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3 years ago

Evaluate the expression 4-2x2+3

Mathematics

1 answer:

lutik1710 [3]3 years ago

8 0

Answer:3

Step-by-step explanation:

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I need with #10, please?

777dan777 [17]

Answer:

h=7ft

Step-by-step explanation:

sinФ=opposite/hypotenuse

sin45°=5/h

0.71=5/h

multiply via by h

0.71h=(5/h)h

h cancel h it remains 5

0.71h=5

divide through by 0.71

0.71h/0.71=5/0.71

h=7.04ft

h=7ft approximately

4 0

3 years ago

Which describes the cris section of the cube that passes through the vertices A, B, and C shown below? A rectangle that is not a

kow [346]

Answer:

Step-by-step explanation:

brainlist is tnx:))

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;

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