Can you please help me understand how to solve this math question? 1. The following is a result of the simplification of which of the following:x'y3 38x*y?,4x*y3O All the answers are correct.2x3

Consider the expression x2y38x4y223. Simplify the expression x2y38x4y223 as follows:…

Answered: 1. The following is a result of the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Can you please help me understand how to solve this math question?

1. The following is a result of the simplification of which of the following:
x'y3 3
8x*y?,
4x*y3
O All the answers are correct.
2x3

Expert Solution

Step 1

Consider the expression ${(\frac{x^{2} y^{3}}{8 x^{4} y^{2}})}^{\frac{2}{3}}$ .

Simplify the expression ${(\frac{x^{2} y^{3}}{8 x^{4} y^{2}})}^{\frac{2}{3}}$ as follows:

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

hence, It is a correct option.

Step 2

Consider the expression ${(\frac{x^{3} y^{2}}{4 x^{4} y^{3}})}^{\frac{3}{2}}$ .

Simplify the expression ${(\frac{x^{3} y^{2}}{4 x^{4} y^{3}})}^{\frac{3}{2}}$ as follows:

${(\frac{x^{3} y^{2}}{4 x^{4} y^{3}})}^{\frac{3}{2}} = (\frac{{(x^{3})}^{\frac{3}{2}} {(y^{2})}^{\frac{3}{2}}}{{(4)}^{\frac{3}{2}} {(x^{4})}^{\frac{3}{2}} {(y^{2})}^{\frac{3}{2}}}) = (\frac{(x^{\frac{9}{2}}) (y^{3})}{{(2^{2})}^{\frac{3}{2}} {(x^{\frac{4}{2}})}^{3} (y^{3})}) = (\frac{(x^{\frac{9}{2}}) (y^{3})}{(8) (x^{6}) (y^{3})}) = \frac{x^{\frac{9}{2}}}{8 x^{6}}$

Step by step

Solved in 3 steps

Knowledge Booster

$Fractions$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions