Question

solve the question...​

Answer 1

Given to find the value of :-

$(6 {}^{ - 1} + 8 {}^{ - 1} ) \times \bigg( \dfrac{5}{2} \bigg) {}^{ - 1}$

Solution:-

As we know from the exponent laws

$\red {a {}^{ - n} = \dfrac{1}{a {}^{n} }}$

$\red{ \bigg( \dfrac{a}{b} \bigg) {}^{ - n} = \bigg( \dfrac{b}{a} \bigg) {}^{n} }$

So, by using this we can simplify the question

$\bigg( \dfrac{1}{6} + \dfrac{1}{8} \bigg) \times \bigg( \dfrac{2}{5} \bigg) {}^{1}$

Simplifying this

L.C.M of 6, 8 is 24

$\bigg( \dfrac{1(4) + 1(3)}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{4 + 3}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{7}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{7}{ \not \not{{24}} } \bigg) \times \bigg( \dfrac{ \not2 \: }{5} \bigg)$

$\bigg( \dfrac{7}{12} \bigg) \times \bigg( \dfrac{1}{5} \bigg)$

$\bigg( \dfrac{7 \times 1}{12 \times 5} \bigg)$

$\bigg( \dfrac{7 }{60} \bigg)$

So,

$\red{(6 {}^{ - 1} + 8 {}^{ - 1} ) \times \bigg( \dfrac{5}{2} \bigg) {}^{ - 1} = \bigg( \dfrac{7}{60} \bigg)}$

Note :-

Dear user this is the correct answer according to me kindly check the options.

Know more exponent and laws:-

${a^m \times a^n = a^{m+n}}$

$\dfrac{a^m}{a^n}= a^{m-n}$

${a^0 =1}$

${1^a = 1}$

If the bases are equal then powers also should be equal .

${(a^m)^n = a^{mn}}$

Answer 2

Given to find the value of :-

$(6 {}^{ - 1} + 8 {}^{ - 1} ) \times \bigg( \dfrac{5}{2} \bigg) {}^{ - 1}$

Solution:-

As we know from the exponent laws

$\red {a {}^{ - n} = \dfrac{1}{a {}^{n} }}$

$\red{ \bigg( \dfrac{a}{b} \bigg) {}^{ - n} = \bigg( \dfrac{b}{a} \bigg) {}^{n} }$

So, by using this we can simplify the question

$\bigg( \dfrac{1}{6} + \dfrac{1}{8} \bigg) \times \bigg( \dfrac{2}{5} \bigg) {}^{1}$

Simplifying this

L.C.M of 6, 8 is 24

$\bigg( \dfrac{1(4) + 1(3)}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{4 + 3}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{7}{24} \bigg) \times \bigg( \dfrac{2}{5} \bigg)$

$\bigg( \dfrac{7}{ \not \not{{24}} } \bigg) \times \bigg( \dfrac{ \not2 \: }{5} \bigg)$

$\bigg( \dfrac{7}{12} \bigg) \times \bigg( \dfrac{1}{5} \bigg)$

$\bigg( \dfrac{7 \times 1}{12 \times 5} \bigg)$

$\bigg( \dfrac{7 }{60} \bigg)$

So,

$\red{(6 {}^{ - 1} + 8 {}^{ - 1} ) \times \bigg( \dfrac{5}{2} \bigg) {}^{ - 1} = \bigg( \dfrac{7}{60} \bigg)}$

Note :-

Dear user this is the correct answer according to me kindly check the options.

Know more exponent and laws:-

${a^m \times a^n = a^{m+n}}$

$\dfrac{a^m}{a^n}= a^{m-n}$

${a^0 =1}$

${1^a = 1}$

If the bases are equal then powers also should be equal .

${(a^m)^n = a^{mn}}$