Answer Key 2.6
- (3) + 1 + (4) − (1)
3 + 1 + 4 − 1
7 - (5)2 + (5) − (1)
25 + 5 − 1
29 - (6) − [(6)(5) ÷ 6]
6 − [30 ÷ 6]
6 − 5
1 - [6 + (4) − (1)] ÷ 3
[6 + 4 − 1] ÷ 3
9 ÷ 3
3 - (5)2 − ((3) − 1)
25 − (3 − 1)
25 − 2
23 - (6) + 6(4) − 4(4)
6 + 24 − 16
14 - 5(4) + (2)(5) ÷ 2
20 + 10 ÷ 2
20 + 5
25 - 5((6) + (2)) + 1 + (5)
5(6 + 2) + 1 + 5
5(8) + 6
40 + 6
46 - [4 − ((6) − (4))] ÷ 2 + (6)
[4 − (6 − 4)] ÷ 2 + 6
[4 − 2] ÷ 2 + 6
2 ÷ 2 + 6
1 + 6
7 - (4) + (5) − 1
4 + 5 − 1
8 - \(\begin{array}{rrl}
\\ \\ \\
\dfrac{ab}{a}&=&\dfrac{c}{a} \\ \\
b&=&\dfrac{c}{a}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(g=\dfrac{h}{i}\right)(i) \\ \\
h=gi
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\
\left(\left(\dfrac{f}{g}\right)x=b\right)\dfrac{g}{f} \\ \\
x=\dfrac{bg}{f}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(p=\dfrac{3y}{q}\right)\left(\dfrac{q}{3}\right) \\ \\
y=\dfrac{pq}{3}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(3x=\dfrac{a}{b}\right)\left(\dfrac{1}{3}\right) \\ \\
x=\dfrac{a}{3b}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(\dfrac{ym}{b}=\dfrac{c}{d}\right)\left(\dfrac{b}{m}\right) \\ \\
y=\dfrac{bc}{dm}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(V=\dfrac{4}{3}\pi r^3\right)\left(\dfrac{3}{4r^3}\right) \\ \\
\pi=\dfrac{3V}{4r^3}
\end{array}\) - \(\begin{array}{l}
\\ \\
\left(E=mv^2\right)\left(\dfrac{1}{v^2}\right) \\ \\
m=\dfrac{E}{v^2}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(c=\dfrac{4y}{m+n}\right)\left(\dfrac{m+n}{4}\right) \\ \\
y=\dfrac{c(m+n)}{4}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(\dfrac{rs}{a-3}=k\right)\left(\dfrac{a-3}{s}\right) \\ \\
r=\dfrac{k(a-3)}{s}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\
\left(V=\dfrac{\pi Dn}{12}\right)\left(\dfrac{12}{\pi n}\right) \\ \\
D=\dfrac{12V}{\pi n}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
F\phantom{+kL}&=&kR-kL \\
+kL&&\phantom{kR}+kL \\
\hline
\dfrac{F+kL}{k}&=&\dfrac{kR}{k} \\ \\
R&=&\dfrac{F+kL}{k}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
P\phantom{-np}&=&\phantom{-}np-nc \\
-np&&-np \\
\hline
\dfrac{P-np}{-n}&=&\dfrac{-nc}{-n} \\ \\
c&=&\dfrac{P-np}{-n}
\end{array}\) - \(\begin{array}{rrrrr}
\\ \\ \\
S\phantom{-2B}&=&L&+&2B \\
-2B&&&-&2B \\
\hline
L&=&S&-&2B
\end{array}\) - \(\phantom{1}\)
\(\left(T=\dfrac{D-d}{L}\right)(L) \\ \)
\(\begin{array}{rrrrr}
TL\phantom{+d}&=&D&-&d \\
+d&&&+&d \\
\hline
D&=&TL&+&d
\end{array}\) - \(\phantom{1}\)
\(\left(I=\dfrac{E_a-E_q}{R}\right)(R) \\ \)
\(\begin{array}{rrrrr}
IR&=&E_a&-&E_q \\
+E_q&&&+&E_q \\
\hline
E_a&=&IR&+&E_q
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
\dfrac{L}{1+at}&=&\dfrac{L_o(1+at)}{1+at} \\ \\
L_o&=&\dfrac{L}{1+at}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
2m+p&=&\phantom{-}4m+q \\
-2m-q&&-2m-q \\
\hline
\dfrac{p-q}{2}&=&\dfrac{2m}{2} \\ \\
m&=&\dfrac{p-q}{2}
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{k-m}{r}=q\right)(r) \\ \)
\(\begin{array}{rrl}
k-m&=&qr \\
+m&&\phantom{qr}+m \\
\hline
k&=&qr+m
\end{array}\) - \(\begin{array}{rrrrr}
\\ \\ \\ \\ \\ \\
R\phantom{-b}&=&aT&+&b \\
-b&&&-&b \\
\hline
\dfrac{R-b}{a}&=&\dfrac{aT}{a}&& \\ \\
T&=&\dfrac{R-b}{a}&&
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
Q_1\phantom{+PQ_1}&=&PQ_2-PQ_1 \\
+PQ_1&&\phantom{PQ_2}+PQ_1 \\
\hline
\dfrac{Q_1+PQ_1}{P}&=&\dfrac{PQ_2}{P} \\ \\
Q_2&=&\dfrac{Q_1+PQ_1}{P}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
L\phantom{-\pi r_2-2d}&=&\pi r_1+\pi r_2+2d \\
-\pi r_2-2d&&\phantom{\pi r_1}-\pi r_2-2d \\
\hline
\dfrac{L-\pi r_2-2d}{\pi}&=&\dfrac{\pi r_1}{\pi} \\ \\
r_1&=&\dfrac{L-\pi r_2-2d}{\pi}
\end{array}\) - \(\phantom{1}\)
\(\left(R=\dfrac{kA(T+T_1)}{d}\right)\left(\dfrac{d}{kA}\right) \\ \)
\(\begin{array}{rrl}
\dfrac{Rd}{kA}\phantom{-T}&=&\phantom{-}T+T_1 \\
-T&&-T \\
\hline
T_1&=&\dfrac{Rd}{kA}-T
\end{array}\) - \(\phantom{1}\)
\(\left(P=\dfrac{V_1(V_2-V_1)}{g}\right)\left(\dfrac{g}{V_1}\right) \\ \)
\(\begin{array}{rrl}
\dfrac{Pg}{V_1}\phantom{+V_1}&=&V_2-V_1 \\
+V_1&&\phantom{V_2}+V_1 \\
\hline
V_2&=&\dfrac{Pg}{V_1}+V_1
\end{array}\)
