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const2013 [10]
3 years ago
15

One end point of a line segment is ( -3, -6 ) The length of the line segment is 7 units find four points that could serve as the

other end point of the given line segment
Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

Answer:

<em>(-3, 1), (-3, -13), (4, -6), (-10, -6)</em>

Step-by-step explanation:

You can think of point (-3, -6) as the center of a circle of radius 7. Then all points on the circle are 7 units from point (-3, -6). There are 4 points 7 units away from point (-6, -3) whose coordinates are easy to find. Think of a vertical line and a horizontal line that go through point (-3, -6). Up and sown from point (-3, -6) are points (-3, 1) and (-3, -13). To the right and left of point (-3, -6) are points (4, -6) and (-10, -6).

Mrac [35]3 years ago
6 0

Hey lassie. Here si your answer for this question.

Answer: (4,-6), (-3,-13), (-10,-6), and (-3,1).

Hope this helps you.

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Allisa [31]
Hey there!

Okay let's get cracking :D

The equation is....  (2x/3) - 6 = 9

2x/3 = 15
2x = 45
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4 0
3 years ago
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A cable television compan is laying cable in an area with underground utilities. Two subdivisions are located on opposite sides
Schach [20]

Answer:

Step-by-step explanation:

Given:

Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.

Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.

Note that 0 < t < 1200

Distance from P to T using Pythagoras theorem:

PT^2 = PQ^2 + QT^2

PT = √(t²+100²) = √(t²+10,000)

Distance from T to Q:

TQ = 1200 - t

From above, Cable from P to T is underwater, and costs $80/m to lay.

Cable from T to Q is underground, and costs $40/m to lay.

Total cost of the cable:

C(t) = 80√(t²+10,000) + 40(1200 - t)

Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0

dC/dt = C'(t)

= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)

dC/dt = C'(t)

= 80t/√(t² + 10,000) - 40 = 0

80t/√(t² + 10,000) = 40

80t = 40√(t² + 10,000)

2t = √(t² + 10,000)

square both sides:

4t² = t² + 10,000

3t² = 10,000

t² = 10,000/3

t = 100/√3

≈ 57.735

d^2C/dt^2 =

C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)

C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)

C''(t) = 800,000 / (t² + 10,000)^(3/2)

C''(t) > 0 for all t

Therefore C(t) is minimized at t = 100/√3

C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)

= 80 √(40,000/3) + 48,000 - 4000/√3

= 16000/√3 + 48,000 - 4000/√3

= 48,000 + 12,000/√3

= 48,000 + 4,000√3

= 4000 (12 + √3)

≈ 54,928.20

Minimum cost is $54,928.20

This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)

= 1142.265 m

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scoray [572]
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Vanyuwa [196]

Answer:

5

Step-by-step explanation:

3-(-2)=5

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mina [271]

Answer:

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