):

=COUNTIF(B3:B7, "<6")

If you're using multiple criteria, such as you want to find all the prices that are greater than $5 and less than $8, the following will accomplish it. (The ABS gives you the absolute value of the result, in case the smaller number is first.)

=ABS(SUM(COUNTIF(B3:B7, ">5") - COUNTIF(B3:B7, "<8")))

=COUNTIF(B3:B7, "<6")

If you're using multiple criteria, such as you want to find all the prices that are greater than $5 and less than $8, the following will accomplish it. (The ABS gives you the absolute value of the result, in case the smaller number is first.)

=ABS(SUM(COUNTIF(B3:B7, ">5") - COUNTIF(B3:B7, "<8")))

Nov 06, 2007 | Oracle 10g Database Standard (ODBSEONUPP0)

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