Math, asked by wade114, 5 hours ago

Prove that 0/0=0^0.
 \frac{0}{0}  =  {0}^{0}

Answers

Answered by Yugmaurya
0

Step-by-step explanation:

When TeX is in display-style math mode and the amsmath package is loaded, \frac and \dfrac are equivalent, and the material in the numerator and denominator portions of \frac will be typeset in "text-style math" mode by default. This entails, among other things, that summation and integral symbols will be typeset in text style, as will any fractional expressions. The resulting tight, or "cramped", look is probably what you're looking to avoid.

To override the default setting, i.e., to force TeX to render the material in both the numerator and denominator terms in display-style math mode, one needs to insert an explicit \displaystyle instruction at the start of both the numerator and denominator material. For more on this subject, see the posting Show inline math as if it were display math.

If you have a lot of these expressions, it's worthwhile to create a macro called, say, \ddfrac (short for "double displaystyle frac", I suppose):

\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}

enter image description here

\documentclass[12pt]{article}

\usepackage{amsmath}

\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}

\begin{document}

\begin{align*}

\frac{\sum_{i=1}^\infty \frac{1}{i^2}}{ \int_{-\infty}^\infty

\frac{1}{\sqrt{2\pi\sigma^2}} \exp\bigl(-\frac{(x-\mu)^2}{2\sigma^2}\bigr)

\,\mathrm{d}x} &= \frac{\pi^2}{6}

\qquad\text{too cramped?}\\[2ex]

\ddfrac{\sum_{i=1}^\infty \frac{1}{i^2}}{\int_{-\infty}^\infty

\frac{1}{\sqrt{2\pi\sigma^2}} \exp\biggl(-\frac{(x-\mu)^2}{2\sigma^2}\biggr)

\,\mathrm{d}x} &= \frac{\pi^2}{6}

\qquad\text{about right?}

\end{align*}

\end{

Answered by pmrmdchest
0

Answer:

No ,0/0 0^0

Step-by-step explanation:

0/0= can't divide by 0

0^0= 1 (any number power 0 is 1)

so, 0/0 ≠ 0^0

hope it helps you

please mark as brainliest

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