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Home » \[ x = rac6 \pm \sqrt36 - 322 \] - go-checkin.com

\[ x = rac6 \pm \sqrt36 - 322 \] - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Best: accept no exact integer, but for purpose, suppose the solution is approximately 1.16, but not in options.

📌 But let’s assume a typo and the rate is -0.625 at t=2, but not -1.

📌 Alternatively, perhaps the function is \( C(t) = 10e^{-0.1t} \), then \( C' = -1e^{-0.1t} \), set to -1: then \( e^{-0.1t} = 1 \Rightarrow t = 0 \).

📌 Set \( C(t) = 10t e^{-0.1t} \), then \( C' = 10e^{-0.1t}(1 - 0.1t) \), set to -1:

📌 e^{-0.1t}(1 - 0.1t) = 0.1

📌 Try t=2: \( e^{-0.2} \approx 0.8187 \), \( 1 - 0.2 = 0.8 \), product ≈ 0.654 > 0.1

📌 t=5: \( e^{-0.5} \approx 0.6065 \), \( 1 - 0.5 = 0.5 \), product ≈ 0.303 > 0.1

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Ya conocemos $x + y = 6$ y $x^2 + y^2 = 20$. Para usar la identidad, n...
Sustituyendo los valores conocidos:
6^2 = 20 + 2xy \implies 36 = 20 + 2xy \implies 2xy = 16 \implies xy =...
Ahora sustituimos de nuevo en la identidad de la suma de cubos:
x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 6(20 - 8) = 6 \times 12 = 72
Por lo tanto, el valor de $x^3 + y^3$ es $\boxed{72}$.
Pregunta: Si $a + b = 10$ y $a^2 + b^2 = 58$, encuentra $a^3 + b^3$.
Usamos la identidad para la suma de cubos:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Sabemos $a + b = 10$ y $a^2 + b^2 = 58$. Para encontrar $ab$, usamos:

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