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Brums [2.3K]
3 years ago
5

Averigua el número que cumple la siguiente condición: si se multiplica su sucesor por el número disminuido en 3 unidades se obti

ene 77
Mathematics
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

X=10

X=-8

Step-by-step explanation:

(x+1)*(x-3)=77

x²-3x+x-3=77

x²-2x-3=77

x²-2x-3-77=0

x²-2x-80=0

(x-10)*(x+8)=0

x-10 = 0

x=10

x+8=0

X=-8

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Part A

The pattern of squares is 1, 4, 9, ... which is the set of perfect squares

  • 1 = 1^2
  • 4 = 2^2
  • 9 = 3^2

and so on

The 7th figure will have 49 squares because 7^2 = 49

<h3>Answer: 49</h3>

==============================================================

Part B

Each pattern has one circle per corner (4 circles so far). In addition, there's one circle per unit side to form the perimeter.

  • Pattern 1 has 4+4(1) = 8 circles
  • Pattern 2 has 4+4(2) = 12 circles
  • Pattern 3 has 4+4(3) = 16 circles

The nth term will have 4+4n circles. The first '4' is the number of circles at the corners. The 4n is the circles along the perimeter. If you wanted, 4+4n factors to 4(1+n).

Plug in n = 20 to find the 20th figure has 4+4n = 4+4(20) = 84 circles

<h3>Answer: 84</h3>

==============================================================

Part C

  • Pattern 1 has 1 square + 8 circles = 9 items total
  • Pattern 2 has 4 squares + 12 circles = 16 items total
  • Pattern 3 has 9 squares + 16 circles = 25 items total

This seems to suggest if the pattern number is odd, then we need an odd number of tiles (square + circular).

Let n be the pattern number. Pattern n needs n^2 square tiles and 4+4n = 4n+4 circular tiles. Overall, n^2+4n+4 tiles are needed.

It turns out that if n is odd, then n^2+4n+4 is always odd. The proof is shown below.

Side note: n^2+4n+4 factors to (n+2)^2

<h3>Answer: B) will always be odd</h3>

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