Ask question

Ask question

All categories

4 years ago

13

What is the absolute value of the complex number Negative 4 minus StartRoot 2 EndRoot i?

2 answers:

Makovka662 [10]4 years ago

8 0

Answer:

Step-by-step explanation:

$-4-\sqrt{2}i \\ absolute ~value=\sqrt{(-4)^2+(-\sqrt{2} )^2} =\sqrt{18}=3\sqrt{2 }$

Natali [406]4 years ago

5 0

Answer : The absolute value of the given complex is, $3\sqrt{2}$

Step-by-step explanation :

As we know that,

The complex number is, a + bi

The absolute value = $\sqrt{a^2+b^2}$

Given :

The complex number $-4-\sqrt{2}i$ .

For this complex number:

a = -4

b = $-\sqrt{2}$

Thus, the absolute value will be:

$\sqrt{a^2+b^2}=\sqrt{(-4)^2+(-\sqrt{2})^2}=\sqrt{16+2}=\sqrt{18}=3\sqrt{2}$

Thus, the absolute value of the given complex is, $3\sqrt{2}$

You might be interested in

Find the slope of each line. <br><br>y=1/4x-4

Likurg_2 [28]

Answer:

1/4

Step-by-step explanation:

y = 1/4x -4

This equation is in the form

y= mx+b where m is the slope and b is the y intercept

The slope is 1/4

6 0

3 years ago

PLEASE ANSWER AS SOON AS YOU SEE THIS

jeka57 [31]

Answer:2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Ralph went on a shopping trip. He bought a pair of shoes and a sweater. Ralph spent 12% more on the sweater than on the shoes.

IgorLugansk [536]

Answer: 112%

Step-by-step explanation:

Let x = the cost of the shoes

If the cost of the sweater is 12% more than the cost of the shoes, then 1.12x is the cost of the sweater.

To convert the 1.12x into a percentage, multiply 1.12 by 100

This gives you: 112%

4 0

2 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago

Please explain including w and z in your equation

victus00 [196]

Answer:

w = 9 × z

Hope this helps...

4 0

3 years ago