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antoniya
3 months ago
13

Sasha is bisecting a segment. First, she places the compass on one endpoint, opens it to a width larger than half of the segment

, and swings an arc on either side of the segment. Then, she keeps the compass the same width and places it on the endpoint. What is her next step?
A) Swing arcs on both sides to intersect the first two arcs created.

B) Swing an arc that intersects the segment

C) Swing arcs that intersect a point that is not on the segment

D) Swing an arc that intersects the opposite endpoint.
Mathematics
2 answers:
AnnZ [12.3K]3 months ago
7 0

Answer:

A) Create arcs on both sides to intersect the first two arcs made.

Step-by-step explanation:

To bisect a segment, it involves dividing a line into two equal sections using a bisector.

The steps include;

  • Positioning a compass on one endpoint
  • Adjusting the compass to a width larger than half of the segment
  • Sweeping an arc on both sides of the segment
  • Maintaining the compass width, placing it on the opposite endpoint
  • Sweeping arcs on both sides to meet the initially drawn arcs
  • Utilize a ruler to draw the line bisector at the intersection points of the arcs.

Sasha had reached step four.

Inessa [12.5K]3 months ago
7 0

Answer:

The correct option is A

Step-by-step explanation:

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Answer:

The tangent plane equation for the hyperboloid

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1.

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We have

The ellipsoid's equation is

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

The equation for the tangent plane at the point \left(x_0,y_0,z_0\right)

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1  (Given)

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\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1

F(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}

F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}

(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)

The tangent plane equation at point \left(x_0,y_0,z_0\right)

\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0

The tangent plane equation for the hyperboloid is

\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0

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2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2

Hence, the required tangent plane equation for the hyperboloid is

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0

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