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lara [203]
3 years ago
13

What are the steps for using a compass and straightedge to construct a square? Drag and drop the steps in order from start to fi

nish. Use a straightedge to draw line a and label a point on the line as point P. Without changing the compass width, place the compass point on point S and draw an arc in the interior of ∠SPT . Use the straightedge to draw QS¯¯¯¯¯ and QT¯¯¯¯¯. Construct a line perpendicular to line a through point P. Label a point on this line as point R. With the compass open to the desired side length of the square, place the compass point on point P and draw an arc on line a and an arc on PR←→ . Label the points of intersection as points S and T. Keeping the same compass width, place the compass on point T and draw an arc in the interior of ∠SPT to intersect the previously drawn arc. Label the point of intersection as point Q.
Mathematics
2 answers:
NeTakaya3 years ago
8 0

Answer: The steps for using a compass and straightedge to construct a square are as follows :-

1. Use a straightedge to draw line a and label a point on the line as point P.

2. Construct a line perpendicular to line a through point P.

3. Label a point on this line as point R.

4. With the compass open to the desired side length of the square, place the compass point on point P and draw an arc on line a and an arc on PR←→.

5. Label the points of intersection as points S and T.

6. Keeping the same compass width, place the compass on point T and draw an arc in the interior of∠SPT to intersect the previously drawn arc.

7. Label the point of intersection as point Q.

8. Without changing the compass width, place the compass point on point S and draw an arc in the interior of∠SPT .

9. Use the straightedge to draw QS ,QT,SR and RT.

Here we have finished with the square RTQS.





Salsk061 [2.6K]3 years ago
6 0
<span>1. Use a straightedge to draw line a and label a point on the line as point P. 2. Construct a line perpendicular to line a through point P. 3. Label a point on this line as point R. 4. With the compass open to the desired side length of the square, place the compass point on point P and draw an arc on line a and an arc on PRâ†â†’ . 5. Label the points of intersection as points S and T. 6. Keeping the same compass width, place the compass on point T and draw an arc in the interior of â SPT to intersect the previously drawn arc. 7. Label the point of intersection as point Q. 8. Without changing the compass width, place the compass point on point S and draw an arc in the interior of â SPT . 9. Use the straightedge to draw QSÂŻÂŻÂŻÂŻÂŻ and QTÂŻÂŻÂŻÂŻÂŻ.</span>
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The costume designer should use 1/8 of the fabric dedicated to sashes for each dress.
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The sum of the ages of Mrs. Spelman and Mr. Murphy are 78 years. Six years from now Mrs. Spelman's age will be equal to 6 times
Studentka2010 [4]

Answer:

  • Mrs Spelman: 71 1/7
  • Mr/Ms Murphy: 6 6/7

Step-by-step explanation:

There are no integer solutions to the problem as posed. (We suspect a typo, that the intention is 3 years from now, or that the ratio will be 5:1.)

Let m represent Murphy's age now. Then 78-m represents Spelman's age now, and the ratio in 6 years will be ...

  78 -m +6 = 6(m +6)

  84 -m = 6m +36 . . . . . collect terms

  48 = 7m . . . . . . . . . . . . add m-36

  48/7 = m = 6 6/7 . . . . . divide by the coefficient of m

Then Spelman's age now is ...

  78 -m = 78 -6 6/7 = 71 1/7

Spelman is 71 1/7; Murphy is 6 6/7.

_____

If the ratio is 6:1 in 3 years, then Murphy is 9 and Spelman is 69.

If the ratio is 5:1 in 6 years, then Murphy is 9 and Spelman is 69.

_____

<em>Alternate solution method</em>

I find it easiest to add 12 years to the total (each ages by 6 years), which will give a total age of 90 in 6 years. At that time, Murphy is 1/7 of the total of ages, so dividing that sum into parts with the appropriate ratio gives m'=90/7=12 6/7; s'=77 1/7. So, m=6 6/7; s=71 1/7.

_____

<em>Comment on the problem</em>

We think there is a "typo" because the ratio of 6:1 means the future total should be a multiple of 7. Of course, 78+6 =84 is a multiple of 7, but adding 6 to the total will occur in 3 years, not 6 years.

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Step-by-step explanation:

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3 years ago
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statuscvo [17]

Answer:

(a) The number of different lock combinations is 216,000.

(b) The probability of getting the correct combination in the first try is \frac{1}{216000}.

Step-by-step explanation:

The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first​ cylinder, then to a second number on the second​ cylinder, and then to a third number on the third cylinder.

The numbers can be repeated.

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There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

So the total number of possible combinations is:

Total number of combinations = 60 × 60 × 60 = 216000.

Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three​-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = \frac{1}{60}.

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = \frac{1}{60}.

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = \frac{1}{60}.

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = \frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}.

Thus, the probability of getting the correct combination in the first try is \frac{1}{216000}.

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