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Ivan
3 years ago
13

What are the steps for using a compass and straightedge to construct a square?

Mathematics
2 answers:
natka813 [3]3 years ago
5 0
For problems of this type, you need to look at the steps that only require elements that are already known. If a step requires something that hasn't been done yet. THEN THAT STEP COMES LATER. So looking at the 7 available steps, only the step "Construct horizontal PQ" is possible since that's just a matter of drawing a straight line and labeling the end points P and Q. In any case, here are the 7 steps in order. The number in parenthesis after the 1st number is the original scrambled order of that step.  
1(2). Construct horizontal PQ 
2(5). Construct a circle with point P as the center and a circle with point Q as a center, each circle having radius PQ 
3(4). Label the point of intersection of the two circles above PQ, point R, and the point of intersection of the two circles below PQ, point S 
4(6). Construct RS, the perpendicular bisector of PQ, intersecting PQ at point T
 5(1). Construct a circle with point T as the center with radius TQ 
6(7). Label the point of intersection of circle T and RS closest to point R, point
U, and the point of intersection of circle T and RS closest to point S, point V. 
7(3). Construct PU, UQ, QV, and VP to complete square PUQV.
iren [92.7K]3 years ago
5 0

Answer:

1) Construct horizontal PQ  

2) Construct a circle with point P as the center and a circle with point Q as a center, each circle having radius PQ  

3) Label the point of intersection of the two circles above PQ, point R, and the point of intersection of the two circles below PQ, point S  

4) Construct RS, the perpendicular bisector of PQ, intersecting PQ at point T

5) Construct a circle with point T as the center with radius TQ  

6) Label the point of intersection of circle T and RS closest to point R, point

U, and the point of intersection of circle T and RS closest to point S, point V.  

7) Construct PU, UQ, QV, and VP to complete square PUQV.

Step-by-step explanation:

Just took the test, hope this helps :)

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Evaluate Combination (6,6)
andrew11 [14]

Answer:

nCx = \frac{n!}{x! (n-x)!}

On this case n =6 and x =6 we got:

6C6 = \frac{6!}{6! (6-6)!} = \frac{6!}{6! 0!}= \frac{6!}{6!}=1

Step-by-step explanation:

The utility for the combination formula is in order to find the number of ways to order a set of elements

For this case we want to find the following expression:

6C6

And the general formula for combination is given by:

nCx = \frac{n!}{x! (n-x)!}

On this case n =6 and x =6 we got:

6C6 = \frac{6!}{6! (6-6)!} = \frac{6!}{6! 0!}= \frac{6!}{6!}=1

8 0
3 years ago
Solve y=3x+22 if x=225
MakcuM [25]

Answer:

Y=697

Step-by-step explanation:

Substitute the x with 225.

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Therefore x equals 697

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3 years ago
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Suppose a company did $3,000,000 in annual maintenance in 2013 and expects 85% of those to renew for 2014. Suppose that product
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Answer:

In year 2013 annual maintenance done = $ 3,000,000

Out of this 85% is expected to renew in 2014 = 0.85\times3000000 = $2,550,000

Now Sales in 2013 = $3,000,000

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4 years ago
The length of a rectangle is 5 units more than the width. The area of the rectangle is 36 units. What is the length, in units, o
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Answer:

11 units

Step-by-step explanation:

Let's do it this way:

Length = l

width = w

initial equation for area is l * w

We know that l = w+5

substitute.

w+5 * w = 36

solve.

6w = 36

w=6

The width is 6 units.

6+5 = 11

length = 11 units

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