Question about Nintendo Professor Layton & the Curious Village Games for DS

Do you need the answer? if so this is it :

41286

- 7953 or

41268

- 7935

each adds up to 33333

Posted on Jan 03, 2009

I can solve this there are 2 answers

1 is 41268- 7935

or 21286-7953.

good luck

Posted on Jan 04, 2009

SOURCE: use each of the numbers one through to nine

41268-7935=33333

Hope this helps

Posted on Aug 18, 2009

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Posted on Jan 02, 2017

Try these 0000, 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, or 00000, 11111, 22222, 33333, 44444, 55555, 66666, 77777, 88888, 99999 if these do not work add more details.

Dec 28, 2015 | Televison & Video

I'm not going to do your homework for you, but I'll explain how it's done in general. When you speak a number, you generally say the integer part in the normal manner, say, "point," and then speak the digits of the decimal portion (or, if the decimal portion is that of an irrational number or endless repeat, speak it to the desired number of decimal places). You write the number in words in the same way. For example, 99.995 is "Ninety-nine point nine nine five."

Nov 25, 2014 | Office Equipment & Supplies

99.9999 repeating is technically equivalent to 100, depending on what level math you're in.

For example, 100 / 3 = 33.3333...

but clearly 33.333 * 3 = 99.9999...

So multiplying both sides of my first equation by 3, we get 100 = 99.9999

If you are not looking for "a number equivalent to 100" but "an equation or set of numbers equivalent to 100", then there are many ways. For example:

99 + 9/9 = 99 + 1

And 99 + 1 = 100

For example, 100 / 3 = 33.3333...

but clearly 33.333 * 3 = 99.9999...

So multiplying both sides of my first equation by 3, we get 100 = 99.9999

If you are not looking for "a number equivalent to 100" but "an equation or set of numbers equivalent to 100", then there are many ways. For example:

99 + 9/9 = 99 + 1

And 99 + 1 = 100

Jan 16, 2014 | RXE160 Equivalent NTE16013-ECG POLYMERIC...

**No solutions:**The system is incoherent, incompatible Example: 2x+3y=8 and 2x+3y= 15. The two lines are parallel and distinct.**One solution:**There exits a pair of values (x,y) that satisfy both linear equations. The two lines on a Cartesian graph have one intersection point.**Infinite number of solutions:**The two equations are one and the same (one is just multiplied by some constant). The graph of the two lines yields the same line. One is superposed on the other. Any ordered pair (x,y) that satify one equation (there is an infinity of such pairs) satisfies the other.**Two solutions: cannot happen**because the two lines can either intersect once, be parallel, or superposed one on the other.

Dec 13, 2012 | Mathsoft StudyWorks! Mathematics Deluxe...

Start by combining like terms on each side of the equation.

Remember, when working with variables the real numbers always come last in the equation(i.e. 5+x-7=x+5-7)

Once you have this completed move all variables to one side and all whole numbers to the other side. Solve from there.

Remember, when working with variables the real numbers always come last in the equation(i.e. 5+x-7=x+5-7)

Once you have this completed move all variables to one side and all whole numbers to the other side. Solve from there.

May 17, 2012 | MathRescue Word Problems Of Algebra Lite

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)

2. Quadratic Equations y= ax^2+bx+c

3. Exponential Equations y= ab^x

4. Cubic Equations y=ax^3+ bx^2+cx+d

5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e

6. Equation of a circle (x-h)^2+(y-k)^2= r^2

7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).

8. Proportional equations y=kx; y= k/x, etc.

Jun 14, 2011 | Computers & Internet

The hints give you most of the solution and what's left can be worked out without much difficulty

It can take a while to work out mathematically without using the hints, so it's probably a good idea to use at least one or two of the hints

Remember that the leftmost top number can be a 4 (which, in turn, limits the number of possible combinations in the second column)

You can see the hints and two possible solutions here

http://professorlaytonwalkthrough.blogspot.com/2008/02/puzzle099.html

This popular walkthrough will also help you with anything else

It can take a while to work out mathematically without using the hints, so it's probably a good idea to use at least one or two of the hints

Remember that the leftmost top number can be a 4 (which, in turn, limits the number of possible combinations in the second column)

You can see the hints and two possible solutions here

http://professorlaytonwalkthrough.blogspot.com/2008/02/puzzle099.html

This popular walkthrough will also help you with anything else

May 20, 2010 | Nintendo Professor Layton & the Curious...

"Is there anything that I can do about this to stop my answers being complex numbers when solving cubic equations?"

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

May 05, 2010 | Casio CFX-9850G Plus Calculator

The best way to solve this is to develop one or more equations and then solve for the unknown numbers.

I solved this problem twice using different sets of equations to make sure I was right, and here is what I found:

1) I used the equations x + y = 623, and x = (2/3)y. In these equations, x is the number of girls and y is the number of boys.

Substituting the second equation into the first, I get (2/3)y + y = 623.

Adding the left side together I get : (5/3)y = 623.

Dividing both sides by (5/3) I get 373.8.

This means that x (the number of girls) is 249.2.

2) For the second attempt, I developed the equation: 2x+3x=623. Here 2x is the number of girls, and 3x is the number of boys.

Adding the left side I get: 5x = 623.

Dividing both sides by 5 I get: x = 124.6.

This means that the number of girls (2x) is 249.2. Just like in the first method.

But since you can't have a 1/5 of a girl, the answer must be 249 girls.

I solved this problem twice using different sets of equations to make sure I was right, and here is what I found:

1) I used the equations x + y = 623, and x = (2/3)y. In these equations, x is the number of girls and y is the number of boys.

Substituting the second equation into the first, I get (2/3)y + y = 623.

Adding the left side together I get : (5/3)y = 623.

Dividing both sides by (5/3) I get 373.8.

This means that x (the number of girls) is 249.2.

2) For the second attempt, I developed the equation: 2x+3x=623. Here 2x is the number of girls, and 3x is the number of boys.

Adding the left side I get: 5x = 623.

Dividing both sides by 5 I get: x = 124.6.

This means that the number of girls (2x) is 249.2. Just like in the first method.

But since you can't have a 1/5 of a girl, the answer must be 249 girls.

Nov 19, 2009 | Office Equipment & Supplies

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