Simplify- a-4- b3- c5- d-7-a-7- b-3- c2- d-9A- a3 b6 c3 d2B- a3 b5 c3 d2C- a2 b6 c3 d2D- a3 b6 c3 d4

The correct option is A a^3 nbsp;b^6c^3d^2 Using the fact that nbsp;k^n/k^m nbsp;= k^n nbsp; m for k≠ 0, nbsp; a^ 4 × b^3 × c^5 × d^ 7/a^ 7 × b^ ...

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Byju's Answer

Standard VII

Mathematics

Quotient Law

Simplify: a-4...

Question

Simplify: $\frac{a^{- 4} \times b^{3} \times c^{5} \times d^{- 7}}{a^{- 7} \times b^{- 3} \times c^{2} \times d^{- 9}}$

$a^{3} b^{6} c^{3} d^{2}$

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$a^{3} b^{5} c^{3} d^{2}$

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$a^{2} b^{6} c^{3} d^{2}$

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$a^{3} b^{6} c^{3} d^{4}$

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Solution

The correct option is A
$a^{3} b^{6} c^{3} d^{2}$

Using the fact that $\frac{k^{n}}{k^{m}} = k^{n - m} f o r k \neq 0$ ,
$\frac{a^{- 4} \times b^{3} \times c^{5} \times d^{- 7}}{a^{- 7} \times b^{- 3} \times c^{2} \times d^{- 9}}$

$= a^{(- 4 + 7)} \times b^{(3 + 3)} \times c^{(5 - 2)} \times d^{(- 7 + 9)}$

$= a^{3} \times b^{6} \times c^{3} \times d^{2}$

$= a^{3} b^{6} c^{3} d^{2}$

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Similar questions

$Simplify: \frac{a^{- 4} \times b^{3} \times c^{5} \times d^{- 7}}{a^{- 7} \times b^{- 3} \times c^{2} \times d^{- 9}}$

Q. If

a + b + c + d = 0

, shew that

\frac{a^{5} + b^{5} + c^{5} + d^{5}}{4} = \frac{a^{3} + b^{3} + c^{3} + d^{3}}{3} \cdot \frac{a^{2} + b^{2} + c^{2} + d^{2}}{2}

Q. If

a + b + c = 0

, show that

6 (a^{5} + b^{5} + c^{5}) = 5 (a^{3} + b^{3} + c^{3}) (a^{2} + b^{2} + c^{2})

Q. Simplify each of the following by rationalizing the denominator

\frac{6 - 4 \sqrt{2}}{6 + 4 \sqrt{2}}

\frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}

\frac{1}{3 \sqrt{2} - 2 \sqrt{3}}

\frac{3 \sqrt{5} - \sqrt{7}}{3 \sqrt{3} + \sqrt{2}}

If
A = {1, 2, 3, 4}

B = {3, 4, 5, 6, 7}

C = {6, 7, 8, 9}

D = {7, 8, 9, 10}

$\begin{matrix} C o l u m n A & C o l u m n B 1. A \cup B & a . {3, 4, 6, 7} 2. A \cap B & b . \emptyset 3. (B \cap C) \cup D & c . {3, 4} 4. B \cup C \cup D & d . {3, 4, 5, 6, 7, 8, 9, 10} 5. (A \cap B) \cap (B \cup C) & e . {1, 2, 3, 4, 5, 6, 7} f . {6, 7, 8, 9, 10} g . {7} \end{matrix}$

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Simplify: a−4 × b3 × c5 × d−7a−7 × b−3 × c2 × d−9

The correct option is A a3b6c3d2 Using the fact that knkm=kn−m for k≠0, a−4 × b3 × c5 × d−7a−7 × b−3 × c2 × d−9 =a(−4+7)×b(3+3)×c(5−2)×d(−7+9) =a3×b6×c3×d2 =a3b6c3d2

Simplify: $\frac{a^{- 4} \times b^{3} \times c^{5} \times d^{- 7}}{a^{- 7} \times b^{- 3} \times c^{2} \times d^{- 9}}$