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PIT_PIT [208]
4 years ago
13

Help pls! Complex numbers :sin(2+3i).......Function Values Compute ( in the form u + iv).?

Mathematics
1 answer:
hjlf4 years ago
6 0
<span>sin(a+bi) = sina·coshb + i·cosa·sinh<span>b
sin(2+3i) = sin2·cosh3 + i·cos2·sinh3
so
u = sin2cosh3
v = cos2sinh3
</span></span>
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2. Solve the equation by completing the square. Show your work.
scoundrel [369]

Answer:

<u>x = 5 or 25</u>

Step-by-step explanation:

I think the method I am about to explain is slightly quicker and easier than the method in your question. This works for any 'complete the square' question.

We begin with x² - 30x = - 125.

First, we are going to factorise the left-hand side of the equation by <u>dividing the 'b' value </u><u>(-30)</u><u> by 2</u> (you'll see why this works in a minute):

(x - 15)²

We want these brackets to multiply out to give x² - 30x, so that they equal the left-hand side of the equation. Unfortunately, if we multiply them out, we get:

(x - 15)(x - 15) =

<u>x² - 30x + 225</u>

There is an <u>unwanted term</u> (the + 225, from 15²)! We only want x² - 30x, so we have to remove this term by <u>subtracting it from the left side of the equation</u>. To do this, let's set up the original equation again:

(x² - 30x + 225) <u>- 225</u> = - 125

<em>Note: </em><em>The reason why we don't have to subtract it from both sides is because the original equation is </em><em>x² - 30x = - 125</em><em>, and so we must make sure the left hand side is still equal to </em><em>x² - 30x</em><em>.</em>

So now we know that (x - 15)² multiplies out to give x² - 30x +225, we can write this as (x - 15)² in our equation:

(x² - 30x + 225) - 225 = - 125

is the same as:

(x - 15)² - 225 = - 125

Now add 225 to <u>both sides</u> of the equation:

(x - 15)² = - 125 + 225 = 100

(x -15)² = 100

The next step is to square root both sides. Be careful here, and remember that √100 can either be 10 or -10, as (-10)² = 100. To indicate both results, write ±10 ("plus or minus 10").

√(x - 15)² = √100

x - 15 = ±10

Because, there are two possible values for the right-hand side of the equation, we need to separate our equation into two equations:

1.     x - 15 = 10

and

2.    x - 15 = -10

Now we solve these two simple linear equations for x:

1.     x = 10 + 15   <- add 15 to both sides

        <u>x = 25</u> This is our first solution.

2.    x = -10 + 15  <- add 15 to both sides again

        <u>x = 5</u> This is our other solution.

<u>So our two solutions are x = 5 and x = 25!</u>

I have attached the quick version of the working out for this question - that is what you would be expected to write down in a test.

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4 years ago
Write four different expressions for the perimeter of a pentagon whose sides are all s-2 units long
igor_vitrenko [27]

Answer:  The answers are given below.

 

Step-by-step explanation:  Given that all the sides of a pentagon are (s-2) units long. This clearly tells us that the pentagon is a regular one with each side equal to (s-2) units.

We are to write the perimeter of the pentagon in four different expressions.

Let 'P' represents the perimeter of the pentagon.

Since it is a regular pentagon, we can write the perimeter in following four expressions.

P = (s-2) + (s-2) + (s-2) + (s-2) + (s-2),

  = 5 × (s-2),

  = 5s - 10,

  = 5s - 5×2.

Thus, these are the four expressions for the perimeter of the pentagon.

   


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