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.