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Evaluate the following, without the use of a calculator:

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For an insight into an index of 0, consider the following table: 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
index form$b^3$$b^2$$b^1$$b^0$
expanded form$b\\times b \\times b$$b \\times b$$b$[[0]]
\n

Notice each time the power decreases by $1$, the result is divided by $b$. Using this idea, fill in the rest of the table.

\n

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Each time you reduce the power by 1 you divide the result by the base, that is $b$. Following this pattern:

\\[b^{0}=\\frac{b}{b}=1\\]

\n

From this we conclude that any non-zero number to the power of 0 is 1.

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$x^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$(\\var{coeff1}r)^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$-t^0$ = [[0]]

\n

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Here $t$ is to the power of zero and so it evaluates to $1$. However, there is still a negative sign in front, so the answer must be $-1$. 

\n

\n

Note $(-t)^0$ is quite different to $-t^0$.

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$\\displaystyle\\left(\\frac{\\var{coeff2}y}{\\var{coeff3}x}\\right)^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$\\displaystyle\\frac{m^0}{z}$ = [[0]]

\n

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Note that the power of zero is only acting on the numerator. As such the numerator is equal to $1$ and the denominator remains untouched.

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$(r \\times s)^0$ = [[0]]

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Regardless of the result of the multiplication the power of zero will force it to evaluate to 1.

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$d\\times e^0$ = [[0]]

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The power of zero is only acting on the second number in the product. It is only this number that will become 1. 

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