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1. A ski lift carries Carol up a slope at a rate of 5 mph. She skis down the slope, parallel to the lift, at a speed of 20 mph. The round trip takes 45 minutes. Write and solve a system of equations to determine how far she skis.

**System:**

Let \( x \) be the distance in miles up the slope.

**Defined variables:**

- \( x \): distance (in miles) Carol travels up the slope.
- \( t_1 \): time taken to travel up the slope.
- \( t_2 \): time taken to travel down the slope.

**Solve:**

1. Equation for going up the slope: \( t_1 = \frac{x}{5} \)
2. Equation for skiing down the slope: \( t_2 = \frac{x}{20} \)
3. Total time: \( t_1 + t_2 = \frac{45}{60} \) (since 45 minutes is 0.75 hours)

Using the equations:
\[
\frac{x}{5} + \frac{x}{20} = 0.75
\]

Multiply through by 20 to eliminate fractions:
\[
4x + x = 15
\]

Combine and solve for \( x \):
\[
5x = 15 \\
x = 3
\]

**Solution as a sentence:**

Carol skis a distance of 3 miles up the slope.

Answer :

Final answer:

Carol skis a total distance of 3 miles. The equations set up for the scenario help to find this solution by considering the speeds at which she moves up and down the slope and the total time she spends.

Explanation:

Let's define the distance Carol skis as d, and the time it takes for her to ski up and down as t. We're given that t = 45 minutes or 0.75 hours. We can write the following system of equations from the information:

  1. d = [tex]5t_u_p[/tex], where [tex]t_u_p[/tex] is the time it takes Carol to go up the slope.
  2. d = 20[tex]t_d_o_w_n[/tex], where [tex]t_d_o_w_n[/tex] is the time it takes Carol to go down the slope.

The total time for the round trip is [tex]t_u_p[/tex] + [tex]t_d_o_w_n[/tex] = t = 0.75. Replacing d from the first equation into the second, we have 5 [tex]t_u_p[/tex] = 20[tex]t_d_o_w_n[/tex]. This simplifies to [tex]t_u_p[/tex] = 4[tex]t_d_o_w_n[/tex]. Plugging this into our total time equation, we have 4[tex]t_d_o_w_n[/tex] + [tex]t_d_o_w_n[/tex] = 0.75 solving for [tex]t_d_o_w_n[/tex], we get [tex]t_d_o_w_n[/tex] = 0.15 hours. Now, replace [tex]t_d_o_w_n[/tex] in our second equation, we have d = 20 * 0.15 solving for d, we get d = 3 miles.

Learn more about Systems of Equations here:

https://brainly.com/question/35467992

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Rewritten by : Barada

Hi sorry for the late response, I know the assignment was due at 5pm today but anyway here is the answer.

System:

d= 5(0.75-t)

d= 20t

20t = 3.75 -5t

Define Variables:

d = how far Carol Skied

t = time it took Carol to go up/down

Solve:

20t = 3.75 -5t

20t + 5t = 3.75 -5t + 5t

25t = 3.75

25t/25 = 3.75/25

t = 0.15

Now we plug in the time into the Distances:

20t = 3.75 - 5t

20(0.15) = 3.75 - 5(0.15)

3 = 3.75 - 0.75

3 = 3

Solution as Sentence:

Carol Skied 3 miles