Question

State whether the following statements are true or false:
−3/4÷(5/9÷−4/11)=(−3/4÷5/9)÷−4/11.​

Answer 1

Answer:

answer is false

Step-by-step explanation:

hope it is helpful

thank you

Answer 2

★ How to do :-

Here, we are given with some of the fractions on LHS of the statement and the same three fractions in the RHS of the statement. On LHS, the last two fractions are grouped by brackets and on the RHS, the first two fractions are grouped in the brackets. If we are given with the fractions by grouping in this format, then this property can be classified as the associative property. This property can only be done with the multiplication and addition of fractions. It cannot be done in the subtraction and division of fractions or integers. This also cannot be done by the negative integers. As in this question, we are given with division, we can directly say that the statement is false. But, we should prove by solving it. So, let's solve!!

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➤ Solution :-

${\sf \leadsto \dfrac{(-3)}{4} \div \bigg( \dfrac{5}{9} \div \dfrac{(-4)}{11} \bigg) \neq \bigg( \dfrac{(-3)}{4} \div \dfrac{5}{9} \bigg) \div \dfrac{(-4)}{11}}$

Let's solve the LHS and RHS separately.

LHS :-

${\sf \leadsto \dfrac{(-3)}{4} \div \bigg( \dfrac{5}{9} \div \dfrac{(-4)}{11} \bigg)}$

Let's solve the fractions in bracket first.

Take the reciprocal of second fraction and multiply both fractions.

${\sf \leadsto \dfrac{(-3)}{4} \div \bigg( \dfrac{5}{9} \times \dfrac{11}{(-4)} \bigg)}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{(-3)}{4} \div \bigg( \dfrac{5 \times 11}{9 \times (-4)} \bigg)}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-3)}{4} \div \dfrac{55}{(-36)}}$

Take the reciprocal of second fraction and multiply both fractions.

${\sf \leadsto \dfrac{(-3)}{4} \times \dfrac{(-36)}{55}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{(-3) \times \cancel{(-36)}}{\cancel{4} \times 55} = \dfrac{(-3) \times (-9)}{1 \times 55}}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{27}{55} \: --- LHS}$

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RHS :-

${\sf \leadsto \bigg( \dfrac{(-3)}{4} \div \dfrac{5}{9} \bigg) \div \dfrac{(-4)}{11}}$

Let's solve the fractions in bracket first.

Take the reciprocal of second fraction and multiply both fractions.

${\sf \leadsto \bigg( \dfrac{(-3)}{4} \times \dfrac{9}{5} \bigg) \div \dfrac{(-4)}{11}}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \bigg( \dfrac{(-3) \times 9}{4 \times 5} \bigg) \div \dfrac{(-4)}{11}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-27)}{20} \div \dfrac{(-4)}{11}}$

Take the reciprocal of second fraction and multiply both fractions.

${\sf \leadsto \dfrac{(-27)}{20} \times \dfrac{11}{(-4)}}$

Multiply the fractions.

${\sf \leadsto \dfrac{(-27) \times 11}{20 \times (-4)} = \dfrac{(-297)}{(-80)}}$

Remove the negative sign as it's common in numerator and denominator.

${\sf \leadsto \dfrac{297}{80} \: --- RHS}$

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Comparison :-

${\sf \leadsto \dfrac{27}{54} \: and \: \dfrac{297}{80}}$

As we can see that those both aren't equal. So,

${\sf \leadsto \dfrac{27}{54} \neq \dfrac{297}{80}}$

So,

${\sf \leadsto LHS \neq RHS}$

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${\red{\underline{\boxed{\bf So, \: the \: given \: statement \: is \: false.}}}}$