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Ilya [14]
3 years ago
15

Given the following prescription formula, what is the ratio strength (nearest whole number) of methylcellulose in the finished p

roduct? As a ratio is typically expressed as 1:some number, put ONLY the number in the space provided and NOT the 1: portion. DO NOT include any units. • Progesterone 3.8 g • Glycerin 7 mL • 2% methylcellulose solution 50 mL • Cherry syrup ad 90 mL Your Answer: Answer
Mathematics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

147

Step-by-step explanation:

Given:

Progesterone =  3.8 g

Glycerin = 7 mL

2% methylcellulose solution 50 mL

Cherry syrup ad = 90 mL

Now,

The total volume of the solution = 7 + 50 + 90 = 147 mL

Also,

2% methylcellulose solution 50 mL is concluded as:

the volume of  methylcellulose in the solution is 2% of the total volume of the solution

thus,

volume of methylcellulose = 0.02 × 50 mL = 1 mL

Therefore,

Ratio strength of methylcellulose in the finished product

=\frac{\textup{volume of methylcellulose}}{\textup{ total volume of the solution}}

or

= \frac{\etxtup{1}}{\textup{ 147}}

Hence, the answer according to the question is 147

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Answer:

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Given

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The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =
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b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year                                          

Step-by-step explanation:

Given:

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t: Is the age of the fish in years.

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(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

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                             dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

                             dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

                            L'(t) = 34.416*e^(-0.18*t)

                            L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

                            L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

                            L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

                             6 cm/yr = 34.416*e^(-0.18*t)

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                             0.18*t = Ln(34.416/6)

                             t = Ln(34.416/6) / 0.18

                             t = 10 year

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