Recent content by Rujaxso

  1. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    https://en.wikipedia.org/wiki/Freshman%27s_dream ****(french) its official
  2. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    I guess I wanted to apply A(B+C) = AB + AC to (B+C)^2 and make B^2 + C^2 which is incorrect I see.
  3. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    Btw Found this under Algebra 1 > polynomials > special products on khan academy, ...I will go over that section
  4. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    Okay so instead of squaring each term I need to think about the quantity as a whole in the parenthesis as being a base for the exponent?
  5. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    Thanks Mark. Nope I haven't, I was just going off of what I know about distributing.
  6. R

    MHB Simplifying f(a+h)=-5(a+h)^2+2(a+h)-1

    Not sure where the 2ah is coming from in the middle step. f(a+h)=-5(a+h)^2+2(a+h)-1 = -5(a^2 + 2ah +h^2) +2a +2h -1 = -5a^2 - 10ah -5h^2 +2a +2h -1 Please enlighten me
  7. R

    MHB How Can I Solve for x of y=cx+dx Using the Textbook Solution?

    Should I be more impressed by the use of the word proviso? =) btw you guys and this textbook is the closest thing to a teacher I have right now. Prepare for an onslaught and flurry of annoying questions in the future.
  8. R

    MHB How Can I Solve for x of y=cx+dx Using the Textbook Solution?

    Thanks, got you the first time, I was just trying to confirm the rule/move in English by my 2nd post.
  9. R

    MHB How Can I Solve for x of y=cx+dx Using the Textbook Solution?

    Thanks Skeeter, basically reverse distribution of multiplication over addition I guess.
  10. R

    MHB How Can I Solve for x of y=cx+dx Using the Textbook Solution?

    $$y = cx + dx$$ solve for x textbook solution:$$x = \frac{y}{c+d}$$my steps: $$\frac{y}{c + d}=\frac{cx + dx}{c + d}$$ here I divide to isolate x $$\frac{y}{c + d}= x + x$$ simplifed $$\frac{y}{c + d}=2x$$ adding the two x'sWhat am I missing here?
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