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anonymous

  • 5 years ago

i have a couple of math problems to solve anyone care to help me one on one plz...i really appreciate it

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    \[h(x)=\sqrt{4x+8}\] 4x+8≥0 4x≥-8 x≥-2 D:(x I x≥-2, x E R)

  3. anonymous
    • 5 years ago
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    I REALLY THANK U FOR HELPING ME OUT

  4. anonymous
    • 5 years ago
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    you're welcome

  5. anonymous
    • 5 years ago
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    I GET CONFUSED BECAUSE I HAVENT DONE THIS MATH IN A LONG WHILE..CAN U FINISH HELPING ME PLZ...ONLY IF U CAN

  6. anonymous
    • 5 years ago
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    ok number 11 a) f(x) - g(x) + h(x) just plug the functions f(x), g(x), and h(x) you'll get the result = -2x^2 + 27x -4

  7. anonymous
    • 5 years ago
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    b) g(x)/h(x) again, just plug g(x) and h(x) (10x^2 - 15x)/5x = 2x-3

  8. anonymous
    • 5 years ago
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    im confused on number 11a

  9. anonymous
    • 5 years ago
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    i inserted f g h

  10. anonymous
    • 5 years ago
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    but i got: \[f(23x-4) - G(85) + h(5X)\]

  11. anonymous
    • 5 years ago
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    f(x) - g(x) + h(x) = (8x^2 +7x-4) - (10x^2 -15x) + (5x) 8x^2 +7x-4 - 10x^2 + 15x + 5x

  12. anonymous
    • 5 years ago
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    where did you get that? you just need to substitute f(x) with 8x^2 +7x-4 substitute g(x) with 10x^2 -15x and substitute h(x) with 5x

  13. anonymous
    • 5 years ago
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    i got confused ...now i see it...ughhhhh *sign*

  14. anonymous
    • 5 years ago
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  15. anonymous
    • 5 years ago
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    number 12 is about function composition g(f(3)) first, find f(3) f(x) = 8x^2 +7x-4 so f(3) = 8(3)^2 + 7(3) -4 f(3) = 8(9) + 21 -4 f(3) = 72+21-4 f(3) = 89 therefore g(f(3)) = g(89) g(x) = 10x^2 -15x g(89) = 10(89)^2 - 15(89)

  16. anonymous
    • 5 years ago
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    number 12 is confusing "O

  17. anonymous
    • 5 years ago
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    13) because M is the midpoint, AM=MB 2x^2 - 10x + 15 = x^2 - 2x + 3 x^2 - 8x + 12 = 0 (x-6)(x-2) x = 6 or x=2

  18. anonymous
    • 5 years ago
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    this is an article from wikipedia , related to number 12 http://en.wikipedia.org/wiki/Function_composition

  19. anonymous
    • 5 years ago
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    thanks

  20. anonymous
    • 5 years ago
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  21. anonymous
    • 5 years ago
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    14) AC is an angle bisector, therefore angle BAD = angle CAD x^2 = 5x + 24 x^2 - 5x - 24=0 (x-8)(x+3) x=8 or x=-3

  22. anonymous
    • 5 years ago
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    15) BAD is a right angle 4x+8+7x+5 = 90 degree 11x + 13 = 90 11x = 77 x=7

  23. anonymous
    • 5 years ago
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    16) 8x+12 + 2x+8 = 180 degree 10x + 20 = 180 10x = 160 x= 16

  24. anonymous
    • 5 years ago
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  25. anonymous
    • 5 years ago
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    17) x^2 = 6x-8

  26. anonymous
    • 5 years ago
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    for number 12 how did we get 8 if the problem is f(x) = 3x -4 and g(x) = x to the 2nd -5 you dont have to redo the problem just need to know how 8 got there

  27. anonymous
    • 5 years ago
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    oh, sorry, i did it wrong, i didnt see the f(x) and g(x) in number 11 and 12 are different, i'll do it again

  28. anonymous
    • 5 years ago
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    ok thanks

  29. anonymous
    • 5 years ago
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    f(x) = 3x -4 and g(x) = x^2 - 5 f(3) = 3(3) - 4 = 9-4 = 5 because f(3) = 5, g(f(3)) = g(5) g(5) = (5)^2 - 5 = 25-5 = 20

  30. anonymous
    • 5 years ago
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    thank you very much for correcting it ")

  31. anonymous
    • 5 years ago
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  32. anonymous
    • 5 years ago
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    19, because it's an iscoceles triangle, angle C must be the same as angle A. a triangle has an angle of 180 degree, so : A + B + C = 180 degree 2x^2 + x^2 + 2x^2 = 180

  33. anonymous
    • 5 years ago
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  34. anonymous
    • 5 years ago
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    god ur good at this

  35. anonymous
    • 5 years ago
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    can i borrow ur brains for my finals...lol ")

  36. anonymous
    • 5 years ago
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    this is for number 20

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  37. anonymous
    • 5 years ago
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    lol, not really good, actually, this is kinda like SAT problems

  38. anonymous
    • 5 years ago
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    damn then i am dumb "(

  39. anonymous
    • 5 years ago
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  40. anonymous
    • 5 years ago
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    you just need to keep practicing, you'll be good at this. sorry i can't continue this, i gotta go. bye

  41. anonymous
    • 5 years ago
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    ok thank u very much for all ur help

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