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3 years ago

13

What is the answer? please I need help

2 answers:

AfilCa [17]3 years ago

3 0

Answer:

it's blurry but I think it says p

Vlad [161]3 years ago

3 0

They’re right. It’s P

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I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago

Simplify the expression

skelet666 [1.2K]

This is the expression simplified:

4 0

3 years ago

In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th

iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ ) = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ; 4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

= 61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from 0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ = 0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

6 0

3 years ago

Find the product of 5/6( 1/2) *<br> O 6/8<br> O 5/12<br> O 5/6

inysia [295]

The anser is 6/8 that what i think

6 0

3 years ago

Read 2 more answers

Skip counting represents 8×4

Blizzard [7]

It represents that because multiplication is like adding continually so it represent 8+8+8+8 which is 8 x 4

5 0

3 years ago