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zysi [14]
4 years ago
5

What are the zeros of f(x) = x(x - 9)?

Mathematics
1 answer:
Lera25 [3.4K]4 years ago
8 0
Answer: D

Explanation:

x(x-9) = 0

=> x = 0

=> x-9 = 0
=> x = 9


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Calculate y as a function of x when dy/dx = 4x3 + 3x2 - 6x + 5
PtichkaEL [24]
Answer:
y = x⁴ + x³ - 3x² + 5x + C

======

Separable differential equations such as these ones can be solved by treating dy/dx as a ratio of differentials. Then move the dx with all the x terms and move the dy with all the y terms. After that, integrate both sides of the equation.

   \begin{aligned}
\dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\
dy &= (4x^3 + 3x^2 - 6x + 5) dx \\
\int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx 
\end{aligned}

In general (understood that +C portions are still there), 

   \int x^{m} = \dfrac{x^{m+1}}{m+1}

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For the right-hand side, we use the sum/difference rule for integrals, which says that

   \int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx

Applying these concepts:

   \begin{aligned} 
 \int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\
y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\
&= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one $C$ is needed)} \\
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\end{aligned}

The answer is y = x⁴ + x³ - 3x² + 5x + C
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