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77julia77 [94]
3 years ago
9

Helpppppp I’m literally going insane

Mathematics
2 answers:
Sliva [168]3 years ago
6 0
X= -1 and X= -1.5 do the last one
Bezzdna [24]3 years ago
4 0
X = -1 and -1.5 so the 4th answer
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What is 5 and 2/3 as an improper fraction?
OLga [1]

Answer:

17/3

Step-by-step explanation:

You would need to convert 5 into a fraction with the denominator 3, so 5 = 15/3. 15/3 + 2/3 = 17/3

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3 years ago
How are the shapes alike
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Answer:

Each circle (A, B, and C) contain shapes that all share at least one characteristic. Some shapes are contained in more than one circle because they share more than one characteristic. For example, shape 3 fits the rule for circles A and B, but not circle C. It lies within circles A and B, but not circle C.

Step-by-step explanation:

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There is a claim from the 1960s that 35% of women had college educations in the 60s. To prove it had increased in the 70s, we ra
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Answer: The National Center for Education Statistics (NCES) collects, analyzes and makes available data related to education in the U.S. and other nations. Hope this helped

Step-by-step explanation: how are you?

7 0
3 years ago
Please help with this quizziz
zloy xaker [14]

Answer:

blue

Step-by-step explanation:

8 0
3 years ago
What is the correct justification for the indicated steps?
yanalaym [24]

The given proof of De Moivre's theorem is related to the operations of

complex numbers.

<h3>The Correct Responses;</h3>
  • Step A: Laws of indices
  • Step C: Expanding and collecting like terms
  • Step D: Trigonometric formula for the cosine and sine of the sum of two numbers

<h3>Reasons that make the above selection correct;</h3>

The given proof is presented as follows;

\mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1}}

  • Step A: By laws of indices, we have;

\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1} = \mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}

\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] =  \mathbf{\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}

  • Step B: By expanding, we have;

\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i  \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right]

  • Step D: From trigonometric addition formula, we have;

cos(A + B) = cos(A)·cos(B) - sin(A)·sin(B)

sin(A + B) = sin(A)·cos(B) + sin(B)·cos(A)

Therefore;

cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i  \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right] = \mathbf{ cos(k \cdot \theta + \theta) + i \cdot sin(k \cdot \theta  + \theta)}

Learn more about  complex numbers here:

brainly.com/question/11000934

4 0
2 years ago
Read 2 more answers
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