Can anyone solve the 8th question given

AY = 10 cm.

Step-by-step explanation:

Given that, ΔABC is similar to ΔAXY

and

⇒  

⇒ BC = XY× \frac{5}{3}" alt="\frac{BC}{XY } = \frac{5}{3}" />

⇒ BC = XY× \frac{5}{3}" /> =  \frac{20}{3}[/tex] (as XY = 4 cm given)

Now, check the attached figure,

given, BY bisects ∠XYC

let ∠XYB = ∠BYC = x

⇒ ∠AYX = 180-2x (angle on a straight line)

and also ∠AYX = ACB (similar triangle properties)

⇒ ∠ACB = 180-2x

Now, sum of angles in ΔBYC = 180°

⇒ ∠YBC = x

⇒ BC = YC (as two sides of equal angles are equal in a triangle)

⇒ YC =  \frac{20}{3}[/tex]

And also

AC = AY + YC

⇒

⇒ }{AY }  = \frac{5}{3}[/tex]

⇒ 5AY = 3AY + 20

⇒ AY = 10 cm

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,