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(dy)/(dx)=6x and
y(2)=8.

y(4)=

$\frac{d y}{d x}=6 x$ and $y(2)=8$ . $\newline$ $y(4)=$

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Evaluate functions

Full solution

Q.

\frac{d y}{d x}=6 x

and

y(2)=8

.

\newline

y(4)=

Integrate $6x$ with respect to $x$ : We know that $\frac{dy}{dx} = 6x$ is the derivative of $y$ with respect to $x$ . To find $y(x)$ , we need to integrate $6x$ with respect to $x$ . $\newline$ Calculation: $\int 6x \, dx = 3x^2 + C$
Find constant $C$ : We have the initial condition $y(2) = 8$ . We can use this to find the constant $C$ . $\newline$ Calculation: $8 = 3(2)^2 + C$ $\newline$ $8 = 12 + C$ $\newline$ $C = 8 - 12$ $\newline$ $C = -4$
Calculate $y(4)$ : Now we have the function $y(x) = 3x^2 - 4$ . We can use this to find $y(4)$ . $\newline$ Calculation: $y(4) = 3(4)^2 - 4$ $\newline$ $y(4) = 3(16) - 4$ $\newline$ $y(4) = 48 - 4$ $\newline$ $y(4) = 44$

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Which pattern can we use to factor the expression?

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1

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(C) We can't use any of the patterns.

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What are the real and imaginary parts of

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\begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=3.14 i \text { and } \\ \operatorname{Im}(z)=-19 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 i \end{array}

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\begin{array}{l} \operatorname{Re}(z)=3.14 \text { and } \\ \operatorname{Im}(z)=-19 \end{array}

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What are the real and imaginary parts of

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\newline

1

\newline

\newline

\begin{array}{l} \operatorname{Re}(z)=-45 \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=-45 i \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 i \end{array}

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What are the real and imaginary parts of

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1

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\begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 \end{array}

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\begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 i \end{array}

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\begin{array}{l} \operatorname{Re}(z)=-12 i \text { and } \\ \operatorname{Im}(z)=11 \end{array}

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