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3 years ago

4.

Mathematics

2 answers:

UkoKoshka [18]3 years ago

5 0

Answer:A

Step-by-step explanation:

RideAnS [48]3 years ago

5 0

The answer is a hope this helps

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When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,

kondaur [170]

Answer:

The combined resistance of a circuit consisting of two resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$

where

R is the combined resistance

$r_1, r_2$ are the two resistors

We can re-write the expression as follows:

$\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}$

$R=\frac{r_1 r_2}{r_1+r_2}$

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

To solve this part, we use again the expression for R written in part a:

$R=\frac{r_1 r_2}{r_1+r_2}$

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

$r_1>0$

So the interval of allowed values for r1 is

$0$

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

$r_1=b$

6 0

3 years ago

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