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Um... None of the square roots do anything for me, and I don't think 0 is the answer. Unless it's 10*0, and since 0 = infinite, and infinite is undefined, you could assume 0 to equal 130... I think...
[EDIT]Wait, can we use matrices?[/EDIT}
[EDIT]Wait, can we use matrices?[/EDIT}
nah these problems lose their novelty once people know how to do them lol...
modulus is basically working with remainders
in modulus 4 3+1=0 because 3/4 gives 3 remainder, 1/4 gives 1 remainder, and 4/4 gives no remainder
modulus is basically working with remainders
in modulus 4 3+1=0 because 3/4 gives 3 remainder, 1/4 gives 1 remainder, and 4/4 gives no remainder
another one? please?:)
and what is modulus?
and what is modulus?
tried to ask this question in game and got alot of verbal assaults about my absolute stupidity with this question, so i'll try it again here, 50k for the 1st 1 to post the answer... no verbal insults please... and dont answer if you heard me say the answer ingame... thanks...
what 2 numbers when added equal 10 and when multiplied equal 130?
what 2 numbers when added equal 10 and when multiplied equal 130?
I'm sorry, but that's impossible.
1*9 = 9
2*8 = 16
3*7 = 21
4*6 = 24
5*5 = 25
6*4 = 24
7*3 = 21
8*2 = 16
1*9 = 9
You can get -130 when using one negative number, but -130 != 130
1*9 = 9
2*8 = 16
3*7 = 21
4*6 = 24
5*5 = 25
6*4 = 24
7*3 = 21
8*2 = 16
1*9 = 9
You can get -130 when using one negative number, but -130 != 130
/me picks up his TI-92 and taps in solve(x+y=10 and x*y=130,{x,y}) and gets an answer of...
"FALSE"
So I have to agree with Sheean. The way you have this explained, it is an impossible equation. However...
In case you were just explaining very poorly and you meant that it could be...
x*x*y=130.. or x*x*y*y or x*y*y.. you get answers such as...
7.9357 and 2.0643... along with an infinite number of other possibilities depending on how many x and y you multiply.
But. From your words.. "What 2 numbers" Leaves it as you can only add or multiply once, because any more would be using more than 2 numbers.
So in conclusion. Sheean is right. Its impossible, and probably everyone who abused you was right as well.
"FALSE"
So I have to agree with Sheean. The way you have this explained, it is an impossible equation. However...
In case you were just explaining very poorly and you meant that it could be...
x*x*y=130.. or x*x*y*y or x*y*y.. you get answers such as...
7.9357 and 2.0643... along with an infinite number of other possibilities depending on how many x and y you multiply.
But. From your words.. "What 2 numbers" Leaves it as you can only add or multiply once, because any more would be using more than 2 numbers.
So in conclusion. Sheean is right. Its impossible, and probably everyone who abused you was right as well.
People may be assuming whole numbers.
I believe it's done in modulus 13 (if you don't know modulus, you failed 6th grade Algebra). Like 10*13 = 130, 10+13 = 10
Modulus?
We never learned that in school...
Edit: FM, even with non-integer numbers, it's still impossible. The highest number possible is half of the first (10) squared (25).
Edit: FM, even with non-integer numbers, it's still impossible. The highest number possible is half of the first (10) squared (25).
Errrr, I didn't fail algebra... I passed with all A's.
What is modulus?
What is modulus?
not modulus, and yes its possible. its not a whole number,
and its what 2 numbers added 2 each other and multiplied ot each other once, not repeatedly, so x*y and x+y.
and i didnt say it was a non-integer number, or even a real number.
and its what 2 numbers added 2 each other and multiplied ot each other once, not repeatedly, so x*y and x+y.
and i didnt say it was a non-integer number, or even a real number.
But the modulus solution works...
lets just say it isnt the answer i'm looking for... i wasnt thinking to mess with modulus math when i made the problem.
Okay... Lemme think about this one...
/me graphs y = 130/x to get all possible values to equal 130 when multiplied
/me graphs y = 10 - x to get all possible values to add to 10
/me intersects
0 intersections. Sorry, but using standard modulus, this problem does not seem to work.
/me graphs y = 130/x to get all possible values to equal 130 when multiplied
/me graphs y = 10 - x to get all possible values to add to 10
/me intersects
0 intersections. Sorry, but using standard modulus, this problem does not seem to work.
"and i didnt say it was a non-integer number, or even a real number."
there's the hint in my last post right there
there's the hint in my last post right there
No matrices, they use more than 2 numbers, unless you do an exponetial matrice (trigonometry) and then you get into n-space.
The answer is= 5cubed+5=130
5+5=10
Try it: (5*5*5)+5=130 Also 5+5=10
5 cubed counts still as a basic number of 5 since the cubed is not attached to the number otherwise following it making the 2nd number cubed as well since you wouldn;t take the sign.
5+5=10
Try it: (5*5*5)+5=130 Also 5+5=10
5 cubed counts still as a basic number of 5 since the cubed is not attached to the number otherwise following it making the 2nd number cubed as well since you wouldn;t take the sign.
No. The question's this:
x * y = 130
x + y = 130
5^3 + 5 = 130
5^3 * 5 = 625
You did:
x^3 + y = 130
x + y = 130
You see, you've boxed yourself in. If the cubed is attached, see above. If not, you've violated the problem by adding another operation (he said you can't use x*x*x*y anyhow).
x * y = 130
x + y = 130
5^3 + 5 = 130
5^3 * 5 = 625
You did:
x^3 + y = 130
x + y = 130
You see, you've boxed yourself in. If the cubed is attached, see above. If not, you've violated the problem by adding another operation (he said you can't use x*x*x*y anyhow).
THE ANSWER!!!
(5^x)+5
(5^(3))+5
(125)+5
5x+5
x=1
5+5=10
Then I am not violating the 2 number rule as setting x as a variable in the equation since the "real" number is 5x under both circumstances, just the x-value changes, thus, since the number is in fact 5x and the other is 5, the variable x counts as a compotent of 5x thus making it's value a mere compotent piece to the 5x nullifying it's standing as a third, illegal number.
The cubed doesn't violate the 2 numbers rule either since technically, any number with a "cube" attached to it is merely a number not yet solved in the overall equation, therefore, all cubes and sqaures are allowed under this problem's current rules.
(5^x)+5
(5^(3))+5
(125)+5
5x+5
x=1
5+5=10
Then I am not violating the 2 number rule as setting x as a variable in the equation since the "real" number is 5x under both circumstances, just the x-value changes, thus, since the number is in fact 5x and the other is 5, the variable x counts as a compotent of 5x thus making it's value a mere compotent piece to the 5x nullifying it's standing as a third, illegal number.
The cubed doesn't violate the 2 numbers rule either since technically, any number with a "cube" attached to it is merely a number not yet solved in the overall equation, therefore, all cubes and sqaures are allowed under this problem's current rules.