Show that (0.8a+0.4b)^2=(0.8a-0.4b)^2+1.28ab< br />
please help me solve this asap

Step-by-step explanation:

first

(0.8a+0.4b)^2= (0.8a+0.4b) *(0.8a+0.4b)

=

0.64 a^2 + 0.32 ab +0.32ab + 0.16 b^2=

0.64 a^2 + 0.64ab + 0.16 b^2

second

(0.8a-0.4b)^2+1.28ab

= 0.64 a^2 - 0.64ab +0.16 b^2 + 1.28ab

= 0.64 a^2 + 0.64ab + 0.16 b^2

since the first (left parameters ) equals the second(right side parameters)

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hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.