Question

If a=3 and b=-2 find the values of: 1)a^a+b^b 2)a^b+b^a answer:1) 27 1/4 2) -7 8/9

Answer 1

Explanation :-

Given :

• a = 3
• b = -2

To find :

• a^a + b^b

Solution :

Putting the values of a & b, we get -

= 3³ + 2-²

= 27 + 1/2²

= 27 + 1/4

= (108 + 1) / 4

= 109/4

= 27 ¼

Hence, the value of a^a + b^b is 27 ¼.

In the next case,

Given :

• a = 3
• b = -2

To find :

• a^b + b^a

Solution :

Putting the value of a & b, we get -

= 3-² + (-2)³

= 1/3² + (-8)

= 1/9 + (-8)

= 1/9 - 8

= (1 - 72) / 9

= -71/9

= -7 8/9

Hence, the value of a^b + b^a is -7 8/9..

Answer 2

$\huge \underline{ \purple{ \boxed{ \bf \orange{Answer:-}}}}$

$\blue{\bold{\underline{\underline{Given:-}}}}$

• a = 3

• b = –2

$\blue{\bold{\underline{\underline{To \: Find:-}}}}$

$= {a}^{a}+{b}^{b}$

$= {a}^{b}+{b}^{a}$

$\small \underline{ \blue{ \boxed{ \bf \green{In \: (1):-}}}}$

$= {3}^{3} + { - 2}^{ - 2}$

$= 27 + (\frac{1}{2} )^{2}$

$= 27 + \frac{1}{4}$

$= \frac{109}{4}$

$= 27 \frac{1}{4}$

$\small \underline{ \blue{ \boxed{ \bf \green{In \: (2):-}}}}$

$= {3}^{ - 2} + ({ - 2}) ^{3}$

$= \frac{1}{ {3}^{2} } - 8$

$= \frac{1}{9} - 8$

$= \frac{1 - 72}{9}$

$= \frac{ - 71}{9}$

$= - 7 \frac{8}{9}$

$\bf\blue{Hope\ it\ helps.}$

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