1,(-4)3

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

2,-52

-5.-5

3,(2/3)2

(2/3).(2/3)

(2/3)2

4,

(2/3).(2/3)

(-3)4

5,

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

So do my answers to this questions seem correct?

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- Sep 8th 2009, 03:58 PMKevin19Am I correct with these problems?
**1,**(-4)3

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

**2,**-52

-5**.**-5

**3,**(2/3)2

(2/3)**.**(2/3)

(2/3)2

4,

(2/3)**.**(2/3)

(-3)4

5,

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

So do my answers to this questions seem correct?

- Sep 8th 2009, 04:08 PMpickslides
No, Yes, Yes, Yes & Yes

$\displaystyle

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

$ - Sep 8th 2009, 05:51 PMKevin19
- Sep 8th 2009, 05:55 PMProve It
First, you won't be "solving". To "solve" means to find the value of a pronumeral.

You will be "simplifying".

Anyway...

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

$\displaystyle = -64$.

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

$\displaystyle = 25$

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

$\displaystyle = \frac{4}{9}$

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

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

$\displaystyle = 81$. - Sep 8th 2009, 06:01 PMpickslides
They are the same question

$\displaystyle

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

$