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प्रश्न
The value of [1–2 + 2–2 + 3–2 ] × 62 is ______.
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उत्तर
The value of [1–2 + 2–2 + 3–2 ] × 62 is 49.
Explanation:
Using law of exponents,
`a^-m = 1/a^m` ...[∵ a is non-zero integer]
∴ `[1^-2 + 2^-2 + 3^-2] xx 6^2 = [1/1^2 + 1/2^2 + 1/3^2] xx 6^2`
= `[1 + 1/4 + 1/9] xx 6^2`
= `((36 + 9 + 4)/36) xx 6^2`
= `(49/36) xx 6^2`
= `(7/6)^2 xx 6^2`
= (7)2 × 6–2 × 62
= (7)2 × 62 – 2
= (7)2 × 60 ...[a0 – 1]
= 49
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संबंधित प्रश्न
3–2 can be written as ______.
a3 × a–10 = ______.
By multiplying `(5/3)^4` by ______ we get 54.
Simplify:
`(49 xx z^-3)/(7^-3 xx 10 xx z^-5) (z ≠ 0)`
Find the value of x–3 if x = (100)1 – 4 ÷ (100)0.
Find x so that `(2/9)^3 xx (2/9)^-6 = (2/9)^(2x - 1)`
Find a single machine that will do the same job as the given hook-up.
a (× 23) machine followed by (× 2–2) machine.
For hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.
Find x.
`((-6)/7)^(x - 7) = 1`
Simplify:
`(1/5)^45 xx (1/5)^-60 - (1/5)^(+28) xx (1/5)^-43`
