Se supone que si es 5.31 x 7.6 los límites son 38.98 ahora si fuera en suma mueves los puntos dos veces a la izquierda la sumatoria seria la siguiente .00531 + .0076 la respuesta seria

.00607

4 0

3 years ago

At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug

Rom4ik [11]

- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

$\tan \theta=\frac{6}{x}$

Now, differentiate with respect to t, we get

$\sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}$

Now, calculate the travel distance from radar station to plane after 12min

Distance, $x=800 \times \frac{12}{60}=160$

Substituting ‘x’ value, we get

$\tan \theta=\frac{6}{160}=\frac{3}{80}$

Find the rate of change of theta after 12 min,

$\frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}$

We know, the formula for,

$\sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}$

So, then, $\frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })$

$\frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{1.0014} \times \frac{6}{25600} \times 800$

$=-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}$

When express the value in km/he, we get, the change in theta as

$=-187.237 \mathrm{km} / \mathrm{hr}$

5 0

3 years ago

Why does the temperature increase as we go underground?

bearhunter [10]

Due to the decay of naturally radioactive substances, or even due to the heat made by the movement of the tectonic plates, and formation of gravitational acceleration, the temperature increases.

6 0

3 years ago

Please explain how you got the answer :(

tensa zangetsu [6.8K]

Answer:

36.5 kJ

Explanation:

Energy by machine = energy to machine × efficiency

E = 52.1 kJ × 0.70

E = 36.5 kJ

3 0

3 years ago