2020-II-04
答案:A
$\begin{array}{cl}
& (3a + 2b)(4a – 5b)
… Read $\begin{array}{cl}
& (3a + 2b)(4a – 5b)
$\begin{array}{rcl}
$\begin{array}{rcl}
bx + y
$\begin{array}{cl}
= &
$\begin{array}{cl}
= & 300 \text
當 $u = 2$ 及 $v = 4$ 時,$w=
$\begin{array}{rcl}
5 – 4x & < & 9 \\
-4x & < & 4 \\
x & >
留意 $PQUT$ 為一長方形,則 $PT = QU
原來的弧長
$\begin{array}
$\begin{array}{rcl}
\pi \times (5a)
$\begin{array}
考慮 $\Delta AEG$ 及 $\Delta DEG$。若分別以 $AE$ 及 $DE$
根據圖像,可得
$\begin{array}{rc
$\begin{array}{rcll}
\because \ang
$\begin{array}{cl}
& BE^2 + BC^
考慮 $\Delta ABC$。由於 $\angle ABC = 90^\cir
留意該燈塔與船的最短距離為 $x\text{ km}$。
$\begin{arra