Question

please q 01 and 04 please

Answer 1

1)=3×12×101+4
   =3×4×3×101+4
   =4(3×3×101+1)
then there is a common factor 4 hence it is a composite no
2)Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles  at points A and B .

To find : ∠APB

Proof: let ∠CAP = x and ∠CBP = y

CA = CP [lengths of the tangents from an external point C]

In a triangle PAC, ∠CAP = ∠APC = x

similarly CB = CP and ∠CPB = ∠PBC = y

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180°   [sum of the interior angles in a triangle]

x + y + (x + y) = 180°

2x + 2y = 180°

x + y = 90°

∴ ∠APB = x + y = 90°

:)Hope this helped u...