Ask question

Ask question

All categories

1 year ago

14

‼️‼️‼️‼️HELP‼️‼️‼️‼️‼️‼️‼️

1 answer:

Vesnalui [34]1 year ago

3 0

Michael took the return trip at a velocity 33.75 miles per hour.

<h3>How fast did Michael drive in his return trip?</h3>

Let suppose that Michael drove in <em>straight line</em> road and at <em>constant</em> velocity. Therefore, the speed of the vehicle (v), in miles per hour, can be defined as distance traveled by the vehicle (d), in miles, divided by travel time (t), in hours.

First trip

45 = s / 3 (1)

Second trip

v = s / 4 (2)

By (1) and (2):

45 · 3 = 4 · v

v = 33.75 mi / h

Michael took the return trip at a velocity 33.75 miles per hour.

To learn more on velocities: brainly.com/question/18084516

#SPJ1

You might be interested in

A cone has a height of 17 years and a diameter of 28 yards. What is it’s volume?

choli [55]

I got 3,487.49 when i solved it out

5 0

3 years ago

Read 2 more answers

The number 844,758,200,030 will be rounded to

cestrela7 [59]

Answer:

Trillion because the 2 test is billion

6 0

2 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago

Mrs. Smith asked her students to make observations about the histogram, Annual Rainfall of Cities.

kakasveta [241]

Answer:

Maria and Antonio

Step-by-step explanation:

sorry if it's wrong

5 0

2 years ago

Determining a Number of Solutions

lbvjy [14]

Answer:

Infinitely Many Solutions

Step-by-step explanation:

4 0

3 years ago