Question

Answer the following.
a.
b.
c.
d.

Answer 1

Answer:

A.25

B. 6

C. 5

D. 2/9

Step-by-step explanation:

please mark me brainest

Answer 2

Answer:

Question :

The value of :

${\implies{\bigg( \dfrac{1}{2} \bigg)^{ - 2} + \bigg( \dfrac{1}{3} \bigg)^{ - 2} + \bigg( \dfrac{1}{4} \bigg)^{ - 2}}}$

$\begin{gathered}\end{gathered}$

Solution :

${\implies{\bigg( \dfrac{1}{2} \bigg)^{ - 2} + \bigg( \dfrac{1}{3} \bigg)^{ - 2} + \bigg( \dfrac{1}{4} \bigg)^{ - 2}}}$

Using law of exponent rule : (a/b)^ᵐ = (aᵐ/bᵐ)

${\implies{\bigg( \dfrac{{1}^{ - 2}}{{2}^{ - 2}} \bigg) + \bigg( \dfrac{{1}^{ - 2}}{{3}^{ - 2}} \bigg) + \bigg( \dfrac{{1}^{ - 2}}{{4}^{ - 2}} \bigg)}}$

As we know that the (1 to the power of any number is 1). Then,

${\implies{\bigg( \dfrac{1}{{2}^{ - 2}} \bigg) + \bigg( \dfrac{1}{{3}^{ - 2}} \bigg) + \bigg( \dfrac{1}{{4}^{ - 2}} \bigg)}}$

Again using law of exponent rule : (a)-ⁿ = (1/aⁿ)

${\implies{\Bigg( \dfrac{1}{ \frac{1}{{2}^{2} } } \Bigg) + \Bigg( \dfrac{1}{\frac{1}{{3}^{2}}} \Bigg) + \Bigg( \dfrac{1}{\frac{1}{{4}^{2}}} \Bigg)}}$

${\implies{\Bigg( \dfrac{1}{1} \times {2}^{2} \Bigg) + \Bigg( \dfrac{1}{1} \times {3}^{2} \Bigg) + \Bigg( \dfrac{1}{1} \times {4}^{2} \Bigg)}}$

${\implies{\Big({2}^{2} \Big) + \Big( {3}^{2} \Big) + \Big({4}^{2}\Big)}}$

${\implies{\Big(2 \times 2\Big) + \Big( 3 \times 3 \Big) + \Big(4 \times 4\Big)}}$

${\implies{\big(4\big) + \big( 9\big) + \big(16\big)}}$

${\implies{4 + 9 + 16}}$

${\implies{\large{\rm{\underline{\underline{\red{29 \: ans.}}}}}}}$

Hence, the option (b) 29 is the correct answer. ✔

$\begin{gathered}\end{gathered}$

Learn More :

☼ EXPONENT :

↝ The exponent of a number says how many times to use the number in a multiplication.

☼ LAW OF EXPONENT :

The important laws of exponents are given below:

⠀⠀↠ ${\rm{{a}^{m} \times {a}^{n} = {a}^{m + n}}}$

⠀⠀↠ ${\rm{{a}^{m}/{a}^{n} = {a}^{m - n}}}$

⠀⠀↠ ${\rm{({a}^{m})^{n} = {a}^{mn}}}$

⠀⠀↠ ${\rm{{a}^{n}/{b}^{n} = ({a/b})^{n} }}$

⠀⠀↠ ${\rm{{a}^{0} = 1}}$

⠀⠀↠ ${\rm{{a}^{ - m} = {1/a}^{m}}}$

⠀⠀↠ ${\rm{{a}^{\frac{1}{n} } = \sqrt[n]{a}}}$

☼ Algebraic identities :

⠀⠀➛ (a+b)²+(a-b)² = 2a²+2b²

⠀⠀➛ (a+b)²-(a-b)² = 4ab

⠀⠀➛ (a+b)(a -b) = a²-b²

⠀⠀➛ (a+b+c)² = a²+b²+c²+2ab+2bc+2ca

⠀⠀➛ (a-b)³ = a³-b³-3ab(a-b)

⠀⠀➛ (a³+b³) = (a+b)(a²-ab+b²)

⠀⠀➛ a²+b² = (a+b)²-2ab

⠀⠀➛ a³-b³ = (a-b)(a²+ab +b²)

⠀⠀➛ If a + b + c = 0 then a³ + b³ + c³ = 3abc

$\underline{\rule{220pt}{3pt}}$