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Serhud [2]
3 years ago
7

Martha is training for a duathlon, which includes biking and running. She knows that yesterday she covered a total distance of o

ver 55.5 miles in more than than 4.5 hours of training. Martha runs at a speed of 6 mph and bikes at a rate of 15.5 mph.

Mathematics
1 answer:
umka21 [38]3 years ago
4 0

<u>Answer:</u>

<u>Distance:</u> 6x + 15.5y > 55.5

<u>Time:</u> x + y > 4.5

<u>Step-by-step explanation:</u>

We are given that Martha is training for a duathlon and she covered a total distance of over 55.5 miles in more than than 4.5 hours of training.

Also, she runs at a speed of 6 mph and bikes at a rate of 15.5 mph.

We are to write inequalities representing the distance she traveled and the total time she spent training.

<u>Distance:</u> 6x + 15.5y > 55.5

(formula for distance = speed x time so speeds for running and biking are multiplied by their number of hours)

<u>Time:</u> x + y > 4.5

(she trained for more than 4.5 hours, x hours for running and y hours for biking.

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Write the roman numeral of the Olympics Games that follow the 28th games
bogdanovich [222]

The  Olympics Games that follow the 28th games will be "XXIX Games."

The Olympic Games that follow the 28th Olympic games is the 29th Olympic Games.

29 in Roman numerals is:

= XXVIII

An X in Roman numerals is 10 so the first two Xs are to signify "20."

An "IX" in Roman numerals is "9" which will then take the number to 29.

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7 0
3 years ago
Jason and Alison are hiking in the woods when they spot a rare owl in a tree. Jason stops and measures an angle of elevation of
Shalnov [3]

Answer:

Owl= 67.14ft

Step-by-step explanation:

<em>See comment for complete question.</em>

The given information is represented in the attached figure.

First convert 22°8'6'' and 30° 40’ 30” to degrees

22^{\circ} 8'6'' = 22 + \frac{8}{60} + \frac{6}{3600}

22^{\circ} 8'6'' = 22.135^{\circ}

30^{\circ} 40'30'' = 30 + \frac{40}{60} + \frac{30}{3600}

30^{\circ} 40'30'' = 30.675^{\circ}

Considering Jason's position:

tan(22.135^{\circ}) = \frac{H}{x + 48}

Where x = distance between the tree and Alison

Make H the subject

H = (x + 48)tan(22.135^{\circ})

Considering Alison's position

tan(30.675^{\circ}) = \frac{H}{x}

Make H the subject

H = xtan(30.675^{\circ})

H = H

(x + 48)tan(22.135^{\circ}) = xtan(30.675^{\circ})

(x + 48) *0.4068 = x* 0.5932

Open bracket

x *0.4068 + 48 *0.4068 = 0.5932x

0.4068x + 19.5264 = 0.5932x

Collect Like Terms

-0.5932x+0.4068x = -19.5264

-0.1864x = -19.5264

0.1864x = 19.5264

Make x the subject

x = \frac{19.5264}{0.1864}

x = 104.76

Substitute 104.76 for x in H = xtan(30.675^{\circ})

H = 104.76 * tan(30.675^{\circ})

H = 104.76 * 0.5932

H = 62.14

The above represents the height of the tree.

The height of the owl is:

Owl= H + 5

Owl= 62.14 + 5

Owl= 67.14ft

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Triangle abc is a regular triangle
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