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2 years ago

A woman jogs for 2 km at 8 km/h and then walks for 2 km

Mathematics

1 answer:

Artemon [7]2 years ago

3 0

Answer:

6 6/7 km/h

Step-by-step explanation:

The relations between speed, time, and distance are ...

time = distance/speed

speed = distance/time

So, to find the average speed, we need to know the total distance and the total time. The distances are given, but we need to compute the times.

time jogging = (2 km)/(8 km/h) = 1/4 h

time walking = (2 km)/(6 km/h) = 1/3 h

Then the woman's average speed is ...

average speed = (total distance)/(total time) = (2 km + 2 km)/(1/4 h + 1/3 h)

= (4 km)/(7/12 h) = 48/7 km/h

= 6 6/7 km/h

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A central angle theta in a circle of radius 7 m is subtended by an arc of length 8 m. Find the measure of theta in degrees. (Rou

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Answer: In degrees , The measure of $\theta=65.5^{\circ}$

In radians , the measure of $\theta =\dfrac{8}{7}\text{ radians}$ .

Step-by-step explanation:

We know that the formula for length of arc is given by :-

$l=\theta r$

, where $\theta$ = Central angle subtended by arc.

r= radius of the circle.

As per given , we have

Radius of circle : r=7 m

Length of arc : l= 8 m

Substitute these values in the above formula , we get

$8=\theta (7)\\\\\Rightarrow\ \theta =\dfrac{8}{7}\text{ radians}$

Hence, the measure of $\theta =\dfrac{8}{7}\text{ radians}$ .

To convert it into degrees we multiply it with $\dfrac{180^{\circ}}{\pi}$

The measure of $\theta =\dfrac{8}{7}\times\dfrac{180^{\circ}}{\pi}$

$=(\dfrac{8\times180}{7\times3.14})^{\circ}=65.5141037307^{\circ}$

$\approx65.5^{\circ}$

Hence, the measure of $\theta=65.5^{\circ}$

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2 years ago

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2 years ago

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2 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

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2 years ago