$Hi \\ \\ x+9\ \textgreater \ -4 \\ \\ x\ \textgreater \ -4-9 \\ \\ \boxed{x\ \textgreater \ -13} \\ \\ S=\{x\in~\mathbb{R} |x\ \textgreater \ -13\} \\ \\ \\$

3 0

3 years ago

Read 2 more answers

Travis jogged to his friend's house 10 miles away and then got a ride back home. It took him 2 hours longer to jog there than ri

spin [16.1K]

Using the relation between velocity, distance and time, it is found that his jogging rate was of 4.5 mph.

<h3>What is the relation between velocity, distance and time?</h3>

Velocity is <u>distance divided by time,</u> hence:

$v = \frac{d}{t}$

Jogging, his velocity was of v, while the time was of t + 2, for a distance of 10 miles, hence:

$v = \frac{10}{t + 2}$

Riding, his velocity was of 16, while the time was of t, also for a distance of 10 miles, hence:

$v + 16 = \frac{10}{t}$

Then, considering the first equation and replacing in the second:

$\frac{10}{t + 2} + 16 = \frac{10}{t}$

$\frac{10}{t} - \frac{10}{t + 2} = 16$

$\frac{10t + 20 - 10t}{t(t + 2)} = 16$

16t² + 32t - 20 = 0 -> t = 0.5.

Then his jogging rate in mph was:

$v = \frac{10}{0.5 + 2} = 4.5$

More can be learned about the relation between velocity, distance and time at brainly.com/question/24316569

#SPJ1

8 0

2 years ago

Of a government by whom money is paid to support services​

Of a government by whom money is paid to support services