wood

05-19-2009, 10:02 PM

simplify using the laws of exponents

x^2 (square root of x + square root of y^3)

Y^3

I'm not sure where to start this question

x^2 (square root of x + square root of y^3)

Y^3

I'm not sure where to start this question

View Full Version : simplify x^2/Y3 (square root of x + square root of y^3)

wood

05-19-2009, 10:02 PM

simplify using the laws of exponents

x^2 (square root of x + square root of y^3)

Y^3

I'm not sure where to start this question

x^2 (square root of x + square root of y^3)

Y^3

I'm not sure where to start this question

BigGlenntheHeavy

05-19-2009, 10:13 PM

Is \ this \ what \ you \ want \ to \ simplify: \ \frac{x^{2}[x^{1/2}+y^{3/2}]}{y^{3}}?

wood

05-19-2009, 11:02 PM

no

X^2 (square root of x + square root of y^3)

y^3

X^2 (square root of x + square root of y^3)

y^3

BigGlenntheHeavy

05-20-2009, 12:00 AM

How \ about \ this: \ \frac{\sqrt x^{2}}{\sqrt y^{3}} \ ?

wood

05-20-2009, 12:53 AM

no

x^2

y^3

Is the first part of the expression, it is all multiplied by ( square root of x + square root of y^3)

Is there a square root sign on the computer?

x^2

y^3

Is the first part of the expression, it is all multiplied by ( square root of x + square root of y^3)

Is there a square root sign on the computer?

BigGlenntheHeavy

05-20-2009, 01:07 AM

Hey \ Wood, \ x^{1/2} = \sqrt x.

wood

05-20-2009, 02:13 AM

ok so,

x^2 all multiplied by( x^1^/^2 + (y^3)^1^/^2

y^3

x^2 all multiplied by( x^1^/^2 + (y^3)^1^/^2

y^3

BigGlenntheHeavy

05-20-2009, 03:17 AM

See my first answer, there is nothing to simplify except combining terms in the numerator.

Subhotosh Khan

05-20-2009, 08:17 AM

If you are using standard MS keyboard then to type square-root sign (?)- type <ALT>251 on the number key-pad with "numlock" key activated.

wood

05-20-2009, 10:23 AM

so the Y^3 that is the denominator in the question is only under X^2 how do I get it to be the denominator for the entire question.

Subhotosh Khan

05-20-2009, 11:02 AM

so the Y^3 that is the denominator in the question is only under X^2 how do I get it to be the denominator for the entire question.

Do you agree with following?

\frac{2}{3}\cdot [2 + 3] \,

=\frac{2}{3}\cdot \frac{[2 + 3]}{1} \,

= \, \frac{2\cdot [2 + 3] }{3\cdot 1}

= \, \frac{2\cdot [2 + 3] }{3}

Do you agree with following?

\frac{2}{3}\cdot [2 + 3] \,

=\frac{2}{3}\cdot \frac{[2 + 3]}{1} \,

= \, \frac{2\cdot [2 + 3] }{3\cdot 1}

= \, \frac{2\cdot [2 + 3] }{3}

wood

05-20-2009, 11:20 PM

yes i agree

Subhotosh Khan

05-21-2009, 07:46 AM

how do I get it to be the denominator for the entire question

Those are the same steps.....

Those are the same steps.....

wood

05-21-2009, 09:25 PM

so

x^2(x^1^/^2+y^3^/^2)

y^3

2/1*1/2=2/2=1

x + x^2y^3^/^2

Y^3

is this my final answer?

x^2(x^1^/^2+y^3^/^2)

y^3

2/1*1/2=2/2=1

x + x^2y^3^/^2

Y^3

is this my final answer?

Denis

05-22-2009, 12:56 AM

so the Y^3 that is the denominator in the question is only under X^2 how do I get it to be the denominator for the entire question.

Woody, you haven't been listening!

If you have a/b * c, the b is already the FULL denominator: same as ac / b

AND what are you doing here: x^2y^3^/^2 ?

Do you mean y^(3/2)?

If your teacher is trying to teach how to simplify, he/she should be shot at sunrise

for picking such a ridiculous expression :shock:

Woody, you haven't been listening!

If you have a/b * c, the b is already the FULL denominator: same as ac / b

AND what are you doing here: x^2y^3^/^2 ?

Do you mean y^(3/2)?

If your teacher is trying to teach how to simplify, he/she should be shot at sunrise

for picking such a ridiculous expression :shock:

Powered by vBulletin® Version 4.2.2 Copyright © 2016 vBulletin Solutions, Inc. All rights reserved.