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Serggg [28]
3 years ago
12

Graph the system of linear inequalities.

Mathematics
1 answer:
Murrr4er [49]3 years ago
7 0

I'm not very sure of your question... Could you please state what is your question?

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There are 3 because 12 divided by 4 is 3.
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18 - 7x = -20.52.5 = 7xx = 5/14 (c)


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How do I answer this?
Agata [3.3K]

Answer:

2k³ is the GCF.

Step-by-step explanation:

GCF of 10k^{3} ,6k^{3}

Finding the factors:

=10k³[Factoring 10k³]

=2×5×k×k×k

=6k³[Factoring 6k³]

=2×3×k×k×k

Common factors in 10k³ and 6k³.

=2×k³

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Evaluate the expression. 9−32÷4
Veseljchak [2.6K]
The answer to the question is -1
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3 years ago
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HELPPPP!!!
Anuta_ua [19.1K]

Answer:

  A = √29

Step-by-step explanation:

The short of it is that ...

  A² = 2² + 5² = 29

  A = √29

_____

<u>Amplitude</u>

If you expand the second form using the sum-of-angles formula, you get ...

  Asin(ωt +φ) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)

Comparing this to the first form, you find ...

  c₂ = 2 = Acos(φ)

  c₁ = 5 = Asin(φ)

The Pythagorean identity can be invoked to simplify the sum of squares:

  (Asin(φ))² + (Acos(φ))² = A²(sin(φ)² +cos(φ)²) = A²·1 = A²

In terms of c₁ and c₂, this is ...

  (c₁)² +(c₂)² = A²

  A = √((c₁)² +(c₂)²) . . . . . . . formula for amplitude

_____

<u>Phase Shift</u>

We know that tan(φ) = sin(φ)/cos(φ) = (Asin(φ))/(Acos(φ)) = 5/2, so ...

  φ = arctan(c₁/c₂) . . . . . . . formula for phase shift*

  φ = arctan(5/2) ≈ 1.19029 radians

___

* remember that c₁ is the coefficient of the cosine term, and c₂ is the coefficient of the sine term.

3 0
3 years ago
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