Question

plz answer anyone who knows the answer​

Answer 1

ANSWERS:

5. debit

6. a copywriter buys a new business laptop but forgets to enter the purchase in the books.

7. Market Price

EXPLANATION:

5. Generally, the asset account balances are debit balances and are increased with a debit entry and decreased with a credit entry.

6. Somebody may forget to enter an invoice you've paid or the sale of a service. Error of omission happens when we forget to enter a transaction in the books.

7. It can be said that the closing price is the last price at which the stock traded during the regular trading day. Hence it can be said that closing stock is market price.

A stock's closing price is the standard benchmark used by investors to track its performance over time.

Answer 2

Answer:

1)

Given that , $\sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x}$.

Need To Find : Value of x ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀Given that ,

$\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: Given \::}}$

$\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:\\\\ \dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} - \dfrac{1}{p + q + x} = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{1}{p} + \dfrac{1}{q}\bigg] +\bigg[ \dfrac{1}{x} - \dfrac{1}{p + q + x} \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{(x + p + q )- x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{ p + q }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{1}{pq} + \dfrac{ 1 }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xp + xq + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xq + xp + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x( x + q ) + p ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = \dfrac{0}{( p + q )} \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \times (pq) (x) ( p + q + x ) \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \:\\\\ \dashrightarrow \sf \:x\:=\:-p \:\:or\:\: x \:=\:-q \:$

$\dashrightarrow \:\underline {\boxed{\purple {\pmb{\frak{\:\:x\:=\:-p \:\:or\:\: \:-q \:}}}}}\:\:\bigstar \:\:$

$\qquad \therefore \underline {\sf Hence, The \:value \:of \: x \:can \: be \:\pmb{\bf{ - p \: or \:-q \:}}.}$

2)

Question:-

If ( 2 - √5 )/( 2 + √5 ) =a√5 + b , then find the value of a and b.

To Find:-

Find the value of a and b.

Solution:-

$\dashrightarrow\sf \: \dfrac { 2 - \sqrt { 5 } } { 2 + \sqrt { 5 } } = a \sqrt { 5 } + b$

Rationalise the denominator

$\dashrightarrow\sf \: \dfrac { 2 - \sqrt { 5 } } { 2 + \sqrt { 5 } } \times \dfrac { 2 - \sqrt { 5 } } { 2 - \sqrt { 5 } } = a \sqrt { 5 } + b$

$\dashrightarrow\sf \: \dfrac { { ( 2 - \sqrt { 5 } ) }^{ 2 } } { ( 2 + \sqrt { 5 } )( 2 - \sqrt { 5 } ) } = a \sqrt { 5 } + b$

$\dashrightarrow\sf \: \dfrac { 4 + 5 - 4 \sqrt { 5 } } { 4 - 5 } = a \sqrt { 5 } + b$

$\dashrightarrow\sf \: \dfrac { - ( 9 + 4 \sqrt { 5 } ) } { - 1 } = a \sqrt { 5 } + b$

$\dashrightarrow\sf \: 9 + 4 \sqrt { 5 } = a \sqrt { 5 } + b$

Hence ,

a = 4 , b = 9

Explanation:

$\cancel \frac{19}{6}$