\begin{tabular}{cl} Drag and drop the steps to show the correct order for multiplying \( 3.5 \times 10^{5} \) with \( 4.0 \times 10^{-2} \). This is the expression \( \left(3.5 \times 10^{5}\right)\left(4.0 \times 10^{-2}\right) \) : \\ \hline Sometimes answers end up NOT being in scientific notation after the multiplication or division. \\ When this happens we have to regroup. Hint: Slide of the 1.2 Study Slides. \\ \hline\end{tabular}

b. \( \begin{array}{l}\frac{1}{2} x+\frac{5}{6} y=2 \\ \frac{3}{4} x-\frac{1}{2} y=3\end{array} \)

First combine the like terms in \( \frac{3}{4} a+8-\frac{1}{4} a-2 \) \( \frac{3}{4} a+8-\frac{1}{4} a-2=\square ?+10 \)

Higher Order Thinking a. Is the value of the expression \( \left(\frac{1}{4^{-2}}\right)^{3} \) greater than 1 , equal to 1 or less than 1 ? b. If the value of the expression is greater than 1 , show how you can change one sign to make the value less than 1 . If the value is less than 1 , show how you can change one sign to make the value greater than 1 . If the value is equal to 1 , show how you can make one change to make the value not equal to 1 .

(b) Considering the state of water dams level in Eastern Cape, South Africa. Assume that the area covered by water weed on Beervlei Dam of Groot River in the Eastern Cape decreases according to the formula below. \[ W(t)=W_{0} e^{-\alpha t} \text {, } \] where \( W \) is the area, in square metres, covered by the water weed after \( t \) days and \( \alpha \) is a constant to be determined. Here \( W_{0}=500 \mathrm{~m}^{2} \) stands for the initial area of water weed. After 10 days, the area is \( 100 \mathrm{~m}^{2} \). (i) Find the value of the constant \( \alpha \) without simplifying your answer. (ii) Using the expressions of \( \alpha \) and the area \( W \) to determine the area covered by water weed after 1 month and 10 days.

e. \( \frac{a}{3}+0,1 k-\frac{a}{2}-0,3 k= \)