Jeffes02

Sorry, but can you tell us the meaning and the formula of calculating these points? I can't find any translation on the web of what those abbreviations mean besides it is some kind of a special word for courts in India, any help and clearance will help us answer your question more easily, cheers.

Source: https://forum.wordreference.com/threads/obc-general-sc-st-pc-university.1806282/

Oh yes XD I messed up the numbers and my vocabulary

(Corrected :P)

Input:-1.5,2,-3,1/3 from least to greatest

Intepretation: List -1.5,2,-3, and 1/3 by ascending order:

By the rule that negative numbers are smaller than numbers: (Note that the higher the negative number is, the smaller it actually is.)

Answer:\(-3,-1.5,1/3,2\)

Fixed :D

Input: 2+2=?

Intepretation: \(2+2=?\)

Answer: \(2+2=5\) Because the Big Brother said so

(Just kidding, it is \(4\))

Input: 12/m=2/4+1x10^33!= x i know the answer!

Intepretation: \(\frac{12}{m}=\frac{2}{4}+1\cdot 10^{33}\)

Simplify: \(\frac{12}{m}=\frac{1}{2}+10^{33}\)

\(m=\frac{8}{666666666666666666666666666666667}\)

\(m≈1.2\cdot10^{-32}\)

\(\)

\(\)

Input: x|x|=2x+1

Intepretation: \(x\left|x\right|=2x+1\)

Rewrite into alternative form:

\(x\sqrt{x^2}=2x+1\)

Simplify the square:

\(x^2=2x+1\)

Move the terms to the left:

\(x^2-2x-1=0\)

This function doesn't look like that it is factorizable, applying formulaic solution:

For \(ax^2+bx+c=0,\space x = {-b \pm \sqrt{b^2-4ac} \over 2a}\),

Plug \(a=1,\space b=-2,\space c=-1\)

\(x=\frac{2\pm \sqrt{4+4}}{2}\)

\(=\frac{2\pm \sqrt{8}}{2}\)

\(=\frac{2\pm 2\sqrt{2}}{2}\)

\(=1±\sqrt2\)

\(x=1+\sqrt{2}\ or\ 1-\sqrt{2}\)

Input: What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

Intepretation: Solve for \(a\) in \(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6\)

Simplify:

\(\sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}}=6\)

We know that \(\sqrt{4+4\sqrt{1+a}}\) is \(\sqrt4=2\) times larger than \(\sqrt{1+\sqrt{1+a}}\)

Merge:

\(3\sqrt{1+\sqrt{1+a}}=6\)

Divide both sides by a factor of 3:

\(\sqrt{1+\sqrt{1+a}}=2\)

Since we know that \(\sqrt4=2\)

Therefore:

\(1+\sqrt{1+a}=4\)

\(\sqrt{1+a}=3\)

Since \(\sqrt9=3\)

\(1+a=9\)

\(a=8\)

Q.E.D.

(For one to solve this question, you just need to know the basic ideas of squares and powers (And some work) :P)

Input:nsolve(y+57+x+16+4x-5,x)

Intepretation:Solve \((y+57+x+16+4x-5=0)\) for \(x\)

\(y+57+x+16+4x-5=0\)

Merge the corresponding terms:

\(y+5x+68=0\)

Move \(y\) and \(68\) to the left:

\(5x=-y-68\)

Divide the entire function by a factor of 5:

\(x=-\frac{1}{5}y-\frac{68}{5}\)

\(=\frac{1}{5}\left(-y-68\right)\)

Q.E.D.