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Juli2301 [7.4K]

3 years ago

8

Why are I/O (input/output operations) important in database environment?

Computers and Technology

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

Database environment is created by storing the data and other related components.The input-output action takes place in the database when the reading and writing of the record units of block is to be done to and from disk unit to memory storage.

The database engine is responsible for reading action of the record before entering it to the memory and then stores it. If the data record unit is to be exported to the disk then output action takes place .It also helps in performing indexing ,vacuuming, portioning etc the information Thus, I/O is important for database.

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the volume of two similar solids are 1080cm and 1715cm .if the curved surface area of the smaller cone is 840cm .fond the curved

gavmur [86]

Answer:

$A_{big} = 1143.33cm^2$

Explanation:

The given parameters are:

$V_{small} = 1080$

$V_{big} = 1715$

$C_{small} = 840$

Required

Determine the curved surface area of the big cone

The volume of a cone is:

$V = \frac{1}{3}\pi r^2h$

For the big cone:

$V_{big} = \frac{1}{3}\pi R^2H$

Where

R = radius of the big cone and H = height of the big cone

For the small cone:

$V_{small} = \frac{1}{3}\pi r^2h$

Where

r = radius of the small cone and H = height of the small cone

Because both cones are similar, then:

$\frac{H}{h} = \frac{R}{r}$

and

$\frac{V_{big}}{V_{small}} = \frac{\frac{1}{3}\pi R^2H}{\frac{1}{3}\pi r^2h}$

$\frac{V_{big}}{V_{small}} = \frac{R^2H}{r^2h}$

Substitute values for Vbig and Vsmall

$\frac{1715}{1080} = \frac{R^2H}{r^2h}$

Recall that: $\frac{H}{h} = \frac{R}{r}$

So, we have:

$\frac{1715}{1080} = \frac{R^2*R}{r^2*r}$

$\frac{1715}{1080} = \frac{R^3}{r^3}$

Take cube roots of both sides

$\sqrt[3]{\frac{1715}{1080}} = \frac{R}{r}$

Factorize

$\sqrt[3]{\frac{343*5}{216*5}} = \frac{R}{r}$

$\sqrt[3]{\frac{343}{216}} = \frac{R}{r}$

$\frac{7}{6} = \frac{R}{r}$

The curved surface area is calculated as:

$Area = \pi rl$

Where

$l = slant\ height$

For the big cone:

$A_{big} = \pi RL$

For the small cone

$A_{small} = \pi rl$

Because both cones are similar, then:

$\frac{L}{l} = \frac{R}{r}$

and

$\frac{A_{big}}{A_{small}} = \frac{\pi RL}{\pi rl}$

$\frac{A_{big}}{A_{small}} = \frac{RL}{rl}$

This gives:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{L}{l}$

Recall that:

$\frac{L}{l} = \frac{R}{r}$

So, we have:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{R}{r}$

$\frac{A_{big}}{A_{small}} = (\frac{R}{r})^2$

Make $A_{big}$ the subject

$A_{big} = (\frac{R}{r})^2 * A_{small}$

Substitute values for $\frac{R}{r}$ and $A_{small}$

$A_{big} = (\frac{7}{6})^2 * 840$

$A_{big} = \frac{49}{36} * 840$

$A_{big} = \frac{49* 840}{36}$

$A_{big} = 1143.33cm^2$

<em>Hence, the curved surface area of the big cone is 1143.33cm^2</em>

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3 years ago