Which equations are true for x = –2 and x = 2? Select two options x2 – 4 = 0 x2 = –4 3x2 + 12 = 0 4x2 = 16 2(x – 2)2 = 0
1 answer:
You might be interested in
Answer: " x = 12 " .
________________________________________________
Step-by-step explanation:
________________________________________________
<u>Note</u>: If 2 (two) triangles are "similar" ; then their corresponding sides are "proportional".
________________________________________________
Since one of the sides [the "hypotenuse"] of one the triangles given ; shows the measurement containing an expression with a value of "x" — specifically; " 6x + 28 " ;
→ And since we want to solve for the value of "x" ; we use this "hypotenuse" of said particular triangle to set up a "proportion" with corresponding sides; so we can solve for "x" :
→ Set up proportions as a "ratio" ; or fraction:
→ 7:28::25:(6x + 28) ; Solve for "x" ;
Let us write this in the form of a "fraction" :
→ ;
________________________________________________
<u>Note</u>:
→ Rewrite: " " ; by simplifying to:
" " ;
________________________________________________
→ Since: " " ;
= " (7÷7) / (28÷7) " ;
= " (1 /4) " ;
= " " .
________________________________________________
Now, rewrite the "proportion" ; as follows:
________________________________________________
;
________________________________________________
Now: "Cross multiply" ; that is:
______________________________________
→ Given: " " ;
and: " b0 " ; " d
0 " ;
→ Then: " a * d " = " b * c " .
________________________________________________
Likewise: " 1(6x + 28) = (4) * (25) " ;
______________________________________________
SImplify: " 6x + 28 = 100 " ; Solve for "x" ;
→ Subtract "28" from each side of the equation:
→ " 6x + 28 - 28 = 100 - 28 " ;
to get: → " 6x = 72 " ;
Now, divide each side of the equation by: " 6 " ;
to isolate "x" on one side of the equation; & to solve for "x" ; as follows:
→ " 6x = 72 " ;
→ 6x / 6 = 72 / 6 ;
to get:
→ " x = 12 " ;
→ which is the answer.
______________________________________________
Hope this helps!
Best wishes in your academic pursuits
— and within the "Brainly" community!
______________________________________________