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Readme [11.4K]
3 years ago
12

Leo is constructing a tangent line from point Q to circle P. What is his next step? circle P and point Q on the outside of circl

e P, a segment is created from points P and Q, a set of arcs are created from point P above and below segment PQ that are greater than half the distance of segment PQ, another set of arcs are created from point Q above and below segment PQ that are greater than half the distance of segment PQ, the intersections of the created arcs are connected with a line creating point R on segment PQ Construct a circle from point Q with the radius PR. Construct the perpendicular bisector of segment PQ. Construct a circle from point R with the radius RP. Construct the perpendicular bisector of segment RQ.
Mathematics
2 answers:
mariarad [96]3 years ago
4 0

Answer:

The answer is "Construct a circle from point R with the radius RP".

Step-by-step explanation:

Please find the diagram to this question in the attached file.

In this question, the line is perpendicular to the PQ and the bisector perpendicular to the PQ throughout the given diagram, and in the next section, it draws the tangent through point A to the circle by drawing 'Draw a circle with radius RP through point R. It circle crosses that given circle P in 2 points. Its tangents could only be achieved with any of these new points by adding point Q.

denpristay [2]3 years ago
3 0

Answer:

The answer is C

Step-by-step explanation:

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A <em>second-degree equation</em> or <em>quadratic equation of a variable</em> is an equation that it has the general expression:

ax^{2} +bx+c=0 with a\neq 0

Where x is the variable, and a, b and c constants; a is the quadratic coefficient (other than 0), b the linear coefficient and c is the independent term. This polynomial can be interpreted by means of the graph of a quadratic function, that is, by a parabola. This graphical representation is useful, because the abscissas of the intersections or point of tangency of this graph, in the case of existing, with the X axis are the real roots of the equation. If the parabola does not cut the X axis the roots are complex numbers, they correspond to a negative discriminant.

Second degree equation solutions

For a quadratic equation with real or complex coefficients there are always two solutions, not necessarily different, called roots, which can be real or complex (if the coefficients are real and there are two non-real solutions, then they must be complex conjugates). General formula for obtaining roots:

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The discriminant serves to analyze the nature of the roots that can be real or complex.

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Solving the problem of the answer.

x^{2} -13x-48=0 with a = 1, b = -13, and c = -48

Substituting the values in the general formula for a quadratic equation.

x=\frac{-(-13)±\sqrt{(-13)^{2} -4(1)(-48)} }{2(1)}

x=\frac{13±\sqrt{169+192} }{2}

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Then, we obtain the roots:

x=\frac{13+\sqrt{361} }{2} and x=\frac{13-\sqrt{361} }{2}

Solving the roots:

x=\frac{13+\sqrt{361} }{2}\\x=\frac{13+19}{2}\\x=\frac{32}{2}\\x=16

x=\frac{13-\sqrt{361} }{2}\\x=\frac{13-19}{2}\\x=\frac{-6}{2}\\x=-3

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