MHB Evaluate 1/(-5x2y3) Answer: Is it Correct?

[mathdrama]

I was to evaluate this: (-5x^2y^3)^-2

This is what I came up with:= 1/(-5x2y3) (-5x2y3)
=1/25x4y6

Is this correct?

 

[Opalg]

I was to evaluate this: (-5x^2y^3)^-2

This is what I came up with:= 1/(-5x2y3) (-5x2y3)
=1/25x4y6

Is this correct?

Don't forget to use parentheses and exponent signs. But if 1/25x4y6 is supposed to mean 1/(25x^4y^6) then yes, you are correct.

 

[sweer6]

Assuming this is the same question as yours, below is what i got
(-5x^((2y)^(3)))^(-2)

Expand the exponent (3) to the expression.
(-5x^((2^(3)y^(3))))^(-2)

Cubing a number is the same as multiplying the number by itself 3 times (2*2*2). In this case, 2 cubed is 8.
(-5x^((8y^(3))))^(-2)

Remove the parentheses around the expression 8y^(3).
(-5x^(8y^(3)))^(-2)

Squaring an expression is the same as multiplying the expression by itself 2 times.
(1)/((-5x^(8y^(3)))(-5x^(8y^(3))))

Multiply -5x^(8y^(3)) by -5x^(8y^(3)) to get -5.
(1)/(-5)

Move the minus sign from the denominator to the front of the expression.
-((1)/(5))

Multiply -1 by the (1)/(5) inside the parentheses.
-(1)/(5)

 

[MarkFL]

Assuming this is the same question as yours, below is what i got
(-5x^((2y)^(3)))^(-2)

Expand the exponent (3) to the expression.
(-5x^((2^(3)y^(3))))^(-2)

Cubing a number is the same as multiplying the number by itself 3 times (2*2*2). In this case, 2 cubed is 8.
(-5x^((8y^(3))))^(-2)

Remove the parentheses around the expression 8y^(3).
(-5x^(8y^(3)))^(-2)

Squaring an expression is the same as multiplying the expression by itself 2 times.
(1)/((-5x^(8y^(3)))(-5x^(8y^(3))))

Multiply -5x^(8y^(3)) by -5x^(8y^(3)) to get -5.
(1)/(-5)

Move the minus sign from the denominator to the front of the expression.
-((1)/(5))

Multiply -1 by the (1)/(5) inside the parentheses.
-(1)/(5)

I believe you have incorrectly interpreted the problem. In any case, your subsequent algebra is flawed, where you state:

Multiply -5x^(8y^(3)) by -5x^(8y^(3)) to get -5.

 

[Deveno]

I was to evaluate this: (-5x^2y^3)^-2

This is what I came up with:= 1/(-5x2y3) (-5x2y3)
=1/25x4y6

Is this correct?

It really helps if you use Latex.

We start with:

$(-5x^2y^3)^{-2}$.

Using the rule:

$(ab)^c = (a^c)(b^c)$

with: $a = -5x^2$ and $b = 2y^3$ and $c = -2$, we get:

$(-5x^2y^3)^{-2} = (-5x^2)^{-2}(y^3)^{-2}$

and applying it again we have $(-5x^2)^{-2} = (-5)^{-2}(x^2)^{-2}$, so the whole thing is:

$= (-5)^{-2}(x^2)^{-2}(y^3)^{-2}$.

Now using the rule:

$(a^b)^c = a^{bc}$ we get:

$ = (-5)^{-2}(x^{-4})(y^{-6})$.

Finally, using the rule:

$a^{-b} = \dfrac{1}{a^b}$ we have our expression is:

$= \dfrac{1}{5^2}\cdot\dfrac{1}{x^4}\cdot\dfrac{1}{y^6} = \dfrac{1}{25x^4y^6}$

As you can see, this is much easier on the eyes than what you posted, although our answers agree.

It's hard to say whether or not this really represents a "simplification", as nothing really "cancelled out".

 

[kmhock]

I was to evaluate this: (-5x^2y^3)^-2

This is what I came up with:= 1/(-5x2y3) (-5x2y3)
=1/25x4y6

Is this correct?

There are already a few replies to this. Just thought I would add a slightly different one. Here is how I would do it:

$$
(-5x^2y^3)^{-2} = \frac{1}{(-5x^2y^3)^2}
= \frac{1}{5^2(x^2)^2(y^3)^2}
= \frac{1}{25x^4y^6}
$$