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aleksklad [387]
2 years ago
14

Assume that lines which appear to be tangent are tangent. Find the value of the?

Mathematics
2 answers:
Makovka662 [10]2 years ago
7 0

Answer:

12

Step-by-step explanation:

Assume that lines which appear to be tangent are tangent.

Therefor this is a right triangle.

Use the pythagorean theorem

c^2 = a^2 + b^2

15^2 = ?^2 + 9^2

225 = ?^2 + 81

subtract 81 from both sides

144 = ?^2

Take the square root of both sides

12 = ?

goldfiish [28.3K]2 years ago
3 0

9514 1404 393

Answer:

  ? = 12

Step-by-step explanation:

The tangent makes a right angle with the radius at the point of tangency. That means the Pythagorean theorem can be used to find the missing length.

  ?² +9² = 12²

  ?² = 225 -81 = 144

  ? = √144 = 12

The value of ? is 12.

__

<em>Additional comment</em>

When you see a right triangle with the ratio of hypotenuse to leg of 15:9 = 5:3, you know immediately that it is a multiple of a 3:4:5 right triangle. Here, the scale factor is 3, so the missing side is 3×4 = 12.

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Find a possible formula for a fourth degree polynomial g that has a double zero at -2, g(4) = 0, g(3) = 0, and g(0) = 12. g(x) =
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<h2>Answer:</h2>

The possible formula for a fourth degree polynomial g is:

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<h2>Step-by-step explanation:</h2>

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Here the polynomial g  has a double zero at -2, g(4) = 0, g(3) = 0.

This means that the polynomial g(x) is given by:

g(x)=m(x-(-2))^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x+2)^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x^2+2^2+2\times 2\times x)(x(x-3)-4(x-3))\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-3x-4x+12)\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-7x+12)\\\\i.e.\\\\g(x)=m[x^2(x^2-7x+12)+4(x^2-7x+12)+4x(x^2-7x+12)]\\\\i.e.\\\\g(x)=m[x^4-7x^3+12x^2+4x^2-28x+48+4x^3-28x^2+48x]\\\\i.e.\\\\g(x)=m[x^4-7x^3+4x^3+12x^2+4x^2-28x^2-28x+48x+48]\\\\i.e.\\\\g(x)=m[x^4-3x^3-12x^2+20x+48]

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Hence, the polynomial g(x) is given by:

g(x)=\dfrac{1}{4}(x^4-3x^3-12x^2+20x+48)

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