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DedPeter [7]
3 years ago
11

What are the steps for using a compass and straightedge to construct the bisector of ∠G? Two rays share a common endpoint labele

d G, forming an obtuse angle. One ray is horizontal and points right. The second ray is inclined to horizontal in upward-left direction. Drag the steps and drop them in order from start to finish. Without changing the opening of the compass, place the point of the compass on point J and draw another arc in the interior of the angle.Use the straightedge to draw GK−→−.Place the point of the compass on point H and draw an arc in the interior of the angle.Label the intersection of the arcs in the interior of the angle as point K.Place the point of the compass on point G and draw an arc that intersects the sides of ∠G. Label the points of intersection as points H and J.

Mathematics
2 answers:
Dahasolnce [82]3 years ago
7 0

Answer:

Ordered steps to bisect angle G is given below.

Step-by-step explanation:

1. Place the point of the compass on point G and draw an arc that intersects the sides of ∠G.

2.Label the points of intersection as points H and J.

3. Place the point of the compass on point H and draw an arc in the interior of the angle.

4. Without changing the opening of the compass, place the point of the compass on point J and draw another arc in the interior of the angle.

5.Label the intersection of the arcs in the interior of the angle as point K.

6. Use the straightedge to draw GK.


notsponge [240]3 years ago
6 0

Answer:

The Quiz Was Updated This Is It Now

Step-by-step explanation:

DO IN ORDER!

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Simora [160]

Answer:

Two Decimal Places Rule #1: If the last digit in 0.167 is less than 5, then remove the last digit. Two Decimal Places Rule #2: If the last digit in 0.167 is 5 or more and the second to the last digit in 0.167 is less than 9, then remove the last digit and add 1 to the second to the last digit.

Step-by-step explanation:

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In the decimal 55.555, which is 1/10 the value of the 5 in the tenths place?
inna [77]

Answer:

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Step-by-step explanation:

1/10 × 1/10 = 1/100

The value in the hundredths place is 1/10 the value in the tenths place.

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7 0
2 years ago
Solve for q. -5(3-q) +4=5q-11
soldier1979 [14.2K]

Perform the indicated multiplication first:  -15 + 5q + 4 = 5q - 11

Note that 5q appears on both sides of this equation.  Cancelling, we get :

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4 0
3 years ago
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

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Therefore, the directrix is;

y=-\frac{3}{2}

3 0
1 year ago
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ch4aika [34]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
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