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3 years ago

Please answer w/ an explanation of number system. Also tell me if you want to do my Khan Academy

Mathematics

2 answers:

blagie [28]3 years ago

4 0

Answer:

B: Integer

Step-by-step explanation:

I also did that khan academy and got it right

Nataliya [291]3 years ago

3 0

Answer:

None of the choices

Irrational, Real, Complex

Step-by-step explanation:

Number system:

(since I'm having image upload trouble I'll just do an outline format)

Complex numbers (all numbers we have)

I. Imaginary

- cannot be plotted on number line

-include i, sqrt of (-) #s, variables, |-#s|, etc.

(This includes pure imaginary and imaginary #s: In terms of i, pure imaginary is only b*i, and imaginary is a+bi, a and b being integers)

II. Real numbers: can be plotted on # line (have a real value)

A. Irrational numbers

- Non-repeating, non-terminating (ending) decimals

E.X. : 1.12343123510971324...

- Cannot be expressed as a fraction of 2 integers

- Has a real value (pi, phi, tau, square roots of non-perfect non-negative squares e.x. \sqrt2)

B. Rational

- Can be expressed as fraction of 2 integers

E.x. 0.3333... = 1/3

- Can be expressed as terminating or repeating decimal

E.x. 0.3232... , 0.32, etc.

1. Includes: Fractions and terminating/repeating decimals

2. Integers: ..., -3, -2, -1, 0, 1, 2, 3, ...

a. Whole numbers: 0, 1, 2, 3, ...

i. Counting/natural numbers: 1, 2, 3, ...

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MA_775_DIABLO [31]

Answer:

$2*log(x)+log(y)$

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

Logarithm of a power:

$log_ba^c=c*log_ba$

So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: $log_ba=x\implies b^x=a$

Now if we raise both sides to the power of c, we get the following equation: $(b^x)^c=a^c$

Using the exponential identity: $(x^a)^c=x^{a*c}$

We get the equation: $b^{xc}=a^c$

If we convert this back into logarithmic form we get: $log_ba^c=x*c$

Since x was the basic logarithm we started with, we substitute it back in, to get the equation: $log_ba^c=c*log_ba$

Now the second logarithmic property you need to know is

The Logarithm of a Product:

$log_b{ac}=log_ba+log_bc$

Now for a quick proof, let's just say: $x=log_ba\text{ and }y=log_bc$

Now rewriting them both in exponential form, we get the equations:

$b^x=a\\b^y=c$

We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

$b^x*b^y=a*c$

Using the exponential identity: $x^{a}*x^b=x^{a+b}$ , we can rewrite the equation as:

$b^{x+y}=ac$

taking the logarithm of both sides, we get:

$log_bac=x+y$

Since x and y are just the logarithms we started with, we can substitute them back in to get: $log_bac=log_ba+log_bc$

Now let's use these identities to rewrite the equation you gave

$log(x^2y)$

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

$log(x^2)+log(y)$

Now using the logarithm of a power to rewrite the log(x^2) we get:

$2*log(x)+log(y)$

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Answer:

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