# Am I correct with these problems?

• Sep 8th 2009, 03:58 PM
Kevin19
Am I correct with these problems?
1, (-4)3

(-4).(-4). (-4).(-4)

2, -52

-5 . -5

3, (2/3)2

(2/3) . (2/3)

4,
(2/3)2

(2/3) . (2/3)

5,
(-3)4

(-3) . (-3) . (-3) . (-3)

So do my answers to this questions seem correct?

• Sep 8th 2009, 04:08 PM
pickslides
No, Yes, Yes, Yes & Yes

$
(-4)^3 = -4\times -4\times -4
$
• Sep 8th 2009, 05:51 PM
Kevin19
Quote:

Originally Posted by pickslides
No, Yes, Yes, Yes & Yes

$
(-4)^3 = -4\times -4\times -4
$

Ok,

How do I solve questions 3 and 4?

Thank you!
• Sep 8th 2009, 05:55 PM
Prove It
Quote:

Originally Posted by Kevin19
Ok,

How do I solve questions 3 and 4?

Thank you!

First, you won't be "solving". To "solve" means to find the value of a pronumeral.

You will be "simplifying".

Anyway...

1. $(-4)^3 = (-4)\cdot(-4)\cdot(-4)$

$= -64$.

2. $(-5)^2 = (-5)\cdot(-5)$

$= 25$

3. $\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2}$

$= \frac{4}{9}$

4. It appears that Q. 4 is the same as Q. 3.

5. $(-3)^4 = (-3)\cdot(-3)\cdot(-3)\cdot(-3)$

$= 81$.
• Sep 8th 2009, 06:01 PM
pickslides
They are the same question

$
\left(\frac{2}{3}\right)^2 = \frac{2}{3} \times \frac{2}{3} = \frac{2\times 2 }{3 \times 3} = \frac{4}{9}
$