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ANEK [815]
4 years ago
11

Determine whether the function is odd, even, neither, or can’t be determined: g(x) = x2 + x Question options:

Mathematics
2 answers:
11111nata11111 [884]4 years ago
8 0

The answer to your question is neither.

Anastasy [175]4 years ago
7 0

I have already done this question. The correct answer is c

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Answer:

y is 29 degrees

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* if my savings of $x grows 10% each year how much money would i have in 1 year
german

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10/100*$x

Step-by-step explanation:

You will multiply the $x by 10%

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Please help me....<br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B22%7D%7B7%7D%20%20%20%5Cdiv%20%20%5Cfrac%7B22%7D%7B7%7D
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<h3><u>Answer</u> :</h3>

\sf \dashrightarrow \dfrac{22}{7} \div \dfrac{22}{7}

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Find the value of 1÷3x when X=1​
Keith_Richards [23]

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3 years ago
How to solve this trigonometric equation cos3x + sin5x = 0
mrs_skeptik [129]

Answer:

  x = {nπ -π/4, (4nπ -π)/16}

Step-by-step explanation:

It can be helpful to make use of the identities for angle sums and differences to rewrite the sum:

  cos(3x) +sin(5x) = cos(4x -x) +sin(4x +x)

  = cos(4x)cos(x) +sin(4x)sin(x) +sin(4x)cos(x) +cos(4x)sin(x)

  = sin(x)(sin(4x) +cos(4x)) +cos(x)(sin(4x) +cos(4x))

  = (sin(x) +cos(x))·(sin(4x) +cos(4x))

Each of the sums in this product is of the same form, so each can be simplified using the identity ...

  sin(x) +cos(x) = √2·sin(x +π/4)

Then the given equation can be rewritten as ...

  cos(3x) +sin(5x) = 0

  2·sin(x +π/4)·sin(4x +π/4) = 0

Of course sin(x) = 0 for x = n·π, so these factors are zero when ...

  sin(x +π/4) = 0   ⇒   x = nπ -π/4

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The solutions are ...

  x ∈ {(n-1)π/4, (4n-1)π/16} . . . . . for any integer n

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3 years ago
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