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3 years ago

The fastest human has ever run is 27 miles per hour. how many miles per minute did the human run?

1 answer:

7 0

<span>If you take twenty seven and divide by sixty you get 0.45. So the equation 27/60=0.45 would show that the fastest human ever was running at a speed of 0.45 miles every minute. .</span>

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Consider the first order differential equation

AURORKA [14]

Answer:

For y(-7) =6.4

The largest interval is between

$-\infty \to -5$

For y(-2.5) = -0.5.

The largest interval is between

$-5 \to 5$

For y(0) = 0

The largest interval is between

$-5 \to 5$

For y(4.5) = -2.1.

The largest interval is between

$-5 \to 5$

For y(14)= 1.7.

The largest interval is between

$9 \to \infty$

Step-by-step explanation:

From m the question we are told that

The first order differential equation is $\frac{y' - t}{ t^2 -25} = \frac{e^t}{t-9}$

Now the first step is to obtain the domain of the differential equation

Now to do that let consider the denominators

Now generally

$t^2 - 25 \ne 0$ side calculation

=> $t\ne \pm5$ $t^2 - 25 = 0$

$t = \pm 5$

Also $t-9\ne 0$ $t -9 = 0$

=> $t\ne 9$ $t= 9$

This means that this first order differential equation is discontinuous at

$t = -5 , \ \ t = 5 \ \ t = 9$

$This \ is \ illustrated \ below \ \\ ------------------\\\. \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ |\\. \ -5 \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ 9$

For y(-7) =6.4

The largest interval is between

$-\infty \to -5$

For y(-2.5) = -0.5.

The largest interval is between

$-5 \to 5$

For y(0) = 0

The largest interval is between

$-5 \to 5$

For y(4.5) = -2.1.

The largest interval is between

$-5 \to 5$

For y(14)= 1.7.

The largest interval is between

$9 \to \infty$

4 0

3 years ago

When lava is expelled from a volcano, it temperature may vary from 700 degrees Celsius to 1200 degrees Celsius. Write an absolut

ValentinkaMS [17]

Answer:

Ix - 950°C I ≤ 250°C

Step-by-step explanation:

We are told that the temperature may vary from 700 degrees Celsius to 1200 degrees Celsius.

And that this temperature is x.

This means that the minimum value of x is 700°C while maximum of x is 1200 °C

Let's find the average of the two temperature limits given:

x_avg = (700 + 1200)/2 =

x_avg = 1900/2

x_avg = 950 °C

Now let's find the distance between the average and either maximum or minimum.

d_avg = (1200 - 700)/2

d_avg = 500/2

d_avg = 250°C.

Now absolute value equation will be in the form of;

Ix - x_avgI ≤ d_avg

Thus;

Ix - 950°C I ≤ 250°C

3 0

3 years ago

Simplify the following : x^2y^-3 / x^5y^2

Likurg_2 [28]

Answer:

Step-by-step explanation:

$\frac{a^{m}}{a^{n}}=a^{m-n}\\\\a^{-m}=\frac{1}{a^{m}}\\\\\\\frac{x^{2}y^{-3}}{x^{5}y^{2}}=x^{(2-5)}*y^{(-3-2)}\\\\ = x^{-3}*y^{-5}\\\\=\frac{1}{x^{3}y^{5}}$