Question

Please solve both the questions with full solution​

Answer 1

Answer:

Step-by-step explanation:

for question 3, just substitute  the values

q4 - by indices rule

(k+1) (3k-2)= k+1+3k-2 = 4k-1

4k - 1 is -9

therefore k = -2

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Answer 2

$\underbrace{ \large \sf Question \: 3 : - }$

• Verify a× (b + c) = a × b + a × c, by taking :-

• a = 3/4
• b = -5/6
• c = 2/3

$\underbrace{ \large \sf Solution \: 3 : - }$

• Simply Substitute the given values in both LHS as well as RHS

$\sf \maltese\underline{LHS :- }$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf a \times (b + c)$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \dfrac{3}{4} \bigg( \dfrac{ - 5}{6} + \dfrac{2}{3} \bigg)$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \frac{3}{4} \bigg( \frac{ - 5 + 4}{6} \bigg)$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \dfrac{ \cancel{3}}{4} \times \dfrac{ - 1}{ \cancel{6} ^{2} }$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \frak{ \frac{ - 1}{8} }} \bigstar$

$\sf \maltese\underline{RHS :- }$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf a \times b + a \times c$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \bigg(\dfrac{ \cancel{3}}{4} \times \frac{ - 5}{ \cancel{6} ^{2} } \bigg) + \bigg( \frac{ \cancel{3}}{ \cancel{4} ^{2} } \times \frac{ \cancel{2}}{ \cancel{3}} \bigg)$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \frac{ - 5}{8} + \frac{1}{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \frac{ - 5 + 4}{8}$

$\\\: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \frak{ \frac{ - 1}{8} }} \bigstar$

LHS = RHS

hence, verified !!

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$\\\underbrace{ \large \sf Question \: 4 : - }$

• Find the value of k so that (-2)^(k+1) × (-2)^(3k-2) = (-2)^(-9)

$\\\underbrace{ \large \sf Solution \: 4 : - }$

$\\ : \implies \sf {( - 2)}^{k + 1} \times {( - 2)}^{3k - 2} = {( - 2)}^{ - 9}$

We know that :-

$\\ \: \: \: \: \: \: \: \: \: \: \: \leadsto \large \underline{ \boxed{ \pink{ \frak{ {a}^{m} + {a}^{n} = {a}^{m + n} }}}} \bigstar$

Applying the same concept in our equation :-

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf {( - 2)}^{k + 1 + 3k - 2} = {( - 2)}^{ - 9}$

$\\ \: \: \: \: \: \: \: \: \: \: \:: \implies \sf {( - 2)}^{4k - 1} = {( - 2)}^{ - 9}$

Since, the bases are same on both LHS as well as RHS, the exponents can be equated as follows :-

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf 4k - 1 = - 9$

$\\ \: \: \: \: \: \: \: \: \: \: \:: \implies \sf 4k = - 9 + 1$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf 4k = - 8$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf k = \dfrac{ - 8}{4}$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{\sf{ k = - 2} } \bigstar$

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hope it's helpful!