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3 years ago

Please help me with this question:)

Mathematics

1 answer:

tester [92]3 years ago

3 0

Answer:

Ada is correct.

Step-by-step explanation:

Problems with Naman:

Naman is supposed to follow PEMDAS, which is the order of operations, and stands for:

Parenthesis

Exponents (& Roots)

Multiplications

Division

Addition

Subtraction

The mistake: Naman moved +6 to the left and added it to the 2 located on the left side of the expression. You cannot do that, as it breaks the law of PEMDAS. On the other hand, Ada did the correct thing, which was to distribute 2 to all terms within the parenthesis (as one term in the parenthesis has a variable, and the other is a constant, meaning that they cannot be combined). This leads to Naman not doing the rest of the question correct.

Next, Ada correctly combined the two constants, which resulted in 0. 0 means nothing, therefore the placeholder is not needed. Therefore, 8x is the final answer.

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Answer:

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Step-by-step explanation:

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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

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<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;

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

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