Peters Olson
08/21/2024 · Elementary School

Multiply the complex numbers and simplify the result. (Express your answer in the form \( (a+b r) \). \( ) \) \( (3+47)(5-2 \pi) \)

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To multiply the complex numbers \( (3 + 47) \) and \( (5 - 2\pi) \), we can treat \( 3 + 47 \) as a complex number where \( 3 \) is the real part and \( 47 \) is the imaginary part. Thus, we can rewrite it as \( 3 + 47i \). Now, we will multiply \( (3 + 47i) \) and \( (5 - 2\pi) \): \[ (3 + 47i)(5 - 2\pi) \] Using the distributive property (also known as the FOIL method for binomials), we have: \[ = 3 \cdot 5 + 3 \cdot (-2\pi) + 47i \cdot 5 + 47i \cdot (-2\pi) \] Calculating each term: 1. \( 3 \cdot 5 = 15 \) 2. \( 3 \cdot (-2\pi) = -6\pi \) 3. \( 47i \cdot 5 = 235i \) 4. \( 47i \cdot (-2\pi) = -94\pi i \) Now, we can combine these results: \[ = 15 - 6\pi + 235i - 94\pi i \] Next, we combine the real parts and the imaginary parts: Real part: \[ 15 - 6\pi \] Imaginary part: \[ (235 - 94\pi)i \] Thus, the final result is: \[ (15 - 6\pi) + (235 - 94\pi)i \] Expressing this in the form \( (a + br) \), we have: \[ (15 - 6\pi) + (235 - 94\pi)i \] So, the answer is: \[ (15 - 6\pi) + (235 - 94\pi)i \]

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\( (15 - 6\pi) + (235 - 94\pi)i \)
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