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Anna007 [38]
2 years ago
13

A paratrooper falls to the ground along a diagonal line. His fall begins 1683 meters above the ground, and the line he follows h

as a slope
of 1.7. That is, he falls 1.7 meters vertically for every 1 meter he moves across horizontally. How far horizontally across the ground does
he land from his initial position in the sky?
Mathematics
1 answer:
iragen [17]2 years ago
4 0

The distance horizontally across the ground the  he land from his initial position is 990 meters.

The slope of a line shows how steep the line is. The slope is calculated using the formula:

Slope = vertical rise / horizontal run

From the question, the slope is 1.7. Hence:

Slope = vertical rise / horizontal run

1.7 = 1683 m / horizontal run

Horizontal run = 990 meters

Hence, the distance horizontally across the ground the  he land from his initial position is 990 meters.

Find out more on slope at: brainly.com/question/3493733

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Please Help. I really don’t get this concept, if you could explain it in detail it would be very appreciated
Dafna11 [192]
<h3>Answer: 24/25</h3>

=================================================

Explanation:

Sine is given to be negative, and so is tangent. This only happens in quadrant Q4

Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.

If tan(theta) < 0, then this means cos(theta) > 0

So we have y < 0 and x > 0 which places the angle somewhere in Q4.

--------------------------

Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24

So this is what your steps may look like

a^2+b^2 = c^2

7^2+b^2 = 25^2

b^2+49 = 625

b^2 = 625-49

b^2 = 576

b = sqrt(576)

b = 24

So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25

---------------------------

As an alternative, you could use the trig identity

sin^2(x) + cos^2(x) = 1

and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.

So,

sin^2(x) + cos^2(x) = 1

(-7/25)^2 + cos^2(x) = 1

(49/625) + cos^2(x) = 1

cos^2(x) = 1 - (49/625)

cos^2(x) = (625/625) - (49/625)

cos^2(x) = (625-49)/625

cos^2(x) = 576/625

cos(x) = sqrt(576/625)

cos(x) = sqrt(576)/sqrt(625)

cos(x) = 24/25

This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)

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4 years ago
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9 and 81.
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3 years ago
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Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair whic
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Answer:

Let f_n be the number of rabbit pairs at the beginning of each month. We start with one pair, that is f_1 = 1. After one month the rabbits still do not produce a new pair, which means f_2 = 1. After two months a new born pair appears, that is f_3 = 2, and so on. Let now n \geq 3 be any natural number. We have that f_n is equal to the previous amount of pairs f_n-1 plus the amount of new born pairs. The last amount is f_n-2, since any two month younger pair produced its first baby pair. Finally we have  

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