\[m\] | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ? | ? | ? |
\[m+10\] | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? |
\[t\] | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | ? | ? | ? | ? | ? |
\[5t\] | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? |
\[z\] | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | ? | ? | ? | ? |
\[\frac{z}{3}\] | \[2\frac{2}{3}\] | 3 | \[3\frac{1}{3}\] | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? |
\[m\] | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | ? | ? |
\[m-7\] | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | ? |
Answer:
(a) The complete table is shown below
By inspection of the above table, we see that \[m=6\] satisfies the equation \[m+10\text{ }=16\]. [\[\because \] at \[m=6,\] LHS = RHS] Hence, \[m=6\] is its solution. (b) The complete table is shown below \[m\] \[m+10\] 1 \[1+10=11\] 2 \[2+10=12\] 3 \[3+10=13\] 4 \[4+10=14\] 5 \[5+10=15\] 6 \[6+10=16\] 7 \[7+10=17\] 8 \[8+10=18\] 9 \[9+10=19\] 10 \[10+10=20\] 11 \[11+10=21\] 12 \[12+10=22\] 13 \[13+10=23\]
By inspection of the above table, we find that \[t=7\] satisfies the equation \[5t=35\]. [\[\because \]at \[t=7,\] LHS = RHS] Hence, \[t=7\] is its solution. (c) The complete table is shown below. \[t\] \[5t\] 3 \[5\times 3=15\] 4 \[5\times 4=20\] 5 \[5\times 5=25\] 6 \[5\times 6=30\] 7 \[5\times 7=35\] 8 \[5\times 8=40\] 9 \[5\times 9=45\] 10 \[5\times 10=50\] 11 \[5\times 11=55\] 12 \[5\times 12=60\] 13 \[5\times 13=65\] 14 \[5\times 14=70\] 15 \[5\times 15=75\] 16 \[5\times 16=80\]
By inspection of the above table, we find that \[t=12\] satisfies the equation\[\frac{z}{3}=4.\] [\[\because \] at \[t=12,\] LHS = EHS] Hence, \[z=12\] is its solution. (d) The complete table is shown below \[z\] \[\frac{z}{3}\] 8 \[\frac{8}{3}=2\frac{2}{3}\] 9 \[\frac{9}{3}=3\] 10 \[\frac{10}{3}=3\frac{1}{3}\] 11 \[\frac{11}{3}=3\frac{2}{3}\] 12 \[\frac{12}{3}=4\] 13 \[\frac{13}{3}=4\frac{1}{3}\] 14 \[\frac{14}{3}=4\frac{2}{3}\] 15 \[\frac{15}{3}=5\] 16 \[\frac{16}{3}=5\frac{1}{3}\] 17 \[\frac{17}{3}=5\frac{2}{3}\] 18 \[\frac{18}{3}=6\] 19 \[\frac{19}{3}=6\frac{1}{3}\] 20 \[\frac{20}{3}=6\frac{2}{3}\].
By inspection of the above table, we find that \[m=10\] satisfies the equation \[m-7=3\]. [\[\because \]at \[m=10,\] LHS = RHS] Hence, \[m=10\] is its solution. \[m\] \[m-7\] 5 \[5-7=-2\] 6 \[6-7=-1\] 7 \[7-7=0\] 8 \[8-7=1\] 9 \[9-7=2\] 10 \[10-7=3\] 11 \[11-7=4\] 12 \[12-7=5\] 13 \[13-7=6\] 14 \[14-7=7\] 15 \[15-7=8\]
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