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Simplify rac15u3v27st3 imes rac28s3t25u3v and state the third term of a geometric progression is 2/9.

Mathematics

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Answer

a

12s2vt\frac{12s^2v}{t}

Step-by-step solution

5 steps

Here are the solutions for the given problems:

Part a· 5 steps

Simplify 15u3v27st3×28s3t25u3v\frac{15u^3v^2}{7st^3} \times \frac{28s^3t^2}{5u^3v}

  1. Combine the numerators and denominators.

    15u3v2×28s3t27st3×5u3v\frac{15u^3v^2 \times 28s^3t^2}{7st^3 \times 5u^3v}

  2. Multiply the numerical coefficients and group like variables.

    (15×28)×u3×v2×s3×t2(7×5)×u3×v×s×t3\frac{(15 \times 28) \times u^3 \times v^2 \times s^3 \times t^2}{(7 \times 5) \times u^3 \times v \times s \times t^3}

  3. Simplify the numerical coefficients.

    42035=12\frac{420}{35} = 12

  4. Simplify the variables using the exponent rule xaxb=xa−b\frac{x^a}{x^b} = x^{a-b}.

    For uu: u3u3=u3−3=u0=1\frac{u^3}{u^3} = u^{3-3} = u^0 = 1
    For vv: v2v1=v2−1=v1=v\frac{v^2}{v^1} = v^{2-1} = v^1 = v
    For ss: s3s1=s3−1=s2\frac{s^3}{s^1} = s^{3-1} = s^2
    For tt: t2t3=t2−3=t−1=1t\frac{t^2}{t^3} = t^{2-3} = t^{-1} = \frac{1}{t}

  5. Combine all simplified parts.

    12×1×v×s2×1t=12s2vt12 \times 1 \times v \times s^2 \times \frac{1}{t} = \frac{12s^2v}{t}

    The simplified expression is 12s2vt\boxed{\frac{12s^2v}{t}}.

Part b

In a geometric progression, the third term is 29\frac{2}{9}.

Let the first term of the geometric progression be aa and the common ratio be rr.
The formula for the nn-th term of a geometric progression is Tn=arn−1T_n = ar^{n-1}.

For the third term (n=3n=3), we have:

T3=ar3−1=ar2T_3 = ar^{3-1} = ar^2

Given that the third term is 29\frac{2}{9}:

ar2=29ar^2 = \frac{2}{9}

This equation relates the first term aa and the common ratio rr of the geometric progression.

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