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Troyanec [42]
3 years ago
9

Write the expression. Then, check all that apply. six times the sum of nine and a number Replace "a number with the variable, n

The two operations are multiplication and addition The two operations are multiplication and subtraction Key Words Replace With six 6 The constants are 6 and 9 times The expression is written as 6n + 9 The expression is written as 6/9 + n) the sum of nine 9 a number​
Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer: A,B,D,F

A. Replace "a number with the variable, n

B. The two operations are multiplication and addition

C. NO

D. The constants are 6 and 9

E. NO

F. The expression is written as 6(9 + n)

Step-by-step explanation:

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I have no idea how to do this. I can’t cooperate with the imaginary number, please help me
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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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The consecutive numbers are...

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