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Xelga [282]
3 years ago
6

To the nearest hundredth, how many cm are in 16 in? (1 in = 2.54 cm)

Mathematics
1 answer:
Murljashka [212]3 years ago
7 0
Well ok all you need to do is multiply 2.54 and 16 then round.

                                   2.54
                                  x  16
                                  -------- 
                                  1524
                                +2540
                               -------------
                             → 40.64 ←
This answer is already rounded to the nearest hundredth. I hope this helps love! :)
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AB is included between--------
nirvana33 [79]

Answer:

B) ∠A and ∠B

Step-by-step explanation:

AB is the line segment formed with endpoints at A and B.  This means it lies between the angle with vertex at A, ∠A, and the angle with vertex at B, ∠B.

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A bus traveled on a level road for 4 hours at an average speed of 20 miles per hour faster than it traveled on a winding road. T
Kruka [31]

To solve this problem, we must recall that the formula for velocity assuming linear motion:

v = d / t

Where,

v = velocity

d = distance

t = time

 

For condition 1: bus travelling on a level road

v1 = d1 / t1

<span>(v2 + 20) = (449 – d2) / 4                        --->1</span>

 

For condition 2: bus travelling on a winding road

v2 = d2 / t2

<span>v2 = d2 / 5                                            --->2</span>

 

Combining equations 1 and 2:

(d2 / 5) + 20 = (449 – d2) / 4

0.8 d2 + 80 = 449 – d2

1.8 d2 = 369

d2 = 205 miles

 

Using equation 2, find for v2:

v2 = 205 / 5

v2 = 41 mph

 

Since v1 = v2 + 20

v1 = 41 + 20

v1 = 61 mph

 

Therefore <span>the average speed on the level road is 61 mph.</span>

4 0
3 years ago
Given a triangle JLM with vertices J(-4,3), L(-2,7), and M(2,7). What would the coordinates be if it was reflected across the li
topjm [15]

Answer: M'(2, - 5), L'(-2, -5), j'(-4, - 1)

Step-by-step explanation:

When we do a reflection over a given line, the distance between all the points (measured perpendicularly to the line) does not change.

The line is y = 1.

Notice that a reflection over a line y = a (for any real value a) only changes the value of the variable y.

Let's reflect the points:

J(-4, 3)

The distance between 3 and 1 is:

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Then the new value of y must also be at a distance 2 of the line y = 1

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The new point is:

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L(-2, 7)

The distance between 7 and 1 is:

7 - 1 = 6.

The new value of y will be:

1 - 6 = -5

The new point is:

L'(-2, -5)

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M'(2, - 5)

3 0
3 years ago
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Alex17521 [72]

Answer:

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2 years ago
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tekilochka [14]

Answer: 294cm

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3 years ago
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