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Delicious77 [7]
2 years ago
15

What is the answer to C?

Mathematics
1 answer:
OleMash [197]2 years ago
4 0

The diagram of the sequence of circular and square tiles gives;

a) 49

b) 84

c) Will always be odd

<h3>Which method can be used to find the order of the sequence in the pattern?</h3>

Pattern 1

Number of square tiles = 1

Number of circular tiles = 8

Total number of tiles = 9

Pattern 2

Number of square tiles = 4

Number of circular tiles = 12

Total number of tiles = 16

Pattern 3

Number of square tiles = 9

Number of circular tiles = 16

Total number of tiles = 25

  • The number of square tiles required to make pattern <em>n </em>= n²

From the diagram, the number of circular tiles on the left or right side of the square tiles are 2 more than the number of square tiles, while the number of circular tiles on the top or bottom are the same as the number of square tiles, therefore;

  • Number of circular tiles = 2•(n + 2) + 2•n

a) The number of square tiles required to make pattern number 7 = 7² = 49

b) The number of circular tiles needed to make pattern number 20 = 2 × 20 + 2 × (20 + 2) = 84

c) The total number of circular tiles is always even while the number of square tiles has the same characteristics (even or odd) as the pattern number

  • Even + Odd = Odd
  • Even + Even = Even

When the pattern number is odd, the total number of tiles will always be odd. The correct option is therefore;

B. Will always be odd

Learn more about sequence and series here:

brainly.com/question/12668338

#SPJ1

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Step-by-step explanation:

You can do long division, which is very very hard to show with typing on a keyboard. You essentially want to divide the leading coefficient for each term. Ill try my best to explain it.

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Now do \frac{3x^2}{x} =3x. Write that down next to your 2x^2. Multiply 3x by (x - 3) to get:

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Multiply -2 by (x - 3) to get:

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Our remainder is 0 so that means (x - 3) goes into that trinomial exactly:

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In the equation 0.75s-5/8 = 44, how do you combine the like terms?
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Step-by-step explanation:

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