Question

hello friends!
a little help please!
a high leveled maths question... and I need a proper answer with steps!!
:)

Answer 1

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Here's your answer...

In the above equation, if you may have noticed, the sum of SQUARES of numbers is equal to 0.

Square of any real number is positive, so the only way for the sum of squares to be 0 is for the squares themselves to be 0.

So (x-1)² = 0 and x = 1

Similarly, y = 3, z = 5, t = 7

xyzt + 16 = 1×3×5×7 + 16
= 105+16
= 121

So (d) is the correct option.

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Answer 2

Given (x - 1)^2 + (y - 3)^2 + (z - 5)^2 + (t - 7)^2 = 0

(i)

⇒ (x - 1)^2 = 0

⇒ x - 1 = 0

⇒ x = 1.

(ii)

⇒ (y - 3)^2 = 0

⇒ y - 3 = 0

⇒ y = 3

(iii)

⇒ (z - 5)^2 = 0

⇒ z - 5 = 0

⇒ z = 5

(iv)

⇒ (t - 7)^2 = 0

⇒ t - 7 = 0

⇒ t = 7.

Now,

⇒ xyzt + 16

⇒ (1)(3)(5)(7) + 16

⇒ 121.

Therefore, the answer is Option(D) - 121.

Hope this helps!