3 frac{1}{4} + 2 frac{1}{2} - 1 frac{5}{6} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the missing value be $x$. The equation is $\frac{x^2}{10} + 1 \frac{5}{12} = 3 \frac{1}{4} + 2 \frac{1}{2} - 1 \frac{5}{6}$.Convert mixed frac

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let the missing value be $x$. The equation is $\frac{x^2}{10} + 1 \frac{5}{12} = 3 \frac{1}{4} + 2 \frac{1}{2} - 1 \frac{5}{6}$.Convert mixed fractions to improper fractions:$\frac{x^2}{10} + \frac{17}{12} = \frac{13}{4} + \frac{5}{2} - \frac{11}{6}$.Find a common denominator $(12)$ for the right side:$\frac{x^2}{10} + \frac{17}{12} = \frac{39}{12} + \frac{30}{12} - \frac{22}{12} = \frac{47}{12}$.Isolate the term with $x$:$\frac{x^2}{10} = \frac{47}{12} - \frac{17}{12} = \frac{30}{12} = \frac{5}{2}$.Solve for $x^2$:$x^2 = \frac{5}{2} 	imes 10 = 25$.Therefore,$x = \sqrt{25} = 5$.

$3 \frac{1}{4} + 2 \frac{1}{2} - 1 \frac{5}{6} = \frac{(?)^{2}}{10} + 1 \frac{5}{12}$

Similar Questions

If $\frac{1}{N} = \frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} - \sqrt{5})},$ what is the value of $N?$

If $n = 7 + 4 \sqrt{3}$,then the value of $\left(\sqrt{n} + \frac{1}{\sqrt{n}}\right)$ is

What approximate value will come in place of the question mark $(?)$ in the given question? (You are not expected to calculate the exact value)
$71.98 \%$ of $1200 + 35.06 \%$ of $270 = ? \%$ of $600$

$108 \div ?$ of $\frac{1}{3} + \frac{2}{5} \times 3 \frac{3}{4} = 10 \frac{1}{2}$

When $52416$ is divided by $312,$ the quotient is $168.$ What will be the quotient when $52.416$ is divided by $0.0168?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module