Question

can you pls tell me why there is 1​

Answer 1

$\huge\bold{{\pink{Q}}{\blue{U}}{\green{E}}{\red{S}}{\purple{T}}{\orange{I}}{\pink{O}}{\blue{N}}{\green{❥}}}$

$\blue{({\red{5^0}}+6^{-1})\times3^2}$

$\blue{=({\red{1}}+\frac{1}{6} )\times9}$

$\blue{=\frac{6+1}{6} \times9}$

$\blue{=\frac{7}{6} \times9}$

$\blue{=\frac{7}{2} \times3}$

$\blue{=\frac{21}{3}}$

$\blue{=10.5}$

Why $\red{5^0}$ is $\red{1}$?

$\huge\bold{{\pink{T}}{\blue{O}}{\green{  }}{\red{F}}{\purple{I}}{\orange{N}}{\pink{D}}{\green{❥}}}$

Why $\red{5^0}$ is $\red{1}$?

$\huge\bold{{\pink{S}}{\blue{O}}{\green{L}}{\red{U}}{\purple{T}}{\orange{I}}{\pink{O}}{\blue{N}}{\green{❥}}}$

Let us take a number $\red{a}$.

$\large\blue{\boxed{\pink{{a}^{0}=1}}}$

Here, we have to find why $\red{5^0=1}$.

We can take $\red{5^0}$ as $\red{5^{(1-1)}}$.

We know that:

$\green{\boxed{\purple{{\frac{a^b}{a^c}} =a^{(b-c)}}}}$

Here, we have to take $\red{a=5}$, $\red{b=1}$ and $\red{c=1}$.

So,

$\red{\frac{5^1}{5^1} =5^{(1-1)}=5^0}$  $......(i)$

But, $\red{5^1=5}$

And, $\red{\frac{5^1}{5^1} =\frac{5}{5} =1}$  $......(ii)$

From, $(i)$ and $(ii)$, we can say:

$\red{5^0=1}$

$\huge\bold{{\pink{H}}{\blue{E}}{\green{N}}{\red{C}}{\purple{E}}{\green{❥}}}$

$\red{5^0}$ is equal to $\red{1}$.

$\huge\bold{{\pink{T}}{\blue{H}}{\green{E}}{\red{R}}{\purple{E}}{\orange{F}}{\pink{O}}{\blue{R}}{\red{E}}{\green{❥}}}$

Since $\red{5^0=1}$, $\red{1}$ is used there.

$\huge\bold{{\pink{D}}{\blue{O}}{\green{N}}{\red{E}}{\purple{࿐}}}$

$\bold\red{<strong>------------------</strong>}$

$\huge\bold{{\pink{K}}{\blue{N}}{\green{O}}{\red{W}}{\purple{  }}{\orange{M}}{\pink{O}}{\blue{R}}{\red{E}}{\green{❥}}}$

$\bold\orange{Exponential  laws-}$

For the integers $a$, $b$, $m$ and $n$,

• $\blue{\boxed{\pink{{a^m}\times{a^n}=a^{(m+n)}}}}$

• $\blue{\boxed{\pink{{a^m}\div{a^n}=a^{(m-n)}}}}$

• $\blue{\boxed{\pink{(a^{m})^n=a^{mn}}}}$

• $\blue{\boxed{\pink{(\frac{a}{b} )^m=\frac{a^m}{b^m}}}}$

• $\blue{\boxed{\pink{a^0=1}}}$

$\bold\red{\boxed{{\blue{✿}}{\pink{H}}{\blue{O}}{\green{P}}{\red{E}}{\purple{  }}{\orange{T}}{\pink{H}}{\blue{I}}{\green{S}}{\red{  }}{\purple{H}}{\orange{E}}{\pink{L}}{\blue{P}}{\green{S}}{\red{  }}{\purple{Y}}{\orange{O}}{\pink{U}}{\blue{✿}}}}$

Answer 2

Answer:

hi actually I am also a great fan of Vijay