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geniusboy [140]
2 years ago
15

Point K is on line segment JL. Given KL = 2x – 2, JL = 4x + 9, and

Mathematics
1 answer:
sergeinik [125]2 years ago
7 0

Answer:

JL = 21

Step-by-step explanation:

Given that K is on line segment JL, therefore:

KL + JK = JL (according to segment addition postulate)

KL = 2x - 2

JK = 5x + 2

JL = 4x + 9

Thus:

(2x - 2) + (5x + 2) = (4x + 9)

Solve for x

2x - 2 + 5x + 2 = 4x + 9

2x +5x - 2 + 2 = 4x + 9

7x = 4x + 9

Subtract 4x from both sides

7x - 4x = 4x + 9 - 4x

3x = 9

Divide both sides by 3

\frac{3x}{3} = \frac{9}{3}

x = 3

Find the numerical length of JL

JL = 4x + 9

Plug in the value of x

JL = 4(3) + 9 = 12 + 9 = 21

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Answer:

dy/dt = 1.02 m/s

Step-by-step explanation:

First of all, let the distance of the dock above the bow of the boat be represented by x.

Also, let the distance of the boat from the dock be represented by y.

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Now, we can use pythagoras theorem to find a relationship between x, y and z since they form a right angle triangle where the vertical part is the dock, the base represents the water and the hypotenuse represents the rope.

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We are told the dock is 5 feet away from the boat. Thus, y = 5

dy/dt is is the speed we are looking for which is the rate at which the boat is approaching the dock.

We are not given the value of z but we can find it by using pythagoras theorem since we know the dock is 1 meter above the bow and the boat is 5 m away.

Thus;

z = √(1² + 5²)

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We are told the rope is being pulled in at the rate of 1 m/s. Thus; dz/dt = 1

Plugging all relevant values into the differentiation equation above gives;

2(1)(0) + 2(5)(dy/dt) = 2(√26)(1)

10(dy/dt) = 2√26

Divide both sides by 10 to get:

dy/dt = (√26)/5

dy/dt = 1.02 m/s

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