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bahrom7893

  • 2 years ago

I guess I'm just being stupid but can someone help me solve these 3 linear equations??? I keep getting no solutions..

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  1. bahrom7893
    • 2 years ago
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    7=25a-5b+c -2=a-b+c 5=9a+3b+c

  2. bahrom7893
    • 2 years ago
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    Find the equation of the parabola y=ax2 (square)+bx+c that passes through the points (-5, 7) (-1,-2) and (3,5) find a, b, and c that was the original question

  3. satellite73
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=7%3D25a-5b%2Bc%2C+-2%3Da-b%2Bc%2C+5%3D9a%2B3b%2Bc

  4. satellite73
    • 2 years ago
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    @amistre64 has a nicer method for doing this ask him

  5. bahrom7893
    • 2 years ago
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    oh cool you can use wolf for this!

  6. bahrom7893
    • 2 years ago
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    i didn't know that

  7. amistre64
    • 2 years ago
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    given a set of points such that the x components create a set of elements: {\(x_1,x_2,...,x_n\)} and the set of y components create a set of elements: {\(y_1,y_2,...,y_n\)} we can create a polynomial by constructing it in this manner:\[y=c_0+c_1(x-x_1)+c_2(x-x_1)(x-x_2)+...+c_k(x-x_1)(x-x_2)...(x-x_k)\] :)

  8. amistre64
    • 2 years ago
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    (-5, 7) (-1,-2) and (3,5) x = {-5,-1,3} ; y={7,-2,5} y = c0 + c1(x+5) + c2(x+5)(x+1) fit the points into it and we can construct the coeffs one by one since the unknown zero out all but one of them 7 = c0 + c1(-5+5) + c2(-5+5)(-5+1) 7 = c0 -2 = 7 + c1(-1+5) + c2(-1+5)(-1+1) -9 = c1(4) -9/4 = c1 3 = 7 - 9(3+5)/4 + c2(3+5)(3+1) -4 = -18 + c2(8)(4) 14 = c2(32) 7/16 = c2 y = 7 - 9(x+5)/4 + 7(x+5)(x+1)/16 and expand if wanted to get: (7x^2 + 6x -33)/16 if i did the mathing right

  9. amistre64
    • 2 years ago
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    now i gotta dbl chk it all lol

  10. amistre64
    • 2 years ago
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    3,5 ; not 3,3 5 = 7 - 9(3+5)/4 + c2(3+5)(3+1) -2 = -18 + c2(8)(4) 16 = c2(32) 1/2 = c2 y = 7 - 9(x+5)/4 + (x+5)(x+1)/2 = (2x^2+3x-7)/4 which now matches the wolf ;)

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