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Home » Thus, $ p(x) = 2x^2 + x $. Then $ p(0) = 0 $. - Appfinity Technologies

Thus, $ p(x) = 2x^2 + x $. Then $ p(0) = 0 $. - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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\( x^2 + 64 = 100 \)
\( x^2 = 36 \)
\( x = 6 \) meters
Evaluate the limit \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \).
\( \frac{(x - 2)(x + 2)}{x - 2} \)
Cancel \( x - 2 \):
\( \lim_{x \to 2} (x + 2) = 4 \)
If \( \sin(\theta) = \frac{3}{5} \) and \( \theta \) is in the first q...
Use \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\( \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 \)

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