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Evgesh-ka [11]
3 years ago
5

A rectangular prism has a length of 4 1/4 in a width of 3 in, and a height of 1 1/4 in.

Mathematics
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

4 1/4 × 3 × 1 1/4

Step-by-step explanation:

To find volume you need to do length×width×height

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12 x -7 = -84 12 + (-7) = 5

Step-by-step explanation:

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5

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It costs $11.18 to buy 13 cans of cranberry sauce. If the cans all cost the same amount, what is the price of each can?
AnnyKZ [126]

Answer:

Each can costs $0.86

Step-by-step explanation:

We find the unit price by dividing the cost by the number of cans.

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8 0
2 years ago
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Write an equation in point-slope form of the line that passes through (3, -8) and (7,-2).
sashaice [31]

Answer:

A line that passes through (3, -8) and (7,-2).

Denote equation of that line: y = ax + b

=> Slope a can be directly determined by:

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=> y = (2/3)x + b

This line passes (3, -8), then: -8 = (2/3)*3 + b => -8 = 2 + b => b = -10

=> y =(2/3)x - 10

There are other ways to find equation of line.

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3 0
2 years ago
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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) =
Dafna1 [17]

Answer:

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

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:

- The length of fish grows with time. It is modeled by the relation:

                                   L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(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:

- The rate of change of length of a fish as it ages each year  can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

                             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)

                             e^(0.18*t) = 34.416 / 6

                             0.18*t = Ln(34.416/6)

                             t = Ln(34.416/6) / 0.18

                             t = 10 year

7 0
3 years ago
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