anonymous
  • anonymous
man i need help!!!!!!!!!!!!!!!!!!!!!! help me with mi math
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AccessDenied
  • AccessDenied
What exactly do you need help with?
hoblos
  • hoblos
question??
anonymous
  • anonymous
Divide (x2 - 9) by (x + 3).

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hoblos
  • hoblos
x2 - 9 = (x-3)(x+3)
hoblos
  • hoblos
so (x-3)(x+3)/(x + 3) = x-3
AccessDenied
  • AccessDenied
We could use a few methods here. Synthetic division (the divisor polynomial is degree 1 and has leading coefficient 1), long division, or just factoring the dividend polynomial and trying to remove a common factor that way.
anonymous
  • anonymous
Find the product of 3c and (4c - 5).
AccessDenied
  • AccessDenied
The product is (3c)(4c - 5) we could just use distributive property here, taking 3c to 4c and -5 for the answer.
anonymous
  • anonymous
so whats the answer to it i still dont understand
anonymous
  • anonymous
nevermind i got it thanks for helping me
AccessDenied
  • AccessDenied
Distributive Property is... a(b + c) = ab + ac so, in this case (3c)(4c - 5) = (3c)(4c) + (3c)(-5) Then, we can reorder and simplify = (3)(4) (c*c) + (3)(-5) c = 12c^2 - 15c okay, no problem. :)
anonymous
  • anonymous
Multiply (2x + 5)(3x - 4). i know its easy but your just so nice
anonymous
  • anonymous
Divide and simplify: x3 x2 + 2
AccessDenied
  • AccessDenied
The usual technique is "FOIL" but I prefer using distribution twice FOIL: First Outer Inner Last (2x + 5)(3x - 4) (2x)(3x) + (2x)(-4) + (5)(3x) + 5(-4) first outer inner last Distributive: (2x + 5)(3x - 4) treating (3x - 4) like the "a" in that distributive property (2x + 5)a 2x(a) + 5(a) (2x)(3x - 4) + 5(3x - 4) (2x)(3x) + (2x)(-4) + (5)(3x) + 5(-4) Both ways work... from there, we can simplify by multiplying together the coefficients and adding like terms... 6x^2 - 8x + 15x - 20 6x^2 + 7x - 20
AccessDenied
  • AccessDenied
Is it \[ \frac{x^3}{x^2 + 2} \] ?
anonymous
  • anonymous
yes bt i got another question 4exponets9 over4 expontents3. n since where here can i get to kno my tutors name?
AccessDenied
  • AccessDenied
there aren't any common factors between x^3 and x^2 + 2, so we end up with a remainder if we divide. I'm not sure if it still wants us to divide it As for next problem, we could just expand it out and cross out the common terms between the numerator and denominator (since 4/4 = 1, multiplying by 1 will not change the number) \[ \frac{4^9}{4^3} = \frac{\cancel{4\times4\times4}\times4\times4\times4\times4\times4\times4}{\cancel{4\times 4\times 4}} \] We would have 6 fours multiplied together as the remaining numbers
anonymous
  • anonymous
whats your name?
AccessDenied
  • AccessDenied
sorry, I prefer to go by this alias here. I do not like to give out my personal information on-line. :P

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