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受欢迎的代数 >

N(t)=550+20t-0.2t^2

解答

N (t) = 550 + 20 t - 0.2 t^{2}

解答

t = - \frac{5 ( N + N ^{2} - 40 N + 840 - 20 )}{2}, t = - \frac{5 ( N - N ^{2} - 40 N + 840 - 20 )}{2}

求解步骤

N (t) = 550 + 20 t - 0.2 t^{2}

在两边乘以

10

Nt \cdot 10 = 5500 + 200 t - 2 t^{2}

交换两边

5500 + 200 t - 2 t^{2} = Nt \cdot 10

将

Nt 10

para o lado esquerdo

5500 + 200 t - 2 t^{2} - Nt \cdot 10 = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 t^{2} + (200 - N \cdot 10) t + 5500 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( 200 - N \cdot 10 ) \pm ( 200 - N \cdot 10 ) ^{2} - 4 ( - 2 ) \cdot 5500}{2 ( - 2 )}

化简

(200 - N \cdot 10)^{2} - 4 (- 2) \cdot 5500 : 10 N^{2} - 40 N + 840

t_{1, 2} = \frac{- ( 200 - N \cdot 10 ) \pm 10 N ^{2} - 40 N + 840}{2 ( - 2 )}

将解分隔开

t_{1} = \frac{- ( 200 - N \cdot 10 ) + 10 N ^{2} - 40 N + 840}{2 ( - 2 )}, t_{2} = \frac{- ( 200 - N \cdot 10 ) - 10 N ^{2} - 40 N + 840}{2 ( - 2 )}

t = \frac{- ( 200 - N 10 ) + 10 N ^{2} - 40 N + 840}{2 ( - 2 )} : - \frac{5 ( N + N ^{2} - 40 N + 840 - 20 )}{2}

t = \frac{- ( 200 - N 10 ) - 10 N ^{2} - 40 N + 840}{2 ( - 2 )} : - \frac{5 ( N - N ^{2} - 40 N + 840 - 20 )}{2}

二次方程组的解是:

t = - \frac{5 ( N + N ^{2} - 40 N + 840 - 20 )}{2}, t = - \frac{5 ( N - N ^{2} - 40 N + 840 - 20 )}{2}

作图

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