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jenyasd209 [6]
2 years ago
5

Use the elimination method to solve the system of equations. Choose the correct ordered pair. 2y = x + 2 x - 3y= -5 O A. (2, 2)

O B. (4,3) O C. (6,4) O D. (8,5)​
Mathematics
1 answer:
Lina20 [59]2 years ago
5 0

Answer:

correct answers: 2 question: Solve the given system of equations using any method and then choose the correct ordered pair from the given choices: x + y = 33x - y ... Select one: O a. (2.1) O b.(-2,-1) O c. (1,2) O d. (-1,2)​. answer. Answers: 2. Get ... Asine function has the following period =4 amplitude = 4 midline y = 1.

PLZ MARK BRAINLIEST I NEED IT ALSO HOPE THIS HELPS

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I have no idea how to do this, can someone please help?
Sveta_85 [38]
Assume your line starts at zero, your first point is (-3,5) meaning your have a slope of 5/-3
[ f(x) = 5/-3x + b] 
4 0
3 years ago
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

5 0
3 years ago
What is the scale factor that takes the large triangle to the small triangle?
Lerok [7]

Step-by-step explanation:

such a scale factor applies to all sides and lines.

we see that DE is 12 units, and AB is 4 units.

so,

12×f = 4

f = 4/12 = 1/3 is the scale factor

control : check the other side

FE = 6 units

6 × 1/3 = 2

so, the corresponding side in the small triangle must be 2 units long.

and indeed, CB is 2 units long. all correct.

5 0
2 years ago
 A store is instructed by corporate headquarters to put a markup of 67​% on all items. An item costing ​$12 is displayed by the
Elis [28]

Answer:

20.04

Step-by-step explanation:

He marked it down to 67%. Not up

7 0
3 years ago
A castle is surrounded by a circular moat which is 5 m wide
Elanso [62]

Answer:

  1414 kL

Step-by-step explanation:

The volume of the donut-shaped moat is the product of its surface area and depth. The area is the product of its centerline length and its width.

<h3>Moat area</h3>

The diameter of the centerline of the moat is (50 m -5 m) = 45 m. The length of that centerline is ...

  C = πd = π(45 m) = 45π m

The area is this length times the width of the moat:

  moat area = (45π m)(5 m) = 225π m²

<h3>Moat volume</h3>

The volume is the product of the area and the depth of the moat:

  V = Ah = (225π m²)(2 m) = 450π m³ ≈ 1413.72 m³

1 cubic meter is 1000 liters, 1 kiloliter.

The volume of the moat is about 1414 kL.

5 0
1 year ago
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