Answer:
(i) 4, 7p \[4\times 7p=(4\times 7)\times p\] \[=28\times p\] = 28p (ii) ? 4p, 7p \[(-4p)\times (7p)=\{(-4)\times 7\}\times (p\times p)\] \[=(-28)\times {{p}^{2}}\] \[=-28{{p}^{2}}\] (iii) ? 4p, 7pq \[(-4p)\times (7pq)=\{(-4)\times 7\}\times \{p\times (pq)\}\] \[=(-28)\times (p\times p\times q)\] \[=(-28)\times ({{p}^{2}}q)\] (iv) \[(4{{p}^{3}})\times (-3p)=\{4\times (-3)\}\times ({{p}^{3}}\times p)\] \[=(-12)\times {{p}^{4}}\] \[=-12{{p}^{4}}\] (v) 4p, 0 \[(4p)\times 0=(4\times 0)\times p\] \[=0\times p\] = 0.
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