ilorty

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Here is a hint:

Change z to a+bi. Since the magnitude is the square root of a^2+b^2, solve for that. 

To make myself more clear, let me show you how to do the left side, and then you can to the right side.

Once we have changed z to a+bi, we can subtract 3 from a. This will give us (a-3)+bi.

Now, squaring a and b, we get

a^2-6a+9+b^2 for the inside of the square root. Therefore, the left side is $\sqrt {a^2-6a+9+b^2}$.

Now, try to do the right side, and solve the equation.

Here are some hints for each angle:

x: In a pentagon, the sum of the angles is 540 degrees. Since this is a regular pentagon, what can you do to get x? In other words, divide 540 by the amount of angles in the pentagon.

y: Since a regular pentagon has all equal sides, the triangle on the top is an isoceles triangle. Therefore, subtract x from 180 and have that equal to 2y. Then, solve for y.

z: Since the pentagon is rotatable, the bottom triangle is the same as the top. Once you have solved for y, subtract that from x, and you will have z. 

Hmm... To me, this looks like maybe homework? Instead of giving you the answer, here are some hints:

Add the last two equations together, what do you get?

Value of '?'? Do you mean c? If so, I can help, but in the future, please make sure there are no typos in your question, to make the answerer's life easier.

To begin, we first need to distribute 2(4+cy). This gives us 8+2cy. From here, we can subtract 8 from both sides and get 2cy=12y. Dividing both sides by y, we get 2c=12. Isolating the c, we get 6. Therefore, c is equal to 6.

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since this is a test, I dont think anyone can help you, as that is cheating. think back to the formulas you've learned in class, and try to figure it out yourself. 

And, make sure they are even

If you need help for getting the outside, all you have to do is list all the numbers from 20 to 50 that does not include numbers from A and B.