All Answers 357332 Answers

Try using Demoivre's Theorem.

The possible values of w + w^2 are -2, -1, 0, 1, 2.

The r-coordinate is the distance between the point and the Earth's axis. In addition, the z coordinate is the distance between the point and the plane that contains the equator. Hence, traveling with constant latitude makes the r and z coordinate constant. Moreover, traveling with constant longitude makes the theta coordinate constant. 

one post = one question

I do not know what intersects perpendicularly means.  Perpendicular to what?

Here are your options:

Courtesy of this site:              http://www.ontrack-media.net/gateway/algebra2/g_a2m3l15s1.html    

66^2/12 = 363R0     (1)

363/12=30R3     It has a remainder so it will have a non-zero number in the 12 place

So 1

I think that method is correct

Hi MPS

I think you might do better if you report this with the report button at the bottom of the screen

I think admin will be much more likely to see it then.

I get 630

Rotations and reflections are classed as different.

What's a radius of a circle that passes through points ADGB? 

If you posted this earlier, please post a link to the original.

I have commented on 2 of your questions before.

You ignored me, or at least did not respond (yet) both times.

So will I think about this newest question? 

Maybe I cannot answer it anyway...  Maybe you will never find out.

Most likely someone else will answer it for you.

Please do not post chunks of your homework here and expect people to do it for you.

One question = one post.

like this

$sin\theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}\\~\\ RHS=\frac{[cos\theta+isin\theta]-[cos(-\theta)+isin(-\theta)]}{2i}\\ RHS=\frac{[cos\theta+isin\theta]-[cos(\theta)-isin(\theta)]}{2i}\\ RHS=\frac{2isin\theta}{2i}\\ RHS=sin\theta\\~\\ RHS=LHS \qquad \qquad Q.E.D.$

Well, how about sharing your solution with other people here. 

nevermind i have solved it

thanks, got it now ;)

!

A(x-h)^2 +k

complete the squares  for both of em

(a+bi)^2 = 2i

a^2 - b^2 + 2abi = 2i

a^2 - b^2 = 0

a^2 = b^2

2abi = 2i

ab = 1

a^2 = b^2 and ab = 1

From ab = 1, we know that either a and b are both positive or both negative, that means we can say that a = b from a^2 = b^2. 

a = b and ab = 1

a = 1, b = 1, or a = -1, b = - 1

(1 + i) or (-1 - i)

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Sorry!

Find all complex numbers $z$ such that $z^2 = 2i$.  Write your solutions in $a+bi$ form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (Don't include quotes in your answer.)  Note: This problem is not about functions.

The question isn't showing for me.

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https://web2.0calc.com/questions/quick-question_36

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f(0) = a*0^2 + b*0 + c = 0

0a + 0b + c = 0

c = 0

f(1) = a*1^2 + b*1 + c = 2

a + b + c = 2

a + b = 2

f(2) = a*2^2 + b*2 + c = 5

4a + 2b + c = 5

4a + 2b = 5

2a = 1

a = .5

b = 1.5

f(x) = .5x^2 + 1.5x + 0

f(3) = .5(3)^2 + 1.5(3) + 0 = 9

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