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

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00000-0000=33333 use the numbers one to nine once to get the equation 33333

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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

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I can solve this there are 2 answers
1 is 41268- 7935

or 21286-7953.

good luck

Posted on Jan 04, 2009

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Posted on Jan 02, 2017

  • 84 Answers

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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Remote control code number


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.

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How do you write 11.2987 in words


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."

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What's a number equivalent to 100 using only the number nine


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

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Explain what it means for a system of linear equations to have no solutions, one solution, two solutions, and infinite solutions.


  1. No solutions: The system is incoherent, incompatible Example: 2x+3y=8 and 2x+3y= 15. The two lines are parallel and distinct.
  2. 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.
  3. 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.
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4x - 2x + 1 = 5 + x - 7


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.

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1 Answer

What are the equations and its kinds?


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
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1 Answer

I cannot get the answer for the 33333 problem. How do you get it


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

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1 Answer

In complex number mode, how to get out?


"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
4591a1b3e15019ae25ad5813d5744ca9.png
If coefficients are complex you should expect some complex roots. Right?
If the coefficients are REAL then depending on the discriminant
80efae5ccea2ac0702e1a9c3b25afd2b.png
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.

d64abc4.jpg










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2 Answers

If there are 623 students and the ratio of girls is 2 to 3 boys. how many girls are there?


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.

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