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Reil [10]
3 years ago
8

GIVING BRAINLIESTTTTT

Mathematics
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

D.Place the compass on a point on the original line.

Complete Question:

A.Create only one arc that intersects the original line.

B.Create two arcs that intersect the original line.

C.Place the compass on a point off the original line.

D.Place the compass on a point on the original line.

Step-by-step explanation:

* Lets revise the steps of constructing parallel lines and a perpendicular line through a point on the line

# Contracting parallel lines

- Given: Line AB and point P not on AB

1. Draw a slant line through point P and intersects AB at point C, this line is the transversal of the parallel lines

2. Place the pin of the compass at point C and draw an arc intersects AB at D and CP at E

3. Without changing the distance of the compass put the pin of the compass on point P and draw an arc intersects CP at point Q

4. Use the compass to measure the distance from D to E and place the pin of the compass at point Q and draw an arc intersects the arc from P to CP at point U

5. Join P and U by a line this line is parallel to AB

# Constructing a perpendicular line through a point on the line

- Given line AB and point P lies on it

1. Place your compass pin at P and draw an arc of any size below AB that crosses the line twice at points C and D

2. Stretch the compass to a larger distance

3. Place the compass pin on point C and draw a small arc above the line

4. Without changing the distance of the compass place the compass pin on point D and draw another arc intersects the arc from C at point E

5. Using a straightedge, join P and E where PE is perpendicular to AB

* Lets find the same steps in the constructions

∵ In first construction we put the pin of the compass at point C which  lies in the original line AB

∵ In second construction we put the pin of the compass at point P which lies in the original line AB

∴ The common step is:  "Place the compass on a point on the original line"

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jasenka [17]
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3 0
3 years ago
Read 2 more answers
Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)
zhenek [66]

ANSWER

The vertical asymptotes are


\Rightarrow x=2\:or\:x=-1

<u>EXPLANATION</u>

We have

f(x)=\frac{x^2+4}{4x^2-4x-8}


For vertical asymptotes we set the denominator to zero and solve the quadratic equation;

4x^2-4x-8=0


\Rightarrow x^2-x-2=0

We split the middle term to obtain,

x^2+x-2x-2=0

\Rightarrow x(x+1)-2(x+1)=0


\Rightarrow (x-2)(x+1)=0


\Rightarrow x=2\:or\:x=-1


Therefore the vertical asymptotes are


x=2\:or\:x=-1





3 0
3 years ago
Read 2 more answers
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