Question

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Answer 1

Answer:

answer 6th is -4 upon 5

Answer 7th is -2 upon 5

Step-by-step explanation:

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Answer 2

$\bf\purple{QuestioN:-}$

1. By what number should   $\bf\dfrac{-12}{35}$  we divided to get  $\bf\dfrac{3}{7}$ ?

2. Divide the sum of $\bf{\dfrac{-12}{7}~~~and~~~\bf\dfrac{13}{5}}$ by the product of  $\bf{\dfrac{-1}{2}~~~and~~~\bf\dfrac{31}{7}}$.

$\bf\pink{AnsweR:-}$

Solution Qs.1 :-

1. According to the question,

$\bf\dfrac{-12}{35}$  is divided by  $\bf\dfrac{x}{y}$ gives  $\bf\dfrac{3}{7}$ .

As we know in division of fractions we need to multiply the multiplicative inverse of the second number. That is,

$:\implies\bf{\dfrac{-12}{35}\times\dfrac{~y~}{x}=\dfrac{~3~}{7}}$

Or in other way, we can also write it as,

$:\implies\bf{\dfrac{~3~}{7}\times\dfrac{x}{y}=\dfrac{-12}{35}}$

$:\implies\bf{\dfrac{~3x~}{7y}=\dfrac{-12}{35}}$

-----------------------------------

$:\implies\bf{3x=-12}$

$:\implies\bf{x=\dfrac{-12}{3}}$

$:\implies\bf{x=-4}$

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$:\implies\bf{7y=35}$

$:\implies\bf{y=\dfrac{35}{7}}$

$:\implies\bf{y=5}$

---------------------------------------------

So the required answer for your question is,

$:\implies\bf{\dfrac{x}{y}=\dfrac{-4}{5}}$

•◘•------------------------------------------------•◘•

Solution Qs.2 :-

2. According to the question, at first we need to find the sum of

$\bf{\dfrac{-12}{7}~~~and~~~\bf\dfrac{13}{5}}$ .

That is,

$:\implies\bf{\dfrac{-12}{7}+\dfrac{13}{5}}$

First need to make the denominators same.

LCM of 7 and 5 is 35

$:\implies\bf{\dfrac{-12}{7}\times\dfrac{5}{5}=\dfrac{-60}{35}}$

$:\implies\bf{\dfrac{13}{5}\times\dfrac{7}{7}=\dfrac{91}{35}}$

Next we can find both of these fraction's sum.

$:\implies\bf{\dfrac{-60}{35}+\dfrac{91}{35}=\dfrac{31}{35}}$ ---------- Eq.1

Next we need to find the product of  $\bf{\dfrac{-1}{2}~~~and~~~\bf\dfrac{31}{7}}$ .

For finding product of fractions, we can directly multiply them.

That is,

$\bf:\implies{\dfrac{-1}{2}\times\bf\dfrac{31}{7}}$

$\bf:\implies{\dfrac{-31}{14}}$ --------------- Eq.2

Next we want to divide Eq.1 by Eq.2 ,

For that we need to multiply Eq.1 by multiplicative inverse of Eq.2,

$\bf:\implies{\dfrac{31}{35}\times\bf\dfrac{7}{31}}$

$\bf:\implies{\dfrac{31\times7}{35\times31}}$

Here, from denominator and numerator 31 get cancelled.

So the required answer is,

$\bf:\implies{\dfrac{7}{35}}$

On cancelling,

$\bf:\implies{\dfrac{1}{5}}$

So,   $\bf\dfrac{1}{5}$   is the correct answer for your question!!

Happy Learning!!