Question

------- becomes the outermost shell on losing valence electrons by --------.​

Answer 1

Hello,Buddy!!

Refer The Attachment ⤴️

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@MrMonarque♡

Hope It Helps You ✌️

Answer 2

SOLUTION :-

In △ABC

Given :-

• $\sf{DE \parallel BC}$
• AD = x
• DB = x - 2
• AE = x + 2
• EC = x - 1

Now, By BPT Theorem,

$\sf{→\frac{AD}{DB} = \frac{AE}{EC}}$

$\sf{→\frac{x}{x - 2} = \frac{x + 2}{x - 1}}$

Now cross multiplying the term,

$\sf{→x(x - 1) = (x - 2)(x + 2)}$

$\sf{→ {x}^{2} - x = {x}^{2} \cancel{+ 2x} \cancel{- 2x} - 4}$

$\sf{→ \cancel{{x}^{2}} - x = \cancel{{x}^{2}} - 4}$

$\sf{→ \cancel{-} x = \cancel{-} 4}$

$\sf→\fbox{x = 4}$

∴ The value of x = 4.