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Relaciona cada polinomio aritmético con su respectiva solución:

Polinomio aritmético
Solución

2+5+(-6)×3+(-12)+8÷(-2)
5

[2+5+(-6)]×3+(-12)+8÷(-2)
-27

2+5+(-6)×3+[(-12)+8]÷(-2)
-17

2+5+(-6)]×3+[(-12)+8]÷(-2)
-13

2+[5+(-6)]×3+(-12)+8÷(-2)
-9

$2$ . Relaciona cada polinomio aritmético con su respectiva solución: $\newline$ \begin{tabular}{|c|c|} $\newline$ \hline Polinomio aritmético & Solución \\ $\newline$ \hline $2+5+(-6) \times 3+(-12)+8 \div(-2)$ & $5$ \\ $\newline$ \hline $[2+5+(-6)] \times 3+(-12)+8 \div(-2)$ & $-27$ \\ $\newline$ \hline $2+5+(-6) \times 3+[(-12)+8] \div(-2)$ & $-17$ \\ $\newline$ \hline $2+5+(-6)] \times 3+[(-12)+8] \div(-2)$ & $-13$ \\ $\newline$ \hline $2+[5+(-6)] \times 3+(-12)+8 \div(-2)$ & $-9$ \\ $\newline$ \hline $\newline$ \end{tabular}

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Transformations of functions

Full solution

Q.

2

. Relaciona cada polinomio aritmético con su respectiva solución:

\newline

\begin{tabular}{|c|c|}

\newline

\hline Polinomio aritmético & Solución \\

\newline

\hline

2+5+(-6) \times 3+(-12)+8 \div(-2)

&

5

\\

\newline

\hline

[2+5+(-6)] \times 3+(-12)+8 \div(-2)

&

-27

\\

\newline

\hline

2+5+(-6) \times 3+[(-12)+8] \div(-2)

&

-17

\\

\newline

\hline

2+5+(-6)] \times 3+[(-12)+8] \div(-2)

&

-13

\\

\newline

\hline

2+[5+(-6)] \times 3+(-12)+8 \div(-2)

&

-9

\\

\newline

\hline

\newline

\end{tabular}

Calculate First Polynomial: Calculate the first polynomial: $2 + 5 + (-6) \times 3 + (-12) + 8 \div (-2)$ . $\newline$ - Perform the operations in the correct order (PEMDAS/BODMAS rules). $\newline$ - Multiplication and division first: $(-6) \times 3 = -18$ , $8 \div (-2) = -4$ . $\newline$ - Then addition and subtraction: $2 + 5 - 18 - 12 - 4 = -27$ .
Calculate Second Polynomial: Calculate the second polynomial: $[2 + 5 + (-6)] \times 3 + (-12) + 8 \div (-2)$ . $\newline$ - First solve inside the brackets: $2 + 5 - 6 = 1$ . $\newline$ - Then multiply by $3$ : $1 \times 3 = 3$ . $\newline$ - Perform remaining operations: $3 - 12 + (-4) = -13$ .

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What kind of transformation converts the graph of

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\text{[A]translation 10 units left}

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\text{[B]translation 10 units right}

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Question

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1

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Question

Tess tried to solve the differential equation

\frac{d y}{d x}=2 x y-6 x

. This is her work:

\newline

\frac{d y}{d x}=2 x y-6 x

\newline

1

\quad \frac{d y}{d x}=(y-3) 2 x

\newline

2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

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3

\quad \ln |y-3|=x^{2}+C_{1}

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4

\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}

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5

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6

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7

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(A) Tess's work is correct.

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4

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\newline

7

is incorrect. Tess didn't account for adding

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Aditya tried to solve the differential equation

\frac{d y}{d x}=6 x-30

. This is his work:

\newline

\frac{d y}{d x}=6 x-30

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1

\quad \int 30 d y=\int 6 x d x

\newline

2

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3

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2

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of the following expressions represents the distance between the sandbag and the target?

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2

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|-250+75|

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f(x) = -(x+1)(x+7)

\newline

The function represents a parabola in the

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\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

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f(x)

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