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svp [43]
3 years ago
14

I’m in desperate need of help

Mathematics
1 answer:
Tomtit [17]3 years ago
4 0
I may not be able to help you but i do feel your pain lol
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Write an equation for the line that passes through the points (0, -2) and (-4, 0).
photoshop1234 [79]
The answer is C
-1/2x - 2
8 0
3 years ago
Kaylee and Olivia go to the movie theater and purchase refreshments for their friends.
KatRina [158]

The price of one bag of popcorn is $3.75

Step-by-step explanation:

Let,

x represent the cost of a bag of popcorn

y represent the cost of a drink.

According to given statement;

4x+12y=30      Eqn 1

x+6y=11.25      Eqn 2

Multplying Eqn 2 by 2

2(x+6y=11.25)\\2x+12y=22.50\ \ \ Eqn\ 3

Subtracting Eqn 3 from Eqn 1

(4x+12y)-(2x+12y)=30-22.50\\4x+12y-2x-12y=7.50\\2x=7.50

Dividing both sides by 2

\frac{2x}{2}=\frac{7.50}{2}\\x=3.75

4x+12y=30 and x+6y=11.25 can be used to find the price of one bag of popcorn and the price of one drink.

The price of one bag of popcorn is $3.75

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/11821063
  • brainly.com/question/11871074

#LearnwithBrainly

3 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 area of a triangle with a base of 1.3cm and height of 0.5 cm?
sammy [17]

Answer:

0.325 cm²

Step-by-step explanation:

A = \frac{1}{2}bh = \frac{1}{2}*1.3*0.5 = 0.325 cm²

6 0
2 years ago
Read 2 more answers
BRAINLIEST! Someone help please!
Papessa [141]

Answer:

60

Step-by-step explanation:

3 0
3 years ago
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