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riadik2000 [5.3K]
2 years ago
9

Will mark brainlest answer the question in the picture linked belowand explain as well

Mathematics
1 answer:
RoseWind [281]2 years ago
7 0

Answer:

¹⁰√4

Step-by-step explanation:

4^(2/5*1/4)

4^(1/10)

¹⁰√4

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On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove
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Velocity, distance and time:

This question is solved using the following formula:

v = \frac{d}{t}

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time t_1, distance of d_1, v = 40. So

v = \frac{d}{t}

40 = \frac{d_1}{t_1}

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is v = 60.

On the first day, he drove 2 more hours, which means that for the second day, the time is: t_1 - 2

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: d_1 - 20

Thus

v = \frac{d}{t}

60 = \frac{d_1 - 20}{t_1 - 2}

System of equations:

Now, from the two equations, a system of equations can be built. So

40 = \frac{d_1}{t_1}

60 = \frac{d_1 - 20}{t_1 - 2}

Find the total distance traveled in the two days:

We solve the system of equation for d_1, which gets the distance on the first day. The distance on the second day is d_2 = d_1 - 20, and the total distance is:

T = d_1 + d_2 = d_1 + d_1 - 20 = 2d_1 - 20

From the first equation:

d_1 = 40t_1

t_1 = \frac{d_1}{40}

Replacing in the second equation:

60 = \frac{d_1 - 20}{t_1 - 2}

d_1 - 20 = 60t_1 - 120

d_1 - 20 = 60\frac{d_1}{40} - 120

d_1 = \frac{3d_1}{2} - 100

d_1 - \frac{3d_1}{2} = -100

-\frac{d_1}{2} = -100

\frac{d_1}{2} = 100

d_1 = 200

Thus, the total distance is:

T = 2d_1 - 20 = 2(200) - 20 = 400 - 20 = 380

The total distance traveled in two days was of 380 miles.

For the relation between velocity, distance and time, you can take a look here: brainly.com/question/14307500

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