Am I correct with these problems?

Printable View

September 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?
September 8th 2009, 04:08 PM
pickslides

No, Yes, Yes, Yes & Yes

$<br /> (-4)^3 = -4\times -4\times -4<br />$
September 8th 2009, 05:51 PM
Kevin19

Quote:

Originally Posted by pickslides

No, Yes, Yes, Yes & Yes

$<br /> (-4)^3 = -4\times -4\times -4<br />$

Ok,

How do I solve questions 3 and 4?

Thank you!
September 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. $(-3)^4 = (-3)\cdot(-3)\cdot(-3)\cdot(-3)$

$= 81$ .
September 8th 2009, 06:01 PM
pickslides

They are the same question

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

All times are GMT -8. The time now is 06:40 AM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.