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Home » Therefore, the minimum value of $ g(x) $ is: - Baxtercollege

Therefore, the minimum value of $ g(x) $ is: - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Solution: Multiply numerator and denominator by the conjugate $ \sqrt{...
\frac{\sqrt{7}(\sqrt{7} + \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} +...
The rationalized form is $ \boxed{\dfrac{7 + \sqrt{21}}{4}} $.
Question: The functions $ f(x) = x^2 - 6x + m $ and $ g(x) = x^2 - 6x...
Solution: Compute $ f(5) = 25 - 30 + m = -5 + m $ and $ g(5) = 25 - 30...
Set $ f(5) = 2g(5) $:
+ m = 2(-5 + 3m)
+ m = -10 + 6m
= 5m \Rightarrow m = 1
The value of $ m $ is $ \boxed{1} $.**Question:** A box contains 8 red...

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R = \frac{17}{2} = 8.5
Thus, the circumradius of the triangle is \(\boxed{8.5}\).
An equilateral triangle has a side length of 10 cm. If each side is in...
The original area \(A_1\) of an equilateral triangle with side \(s = 1...
A_1 = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 10^2 = 25\sqr...
The new side length is \(12\) cm, so the new area \(A_2\) is:
A_2 = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3}
\Delta A = A_2 - A_1 = 36\sqrt{3} - 25\sqrt{3} = 11\sqrt{3}
Thus, the area increases by \(\boxed{11\sqrt{3}}\) square centimeters.
Compute \(\cos 60^\circ\) using the unit circle and verify using the p...

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