Tag: ways to multiply and divide

$\displaystyle\frac{2^{n+4}-2\times2^{n}}{2\times2^{(n+3)}}+2^{-3}$
$\frac { 2^{ n+4 }-2^{ n+1 } }{ 2^{ (n+4) } } +2^{ -3 }$
$\frac { 2^{ n }(2^{ 4 }-2^{ 1 }) }{ 2^{ n }\times { 2 }^{ 4 } } +2^{ -3 }$
$\frac { (2^{ 4 }-2^{ 1 }) }{ { 2 }^{ 4 } } +2^{ -3 }$
$\frac { (16-2) }{ 16 } +\frac { 1 }{ 8 } $
$\frac { 14 }{ 16 } +\frac { 1 }{ 8 } $
$\frac { 14+2 }{ 16 } =1$
Answer (D) 1

As t term is 3t/4,So to find t we need to divide is by 3/4.
Answer (B) Divide both sides by 3/4

average = 12.4 runs per wicket, 
Let's assume total  number of wickets taken by him till the last match=x
 total  number of runs taken by him till the last match=y
$y/x=12.4$
$y=12.4x
$(y+26)/(x+5)=12$
$12.4x+26=12x+60$
$0.4x=34$
$x=85$
The number of wickets taken by him till the last match =85
Answer (D) 85

Given that $4m=5K$ and $6n=7K$
Now divide those two equations, we get $\dfrac { 4m }{ 6n } =\dfrac { 5K }{ 7K } $
Which gives $ \dfrac { m }{ n } =\dfrac { 5 }{ 7 } \times \dfrac { 6 }{ 4 } =\dfrac { 15 }{ 14 } $

$3i\left( 4+3i \right) \left( 52i \right) \ =\quad 156{ i }^{ 2 }\left( 4+3i \right) \ =\quad -156\left( 4+3i \right) \ =\quad -624-468i$

We know that any number multiplied by $0$ will always result in $0$. For example: The equation $0\times 2 = 0$ can be read as 'Zero times Two equals Zero'. That means, you are taking $'2'$ zero times, effectively taking nothing. So, multiplying any number by zero will always be zero.

Option $B$ is correct.
The multiplicative inverse will include $1$ as numerator.

We know that any number multiplied by $1$ will always result in that number itself. For example: The equation $2\times 1 = 2$ can be read as 'two times one equals one'. That means, you are taking $'2'$ one times, effectively taking nothing but multiplying by its unity. So, multiplying any number by one will always be the number itself.

By calculations,

We know that any number divided by $1$ will always result in that number itself. For example: $2\div 1 = 2$ and