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docker41 [41]
1 year ago
5

Beth left school driving toward the lake one hour before tim. tim drove in the opposite direction going 6 mph slower than beth f

or one hour after which time they were 174 miles apart. what was beths speed? and please explain thoroughly and take me through the process.
Mathematics
1 answer:
m_a_m_a [10]1 year ago
3 0

Beth's speed was 60 mph.

<h3>What is speed?</h3>
  • The speed (commonly referred to as v) of an object in everyday use and kinematics is the magnitude of the change in its position over time or the magnitude of the change in its position per unit of time; it is thus a scalar quantity.
  • The distance traveled by an object in a time interval is divided by the duration of the interval; the instantaneous speed is the limit of the average speed as the duration of the time interval approaches zero.
  • Velocity is not the same as speed.

To find what was Beth's speed:

Beth's data:

  • time = 2 hrs; rate = r mph; distance = r×t = 2r miles

Tim's data:

  • time = 1 hr; rate = r-6 mph; distance = r-6 miles

Equation:

  • distance + distance = 174 miles

So,

  • 2r + r - 6 = 174
  • 3r - 6 = 174
  • 3r = 180
  • r = 60 mph (Beth's rate)
  • r-6 = 54 mph (Tims's rate)

Therefore, Beth's speed was 60 mph.

Know more about speed here:

brainly.com/question/4931057

#SPJ4

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10/72 or 0.14

Step-by-step explanation:

There are 72 marbles in total and 10 of them are green. So, it is 10/72 or 1.388 rounded to the nearest hundredth is 0.14.

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Step-by-step explanation:

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Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

5 0
2 years ago
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