Can someone help me with this?? If you can't see the picture don't bother to answer.

Mathematics

1 answer:

Georgia [21]2 years ago

7 0

J is correct. the scale is balanced, so x = 1. there are 9 1’s on the right, and 6 1’s on the left. there are 3 more blocks, and they would have to each be 1.

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$(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Solution:

Given that,

$(8a^{-3})^{\frac{-2}{3}$

We have to write in simplest form

Use the following law of exponent

$(a^m)^n = a^{mn}$

Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$