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Vinvika [58]
4 years ago
9

Let statements p, q, r, and s be as follow:

Mathematics
1 answer:
Bess [88]4 years ago
7 0
The line up should go in to this order.
qp
pq
pq

qp
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A tube-feeding formula is administered to a patient via Gastrostomy at 75
melomori [17]

Answer:

97.5ml protein

Step-by-step explanation:

20hr * 75ml/hr = 1500ml

1500ml * .065 = 97.5ml protein

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What is the constant of variation?
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4 is the constant varriation

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3 years ago
Plss it due today plssss don’t just take the points
Murljashka [212]

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Step-by-step explanation:

3x + 11 + x - 9 + 56 = 180

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6 0
3 years ago
Solve the equation.
djverab [1.8K]
(y - 5)/3 = 1
y - 5 = 3 x 1 = 3
y = 3 + 5 = 8

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7 0
3 years ago
Read 2 more answers
An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0
3 years ago
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