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Nadusha1986 [10]
3 years ago
12

A nut mixture of almonds and cashew nuts at a small fair is $1.00 per pound of almonds and $3.55 per pound of cashew nuts. Over

the entire day, 65 pounds of the nut mixture were sold for $189.95. If p is the number almonds and n is the number of cashew nuts, then the system of equations that models this scenario is: p+n=65p+3.55n=189.95p+n=65p+3.55n=189.95 Determine the correct description and amount of pounds for almonds and cashew nuts that were sold. A (26.0, 39.0)There were 26.0 pounds of almonds and 39.0 pounds of cashew nuts sold at the fair. B (16.0, 49.0)There were 16.0 pounds of almonds and 49.0 pounds of cashew nuts sold at the fair. C (49.0, 16.0)There were 49.0 pounds of almonds and 16.0 pounds of cashew nuts sold at the fair. D (39.0, 26.0)There were 39.0 pounds of almonds and 26.0 pounds of cashew nuts sold at the fair.
Mathematics
2 answers:
RSB [31]3 years ago
5 0

Answer:

B. (16.0, 49.0) There were 16.0 pounds of almonds and 49.0 pounds of cashew nuts sold at the fair.

Step-by-step explanation:

Here, p represents the number of pounds of almond and n be the number of pounds of cashew nuts,

Since, total nuts = 65

⇒ p + n = 65 ----(1)

Also, the total spending is $ 189.95 in which almond costs $ 1.00 per pounds and cashew costs $ 3.55 per pound,

⇒ p + 3.55n = 189.95 -----(2)

Equation (1) - equation (2),

2.55n = 124.95

⇒ n = 49

From equation (1),

p + 49 = 65 ⇒ p = 16

Hence, there are 16 pounds of almonds and 49 pounds of cashew.

Cloud [144]3 years ago
3 0
The answer would be ( C )
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alexandr402 [8]

Answer:

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Step-by-step explanation:

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<em>With values substituted</em>: 5 dm³/s = 2π·(120 dm)·(0.2 dm)·r'(t) . . . . 1 L = 1 dm³

<em>Final answer</em>:

... r'(t) = (5 dm³/s)/(2π·24 dm²) = 5/(48π) dm/s ≈ 0.0331573 dm/s ≈ 3.32 mm/s

b) <em>Relation</em>: In addition to the above relation between volume and radius, we also have the relation between volume and time:

... V(t) = ∫V'(t)·dt = V'(t)·t . . . . . since V'(t) is a constant

... r(t) = √(V(t)/(πh)) = √(V'(t)t/(πh))

<em>Derivative</em>:

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Then

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<em>With values substituted</em>: k = ((5 dm³/s)/(π·0.2 dm)) = √((25/π dm²/s)

... k = (5/√π) (dm/√s)

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_____

<em>Comment on liters in volume calculations</em>

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Here, we're given a radius in meters. The units of radius rate seem best presented in mm/s. Messing with meters or centimeters or millimeters would add powers of 10 that would only confuse the calculations.

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3 years ago
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Step-by-step explanation:

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Answer:

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