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mezya [45]
3 years ago
13

A portion of a hiking trail slopes upward at about a 6 angle. To the nearest tenth of a foot, what is the value of x, the distan

ce the hiker traveled along the path, if he has traveled a horizontal distance of 120 feet?
Mathematics
1 answer:
Virty [35]3 years ago
6 0
The distance covered by the hiker if he traveled a horizontal distance of 120 ft at and angle of 6° will be given by:
cos θ=adjacent/hypotenuse
let the hypotenuse be h, adjacent =120 ft, θ=6°, thus plugging in the values we shall have:
cos 6=120/h
h=120/cos 6
h=120.661 ft
Answer: 120.661 ft
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Answer:

\boxed{0.2}

Step-by-step explanation:

Convert \dfrac{12}{60} to a decimal

Step 1. Reduce the fraction to its lowest terms.

Divide both numerator and denominator by their greatest common factor (12)

\dfrac{12}{60} =\dfrac{1}{5}

Step 2. Convert the denominator to a power of 10

Multiply both numerator and denominator by 2.

\dfrac{1}{5} \times \dfrac{2}{2} = \dfrac{2}{10}

Step 3. Divide the numerator by the denominator

Dividing by 10 moves the decimal point one place to the left.

\dfrac{2}{10} = \boxed{0.2}

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If Denise has 3000 water bottles and gave 4x345 away, how many would she have left?
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Ava and Kelly ran a road race, starting from the same place at the same time. Ava ran at an average speed of 6 miles per hour. K
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<h2>Hello!</h2>

The answer is:

Ava and Kelly will be 3/4 mile apart after 0.375 hours.

<h2>Why?</h2>

To calculate when will Ava and Kelly be 3/4 mile apart, we need to write the equations for both Ava's and Kelly's positions.

Writing the equations we have:

For Ava:

x_A=x_o+v_ot\\\\x_A=0+v_o*t\\\\x_A=v_{o(ava)*t

For Kelly:

We need to calculate when Kelly will be 3/4 mile apart becase is running faster than Ava, so, writing the equation we have:

x_K=x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t

Now, substituting the equation for Ava into the equation for Kelly, we have:

x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t

v_{o(ava)*t+\frac{3}{4}mile}=x_o+v_{o(Kelly)}*t

v_{o(ava)*t+0.75mile}=x_{o}+v_{o(Kelly)}*t\\\\6mph*t+0.75mile=8mph*t\\\\0.75mile=2mph*t\\\\t=\frac{0.75miles}{2mph}=0.375hours

To prove  that the result is correct, we just need to substitute the obtained value for time into both equations, so, substutiting we have:

For Ava:

x_A=v_{o(ava)*t=6mph*0.375hours=2.25miles

For Kelly:

x_K=v_{o(Kelly)}*t=8mph*0.375=3miles

There is a difference of 0.75 miles or 3/4 mile between Ava and Kelly, so, the obtained value for time is correct.

Therefore, we have that Ava and Kelly will be 3/4 mile apart after 0.375 hours.

Have a nice day!

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