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3 years ago

Is this right? Please help!

Mathematics

1 answer:

stepan [7]3 years ago

3 0

Looks right to me all you need to do is move the decimal place over

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NEED HELP PLEASE!

USPshnik [31]

Answer:

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

Step-by-step explanation:

Complex numbers identity:

The complex numbers identity is:

$i^2 = -1$

Square of the sum:

$(x + y)^2 = x^2 + 2xy + y^2$

In this question:

$(x + (3+5i))^2$

$x^2 + 2x(3+5i) + (3+5i)^2$

$x^2 + 6x + 10xi + 9 + 30i + 25i^2$

$x^2 + 6x + 9 + (10x+30)i - 25$

$x^2 + 6x + (10x+30)i - 16$

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

5 0

3 years ago

List the following numbers from least to greatest -5/3, 7/3, -3/4, 7/5.

sukhopar [10]

-3/4, -5/3, 7/5, 7/3

6 0

3 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

3 years ago

WHAT IS THE SQUARE ROOT OF POTATO?

SIZIF [17.4K]

Answer:

Do you mean square root of pi?

Step-by-step explanation:

Because if it is then the answer is 1.772453850905516

8 0

4 years ago

Suppose there are 20 questions on a multiple choice test.if 25% of the answers are choice B how many of the answers are not choi

victus00 [196]

Answer:

Step-by-step explanation:

25% of 20 is 5 5*3=15

4 0

3 years ago