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dlinn [17]
2 years ago
12

Need Help. The question is down below

Mathematics
1 answer:
madreJ [45]2 years ago
4 0

Useful Log Rules:

  • log(A*B) = log(A)+log(B)  .......... log rule 1
  • log(A/B) = log(A) - log(B) .......... log rule 2
  • log(A^B) = B*log(A)  ................... log rule 3

=========================================================

Part (a)

All logs shown below are base 3.

log(500) = log(5*100)

log(500) = log(5*10^2)

log(500) = log(5)+log(10^2) .... use log rule 1

log(500) = log(5) + 2*log(10) .... use log rule 3

log(500) = 1.4650 + 2*2.096 ....... substitution

log(500) = 5.657

<h3>Answer: 5.657</h3>

=========================================================

Part (b)

All logs shown below are base 3.

log(2) = log(10/5)

log(2) = log(10) - log(5) .... use log rule 2

log(2) = 2.096 - 1.4650  ....... substitution

log(2) = 0.631

<h3>Answer: 0.631</h3>
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A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each
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Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

r = Radius of cylinder,

h = height of the cylinder.

Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


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