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3 years ago

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A non-stop train leaves Lahore at 4:00 pm and reaches Karachi at 10:00 am the next day. The speed of the train was 70 km/hour. F

ind the distance between Lahore and Karachi?

1 answer:

denis23 [38]3 years ago

4 0

Calculate the number of hours between 4 PM and 10 AM:

4 PM to 4 AM is 12 hours.

4 AM to 10 AM is 6 hours

Total hours is 12 + 6 = 18 hours.

18 hours x 70 km/hr = 1,260 km total.

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Answer:

$x^2 + 10x + y^2 -4y + -7 = 0$

Step-by-step explanation:

For any circle point with coordinate of center (a,b) and radius r

equation of circle is given by

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Given

coordinate of center = (-5,2)

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$(x-a)^2 + (y-b)^2 = r^2\\(x-(-5))^2 + (y-2)^2 = 6^2\\x^2 + 10x + 25 + y^2 -4y + 4 = 36\\=>x^2 + 10x + y^2 -4y + 29 = 36\\=>x^2 + 10x + y^2 -4y + 29 -36 = 36 -36\\=>x^2 + 10x + y^2 -4y + -7 = 0$

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3 years ago

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3 years ago

A water wheel has a radius measuring between 13 feet and 24 feet. The wheel is able to turn 7π9 radians from its starting positi

madam [21]

Answer:

31.76 ft and 58.64 ft

Step-by-step explanation:

The radius measures between 13 feet and 24 feet.

The wheel is able to turn 7π/9 radians before getting stuck.

We need to find the range of distances that the wheel could spin before getting stuck. That is, the length of arc.

Length of an arc is given as:

$L = \frac{\theta}{2\pi} * 2\pi r$

where θ = central angle = 7π/9 radians

r = radius of the circle

Therefore, for 13 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 13\\\\L = 31.76 ft$

For 24 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 24\\\\L = 58.64 ft$

The wheel could spin between 31.76 ft and 58.64 ft before getting stuck.

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3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$ .

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$ .

Solve this equation for $u$ :

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$ .

$u^{2} + 1 = 2 u^{2}$ .

$u^{2} = 1$ .

Given that $0 \le u \le 1$ , $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$ ,

$x = k \pi$ , where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$ ,

$0 \le k$ .

$k = 0$ or $k = 1$ . Accordingly,

$x = 0$ or $x = \pi$ .

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