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Dvinal [7]
3 years ago
7

Evaluate the function

Mathematics
1 answer:
Sedaia [141]3 years ago
4 0

f(0) =  {4}^{3 \times 0  + 3}  =  {4}^{0 + 3}  =  {4}^{3}  = 64
The answer is above.
Additional note 1:
{4}^{3}  = 4 \times 4 \times 4 = 16 \times 4 = 64
Additional note 2:
In the answer I filled in "0" at the place where "n" was. This is because the question tells us n=0
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I have no idea how to do this. I can’t cooperate with the imaginary number, please help me
NeTakaya

Answer:

Step-by-step explanation:

This is a third degree polynomial because we are given three roots to multiply together to get it.  Even though we only see "2 + i" the conjugate rule tells us that 2 - i MUST also be a root.  Thus, the 3 roots are x = -4, x = 2 + i, x = 2 - i.

Setting those up as factors looks like this (keep in mind that the standard form for the imaginary unit in factor form is ALWAYS "x -"):

If x = -4, then the factor is (x + 4)

If x = 2 + i, then the factor is (x - (2 + i)) which simplifies to (x - 2 - i)

If x = 2 - i, then the factor is (x - (2 - i)) which simplifies to (x - 2 + i)

Now we can FOIL all three of those together, starting with the 2 imaginary factors first (it's just easier that way!):

(x - 2 - i)(x - 2 + i) = x^2-2x+ix-2x+4-2i-ix+2i-i^2

Combining like terms and canceling out the things that cancel out leaves us with:

x^2-4x+4-i^2

Remembr that i^2=-1, so we can rewrite that as

x^2-4x+4-(-1) and

x^2-4x+4+1=x^2-4x+5

That's the product of the 2 imaginary factors.  Now we need to FOIL in the real factor:

(x+4)(x^2-4x+5)

That product is

x^3-4x^2+5x+4x^2-16x+20

which simplifies down to

x^3-11x+20

And there you go!

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4 years ago
How many solutions are in the equation below?
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Only one solution.

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