The value of (5power0)+(6power 2) =
Answers
Answer:
(5⁰ + 6⁻¹) × 3²
To solve this we must apply laws of exponents. First we will apply these two laws of exponents for the brackets.
a⁰ = 1
a^{-m} = \dfrac{1}{a^{m}}a
−m
=
a
m
1
⇒ (5⁰ + 6⁻¹) × 3²
⇒ (1+\dfrac{1}{6})\times 3^{2}(1+
6
1
)×3
2
Now we will find the LCM of the denominators in the fractions in the bracket. The LCM of 1 and 6 is 6 so using this value we will make the denominators equal.
⇒ (1+\dfrac{1}{6})\times 3^{2}(1+
6
1
)×3
2
⇒ (\dfrac{1\times 6 }{1\times 6 } +\dfrac{1}{6})\times 3^{2}(
1×6
1×6
+
6
1
)×3
2
⇒ (\dfrac{6 }{6 } +\dfrac{1}{6})\times 3^{2}(
6
6
+
6
1
)×3
2
Now add the denominators to obtain the answer of the brackets.
⇒ (\dfrac{6 }{6 } +\dfrac{1}{6})\times 3^{2}(
6
6
+
6
1
)×3
2
⇒ \frac{7 }{6 }\times 3^{2}
6
7
×3
2
Now we'll find the value of 3².
3² = 3×3 = 9
⇒ \dfrac{7 }{6 }\times 9
6
7
×9
⇒ \dfrac{7 }{6 \div 3 }\times 9\div 3
6÷3
7
×9÷3
⇒ \dfrac{7 }{2}\times 3
2
7
×3
⇒ \dfrac{21}{2}
2
21
\bf \therefore (5^{0} + 6^{-1}) \times 3^{2} = \frac{21}{2}∴(5
0
+6
−1
)×3
2
=
2
21
_____________________________________
Answer:
The value of 5⁰ + 6² is
= 1+36
= 37
Hope you understand bhai