Question

6. Verify that a + (b + c)(a + b)+(a + c) for each of the following values of a, bande
ii. a = -8,b=-1, -2
i. a = 20, b = -2, c = 4​

Answer 1

Answer:

I don't know this answer

Step-by-step explanation:

sorry sorry sorry sorry sorry

Answer 2

$\tt{ a \div (b + c) \not = (a \div b) + (a \div c)}$

$\tt{now\:put\:the\:values\:for\:a,b\:and\:c}$

$\tt{4 \div ( - 3 + 5) \not = (4 \div - (3)) + (4 \div 5)}$

$\tt{ calculate \: the \: sum}$

$\tt{4 \div \red2 \not = (4 \div - (3)) + (4 \div 5)}$

$\small{ \tt divide \: a \: positive \: and \: a \: negative \: equals \: a \: negative} \ratio( + ) \div ( - ) = ( - )$

$\tt{4 \div 2 \not = ( \red - 4 \div 3+ (4 \div 5)}$

$\tt write \: {\orange {\underline{divison \: as \: a \: fraction.}}}$

$</p><p>⬇⬇⬇⬇⬇⬇$

$\tt4 \div 2 \not = ( \red{ \frac{ - 4}{3}} ) + (4 \div 5)$

$\tt divide \: no.$

$\tt ⬇⬇⬇⬇⬇$

$\tt4 \div 2 \not = ( \frac{ - 4}{3} ) + \red{0.8}$

$\tt \red{ 4 \div 2 }\not = ( \frac{ - 4}{3} ) + {0.8}$

$\tt divide \: no.$

$\tt ⬇⬇⬇⬇⬇$

$\tt \red{2 }\not = ( \frac{ - 4}{3} ) + {0.8}$

$4 \div 2 \not = \red ( - \frac{4}{3} \red) + 0.8$

$\tt remove \: unnecessary \: parenthless$

$\tt ⬇⬇⬇⬇⬇$

$4 \div 2 \not = - \frac{4}{3} + 0.8$

$\tt convert \: decimal \: no. \: into \: a \: \underline \orange{fraction.}$

$\tt ⬇⬇⬇⬇⬇$

$2 \not = - \frac{4}{3} + \frac{4}{5}$

$2 \not = \red{ - \frac{4}{3} + \frac{4}{5} }$

$\tt{calculate \: sum}$

$2 \not= \red{ \frac{ - 8}{15} }$

$\tt{here \: l.h.s \: not \: equal \: to \: r.h.s}$

Hence Verified