The value of 4-frac{5}{1+frac{1}{3+frac{1}{2+ | vedclass

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

Question

(D) To solve the expression $4-\frac{5}{1+\frac{1}{3+\frac{1}{2+\frac{1}{4}}}}$,we simplify from the bottom up.First,simplify the lowest fraction: $2+

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(D) To solve the expression $4-\frac{5}{1+\frac{1}{3+\frac{1}{2+\frac{1}{4}}}}$,we simplify from the bottom up.First,simplify the lowest fraction: $2+\frac{1}{4} = \frac{8+1}{4} = \frac{9}{4}$.Next,substitute this back: $3+\frac{1}{9/4} = 3+\frac{4}{9} = \frac{27+4}{9} = \frac{31}{9}$.Then,substitute this back: $1+\frac{1}{31/9} = 1+\frac{9}{31} = \frac{31+9}{31} = \frac{40}{31}$.Finally,substitute this into the main expression: $4-\frac{5}{40/31} = 4-\frac{5 	imes 31}{40} = 4-\frac{155}{40} = 4-\frac{31}{8}$.Calculating the final value: $\frac{32-31}{8} = \frac{1}{8}$.

The value of $4-\frac{5}{1+\frac{1}{3+\frac{1}{2+\frac{1}{4}}}}$ is:

Similar Questions

If $a+b+c=6, a^{2}+b^{2}+c^{2}=14$ and $a^{3}+b^{3}+c^{3}=36,$ then the value of $abc$ is

The value of $\frac{(2.3)^{3}-0.027}{(2.3)^{2}+0.69+0.09}$ is

The value of $\frac{1}{a^{2}+a x+x^{2}}-\frac{1}{a^{2}-a x+x^{2}}+\frac{2 a x}{a^{4}+a^{2} x^{2}+x^{4}}$ is

$\frac{69842 \times 69842 - 30158 \times 30158}{69842 - 30158} = ?$

If the numerator of a fraction is increased by $2$ and the denominator is increased by $3,$ the fraction becomes $\frac{7}{9},$ and if both the numerator and the denominator are decreased by $1,$ the fraction becomes $\frac{4}{5}.$ What is the original fraction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module