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Home » But original problem said 64,000 — so proceed numerically: - Get link 4share

But original problem said 64,000 — so proceed numerically: - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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The vertex form is \( f(x) = a(x-2)^2 - 3 \).
Substitute \( (1, 0) \): \( 0 = a(1-2)^2 - 3 \).
Simplifying, \( 0 = a(1) - 3 \), so \( a = 3 \).
Expand to standard form: \( f(x) = 3(x-2)^2 - 3 = 3(x^2 - 4x + 4) - 3...
Thus, \( f(x) = 3x^2 - 12x + 9 \), so \( a = 3 \), \( b = -12 \), \( c...
#### \( a = 3, b = -12, c = 9 \)
The sum of an infinite geometric series is 5, and the first term is 2....
The sum \( S \) of an infinite geometric series is \( S = rac{a}{1-r}...
Given \( S = 5 \) and \( a = 2 \), we have \( rac{2}{1-r} = 5 \).
Solving, \( 2 = 5(1-r) \), \( 2 = 5 - 5r \).

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