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erastovalidia [21]
3 years ago
13

write an expression to describe the sequence below.Use n to represent the position of a term in the sequence, where n=1 for the

first term
Mathematics
1 answer:
pashok25 [27]3 years ago
3 0
Ez 400 pionts isi quadrant to tan x
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Elodia [21]
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5 0
3 years ago
What is the factored form of the polynomal? x2-12x+27
lianna [129]

Answer:

The answer is (x-9) (x-3).

Step-by-step explanation:

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When you add the like terms you get,

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6 0
3 years ago
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AleksAgata [21]
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6 0
3 years ago
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7x + 3y + 2x<br> Please help me out!!<br> (simplify please)
Dmitry [639]

Answer:

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6 0
3 years ago
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Select True or False for each statement.
SSSSS [86.1K]

\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32

TRUE

\sqrt[5]{36^4}=36^{4/5}

which surely isn't 36.  FALSE

\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0

FALSE

For the fourth one we have a

\sqrt{98b} + \sqrt{2b}

which isnt

10\sqrt{b}

so this is FALSE.

\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}

= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}

=(\sqrt 5 + \sqrt 6)^2

No fractions in that one so FALSE.

3 0
2 years ago
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