PLEASE HHELP!!! Bring the fraction: b/7a^2c to a denominator of 35a^3c^3 a/a-4 to a denominator of

Question

PLEASE HHELP!!! Bring the fraction:

b/7a^2c to a denominator of 35a^3c^3

a/a-4 to a denominator of 16-a^2

in progress 0
Acacia 4 years 2021-07-22T03:06:02+00:00 1 Answers 23 views 0

Answers ( )

  1. Vodka
    0
    2021-07-22T03:07:53+00:00
    Reply

    Answer:

    \frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}

    \frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}

    Step-by-step explanation:

    Given

    \frac{b}{7a^2c}

    Express the denominator as 35a^3c^3

    To do this, we divide35a^3c^3 by the denominator

    \frac{35a^3c^3}{7a^2c} = 5ac^2

    So, the required fraction is:

    \frac{b}{7a^2c} * \frac{5ac^2}{5ac^2}

    \frac{5abc^2}{35a^3c^3}

    Hence:

    \frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}

    Given

    \frac{a}{a - 4}

    Express the denominator as 16 - a^2

    Multiply the fraction a+4/a+4

    So, we have:

    \frac{a}{a - 4} * \frac{(a + 4)}{(a + 4)}

    Apply difference of two squares to the denominator

    \frac{a^2 + 4a}{a^2 - 16}

    Take the additive inverse of the numerator and denominator

    \frac{-(a^2 + 4a)}{-(a^2 - 16)}

    \frac{-a^2 - 4a}{16 -a^2}

    Hence:

    \frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )