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Butoxors [25]
3 years ago
10

Given the following perfect square trinomial, fill in the missing term. x2 − 16x + ____

Mathematics
2 answers:
pychu [463]3 years ago
6 0
The pattern is:
( a - b )² = a² - 2 a b + b² ( square of last term of binomial - the missing term)
x² - 2 · 8 · x + 8² = x² - 16 x + 64 = ( x - 8 )²
The missing term is: 64
Ymorist [56]3 years ago
6 0

Your Answer:

<em>64</em>

Explanation:

Your answer would be 64 because the middle term of a perfect square trinomial must have the following:

  • The first term must be a perfect square
  • The last term must be a perfect square
  • The middle term must be twice the product of the square roots of the first and last terms

All of these requirements are present in this trinomial. We already know that this is a perfect square trinomial because the question says it. All we need to find is the square root of the missing number that would give you 64 if multiplied by "x" and then "2". Since 8 · 2 = 16, we must find a number that has a square root of 8, in which this case, would be 64. So your final completed perfect square trinomial would be: x^{2} - 16x + 64

Hope this helps y'all :D

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The area of a square is 64 n Superscript 36square units. What is the side length of one side of the square?
arsen [322]

The correct answer is option D which is the side length will be (64n)¹⁸.

<h3>What is the area of the square?</h3>

The square is defined as a quadrilateral having all four sides equal to each other and the area of the square is the product of its sides.

Given that:-

  • The area of the square is given as- A = (64n)³⁶.

The sides of the square will be calculated as follows:-

A = (64n)³⁶

a² = (64n)³⁶    Here a = Side of the square.

a = √ (64n)³⁶

a = 64n¹⁸

Therefore the correct answer is option D which is the side length will be (64n)¹⁸.

To know more about an area of square follow

brainly.com/question/13389992

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Step-by-step explanation:

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Step-by-step explanation: how i got the answer was to multiply everything by two. then add everything together.

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