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3 years ago

13

Difference between system and software development life cycle

Computers and Technology

1 answer:

umka2103 [35]3 years ago

3 0

Answer:

Software development life cycle is about building a software (“only”) in a phased approach systematically.

System development life cycle is about implementing hardware and software in a phased manner systematically.

Explanation:

Here both the hardware and the software is considered as a system.

Hope this helps!

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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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What would happen to a eukaryotic cell if all its mitochondriawere destroyed

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The cell won't reproduce ATP.

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ra1l [238]

Answer:

1 c

2 a

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4 b

5 c

6 b

7 d

8 a

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Where a database is stored

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Answer:

A

Explanation:

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