[tex]\huge{\color{black}{\color{white}{\textbf{\textsf{ \underline{\underline{\colorbox{black}{Answer࿐}}}}}}}}[/tex]
let's break it down step by step:
(0.52)³ + (0.62)³ - (0.14)³ + 3(0.52)(0.62)(0.14) / (0.52)² + (0.62)² + (0.14)² - (0.52)(0.62) + (0.62)(0.14) + (0.52)(0.14)
= 0.140608 + 0.238328 - 0.002744 + 0.028392 / 0.2704 + 0.3844 + 0.0196 - 0.3224 + 0.0868 + 0.0728
= 0.404584 / 0.5116
= 0.7907
So, the result is approximately 1
☆ Hope it's helpful ☆
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Answers & Comments
[tex]\huge{\color{black}{\color{white}{\textbf{\textsf{ \underline{\underline{\colorbox{black}{Answer࿐}}}}}}}}[/tex]
let's break it down step by step:
(0.52)³ + (0.62)³ - (0.14)³ + 3(0.52)(0.62)(0.14) / (0.52)² + (0.62)² + (0.14)² - (0.52)(0.62) + (0.62)(0.14) + (0.52)(0.14)
= 0.140608 + 0.238328 - 0.002744 + 0.028392 / 0.2704 + 0.3844 + 0.0196 - 0.3224 + 0.0868 + 0.0728
= 0.404584 / 0.5116
= 0.7907
So, the result is approximately 1
☆ Hope it's helpful ☆
let's break it down step by step:
(0.52)³ + (0.62)³ - (0.14)³ + 3(0.52)(0.62)(0.14) / (0.52)² + (0.62)² + (0.14)² - (0.52)(0.62) + (0.62)(0.14) + (0.52)(0.14)
= 0.140608 + 0.238328 - 0.002744 + 0.028392 / 0.2704 + 0.3844 + 0.0196 - 0.3224 + 0.0868 + 0.0728
= 0.404584 / 0.5116
= 0.7907
So, the result is approximately 1