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grigory [225]
2 years ago
15

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.
1.Construct a line perpendicular to line m through point F. Label a point on this line as point G.
2.Use a straightedge to draw line m and label a point on the line as point F.
3.Without changing the compass width, place the compass point on point H and draw an arc in the interior of ∠HFK .
4.Use the straightedge to draw JH¯¯¯¯¯ and JK¯¯¯¯¯.
5.With the compass open to the desired side length of the square, place the compass point on point F and draw an arc on line m and an arc on FG←→ . Label the points of intersection as points H and K.
6.Keeping the same compass width, place the compass on point K and draw an arc in the interior of ∠HFK to intersect the previously drawn arc. Label the point of intersection as point J.

Mathematics
2 answers:
Mazyrski [523]2 years ago
5 0

For those of you who are doin the test rn here are the answers to all of them. Hope this helps! ;) I did the quiz

Aleks04 [339]2 years ago
3 0
<span>The steps for using a compass and straightedge to construct a square are given as follows:

1. </span><span>Use a straightedge to draw line m and label a point on the line as point F

2. </span><span>Construct a line perpendicular to line m through point F. Label a point on this line as point G.

3. </span><span>With the compass open to the desired side length of the square, place the compass point on point F and draw an arc on line m and an arc on FG←→ . Label the points of intersection as points H and K.

4. </span><span>Without changing the compass width, place the compass point on point H and draw an arc in the interior of ∠HFK.

5. </span><span>Keeping the same compass width, place the compass on point K and draw an arc in the interior of ∠HFK to intersect the previously drawn arc. Label the point of intersection as point J.

6. </span><span>Use the straightedge to draw JH¯¯¯¯¯ and JK¯¯¯¯¯.
</span>
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Joaquin and Trisha are playing a game in which the lower median wins the game. Their scores are shown below.
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Median is defined as the middle term when the scores are arranged in increasing/decreasing order.

Joaquin's scores: 58, 62, 72, 75, 85, 91        Median: 73.5
Trisha's scores: 55, 74, 76, 90, 91, 92          Median: 83

The answer is letter A. 
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1 year ago
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Craig ran the first part of a race with an average speed of 8 miles per hour and biked the second part of a race with an average
lutik1710 [3]
Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5

Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.
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1 year ago
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Admission to the zoo costs $15 per person. Which graph correctly represents the total cost for a group to visit the zoo?
Maurinko [17]
Y=15x would be the correct graph though if there's more info it might not be
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Alice has a 4-inch by 5-inch photo of your school’s championship girls’ ice hockey team. To celebrate their recent
Pachacha [2.7K]

Alice should pick the enlarged-photo with dimensions of 8-inch by 10-inch.

Step-by-step explanation:

Step 1:

In order for a part of the photo to not be cut off, the enlarged photo's dimensions should be of a constant ratio with the original photo's dimensions.

We divide the dimensions of the enlarged-photo with the dimensions of the original photo to check which has a constant ratio.

Step 2:

The original photo was a 4-inch by 5-inch photo.

Option 1 is 7-inch by 9-inch, so the ratios are

\frac{7}{4} = 1.75, \frac{9}{5} = 1.8. The ratios are different so this cannot be the enlarged photo's dimensions.

Option 2 is 8-inch by 10-inch, so the ratios are

\frac{8}{4} = 2, \frac{10}{5} = 2. The ratios are the same so this can be the enlarged photo's dimensions.

Option 3 is 12-inch by 16-inch, so the ratios are

\frac{12}{4} = 3, \frac{16}{5} = 3.2. The ratios are different so this cannot be the enlarged photo's dimensions.

So the enlarged-photo with dimensions of 8-inch by 10-inch should be picked.

7 0
2 years ago
Sue’s Corner Market has a markup of 60% on bottled water. If the market sells a bottle of water for $2, find the original amount
lakkis [162]

Markup is the amount added to the cost price of goods to cover overhead and profit.

Sue’s Corner Market has a markup of 60% on bottled water.

Let us say original price was $x.

Now price after markup is $2.

So we can make an equation like:

original price + markup price = price after markup

x + 60% of x =2

x + 0.6x =2

1.6x =2

dividing both sides by 1.6

x= 1.25

So original price was 1.25 dollars.

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2 years ago
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