Explain how using systems of equations might help you find a better deal on renting a car.

Mathematics

1 answer:

Alona [7]2 years ago

5 0

Answer:

See explanation

Step-by-step explanation:

Using systems of equations will help you find a better deal on renting a car.

Let's use these two equations as an example.

Company A. y=50x+120 where 120 is how much it costs to rent the vehicle and 50 is the rate per day. If you use company A, you will pay $50+$120= $170 for the first day, $220 for two days, etc...

Company B. y=40x+150 where 150 is how much it costs to rent the vehicle and 40 is the rate per day. If you use company B, you will pay $40+$150= $190 for the first day, $230 for two days, etc...

Now you have to take this into consideration - how long you're renting the vehicle for.

If you rent a car for two days, it'll cost $220 with Company A and $230 with Company B
If you rent a car for three days, it'll cost $270 with Company A and $270 with Company B.
If you rent a car for four days, it'll cost $320 with Company A and $310 with Company B.

As you can see, at first Company A seems to have the better deal because of its low cost to rent a car, but it has a higher fee per day. Company B had a higher cost to rent a car initially, but the driver will pay a lower fee a day. Using Company A is beneficial for those who desire to rent a car for less than three days and Company B would be cheaper for those renting more than three days.

By using system of equation, you can determine the better deal according to the length of rental by comparing the initial cost and daily rental fee.

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If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

$v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2$

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

$v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}$

When we use this to replace the 't" in the displacement formula, we et:

$y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}$

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity $v_0$ , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)