Fill in the blanks-1-6-1-3

Step 1: Write all the fractions with the least common denominator (LCD), 6.___ 1/6=1× 2/3× 2 ___ 1/6=2/6Step 2: Transpose the fractions to the right side of ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of Rational Numbers

Fill in the b...

Question

Fill in the blanks:
$___- \frac{1}{6} = \frac{1}{3}$

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Solution

Step 1: Write all the fractions with the least common denominator (LCD), $6$ .
$_- \frac{1}{6} = \frac{1 \times 2}{3 \times 2}$
$_- \frac{1}{6} = \frac{2}{6}$
Step 2: Transpose the fractions to the right side of the equation and solve for blank.
$_= \frac{2}{6} + \frac{1}{6}$
Step 3: Perform the addition keeping the common denominator.
$_= \frac{2 + 1}{6}___= \frac{3}{6}$
Step 4: Divide both the numerator and denominator by the common factor $3$ .
$\frac{3 \div 3}{6 \div 3} = \frac{1}{2}$
Hence, the required number is $\frac{1}{2}$ .

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Similar questions

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

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Fill in the blanks:___-16=13

Fill in the blanks:
$___- \frac{1}{6} = \frac{1}{3}$