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tia_tia [17]
2 years ago
13

joseph has a new rectangular aquarium. the aquarium has a length of four feet, width of two feet and height of two feet. What is

the volume of Joseph's new aquarium? An aquarium holds one inch in lenght of fish for each twelve square inches of the area of the base of the aquarium. The fish Joseph wants to put in the new aquarium are about three inches in length. About how many fish can Joseph put in the new aquarium?
Mathematics
1 answer:
Darina [25.2K]2 years ago
3 0
12 inches go into one foot, so we can calculate the volume of the tank in inches to make the calculations that follow easier. Therefore, to calculate the volume of the tank, we use length x breadth x height = 4 x 2 x 2 = 16 square feet x 12 for square inches = 192 square inches. 
Every 12 square inches Joseph can fit a one inch fish. The fish that he has are 3 inches long, therefore he can only fit one fish every 36 square inches. 
That means that if we take the total volume of the tank and divide it by the space that a 3 inch fish will take up, we are left with 192/36 = 5.3 fish.
You cannot have a third of a fish, so we round off to the nearest whole number, and we determine that Joseph can put 5 fish in his new aquarium.
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\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}.

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\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}.

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\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}.

(Round to at least one more decimal places than the values in the choices.)

Likewise, with \angle {\rm N} is opposite to side {\rm SB}, the following would also hold:

\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}.

\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}.

\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}.

In other words, the distance between the northern lighthouse and the boat is approximately 30.73\; \rm mi, whereas the distance between the southern lighthouse and the boat is approximately 55.12\; \rm mi. Hence the conclusion.

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