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Rudiy27
3 years ago
15

Find the sum: (3x^2+2x+3)+(x^2+x+1)

Mathematics
1 answer:
elena-s [515]3 years ago
3 0

Answer:

 4x2 + 3x + 4

Step-by-step explanation:

Step-1 : Multiply the coefficient of the first term by the constant   4 • 4 = 16

Step-2 : Find two factors of  16  whose sum equals the coefficient of the middle term, which is   3 .

-16+-1=-17

-8+-2=-10

-4+-4=-8

-2+-8=-10

-1+-16=-17

1+16=17

2+8=10

4+4=8

8+2=10

16+1=17

Final result :

4x2 + 3x + 4

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Answer:

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Step-by-step explanation:

First we need to know that the length of the side of the square is equal to the diameter of the inscribed circle i.e

L = di

Given the area of the square to be 9in², we can get the length of the square.

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L = √9

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Hence the length of one side of the square is 3in

This means that the diameter of the inscribed circle di is also 3in.

Circumference of a circle = π×diameter of the circle(di)

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For the circumscribed circumscribed circle, diameter of the outer circle will be equivalent to the diagonal of the square.

To get the diagonal d0, we will apply the Pythagoras theorem.

d0² = L²+L²

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d0² = 9+9

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= π(3√2)

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Hence, ratio of the circumference of the circumscribed circle to the one of the inscribed will be 3√2 π/3π = √2:1

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Hope this helps:)

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