Simplify the expression sqrt7!
the ! stands for a factorial
this is worth 24 points
be fast my homework is due in 10 minutes

answer:   6 √35 ;   6* sqrt 35

step-by-step explanation:

√(7! ) =

√(7 * 6 * 5 *4 * 3 * 2 * 1 ) ;

        →   {since:   " 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 " .}.

  =   √7 *√6 * √5*√4*√3*√2*√1 = ?

note: "√1 = 1 " ;   and any value, multiplied by "1" , results in that same value;

so we can eliminate the "√1" ;

the "√4" can be simplied to "2" ;   since it is a "perfect square:

note of the other radicands are "perfect squares" .

so, we have:

√(7! )   =   √(7 * 6 * 5 *4 * 3 * 2 * 1 )   ;

        =   2 * √(7 * 6 * 5 * 3 * 2 ) ;

now, let us multiply:   "(7 * 6 * 5 * 3 * 2) " ;

  →   7 * 6 * 5 * 3 * 2 =   1260 ;  

as such:   √(7 * 6 * 5 * 3 * 2 ) = √(1260) ;

now, can the radicand:

  "1260" ;   be factored —

[note:   hereinafter, by "factor(s) //factor/factoring" , we will assume "factorable/factor(s)/factoring" as "regards to "non-zero, positive integers"] —into any perfect squares? ;

yes;  

→   "100" is a "perfect square" since:   " √100 = 10 " ;   and:   ↔ " 10² = 100 " .

we now that "100" is a factor of 1260;   since "1260" is a non-zero positive integer of three (3) or more digits that ends in zero.

since:   " 1260 = 126 * 100 "   ;     →

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632