$\begin{array}{l} 3 \frac{1}{2}+4 \frac{1}{4} \Rightarrow 3+4=7- \\ 2 \frac{2}{7}+3 \frac{1}{2} \Rightarrow 2+3=5- \\ 4 \frac{1}{8}+6 \frac{3}{4} \Rightarrow 4+6=10 \end{array}$ $\newline$ $30$ Understanding mixed numbers

View step-by-step help

Home

Math Problems

Geometry

Sine and cosine of complementary angles

Full solution

\begin{array}{l} 3 \frac{1}{2}+4 \frac{1}{4} \Rightarrow 3+4=7- \\ 2 \frac{2}{7}+3 \frac{1}{2} \Rightarrow 2+3=5- \\ 4 \frac{1}{8}+6 \frac{3}{4} \Rightarrow 4+6=10 \end{array}

\newline

30

Understanding mixed numbers

Add Mixed Numbers: We are given three mixed numbers to add. Let's start with the first one: $3\frac{1}{2} + 4\frac{1}{4}$ . First, we add the whole numbers: $3 + 4 = 7$ . Then we add the fractions: $\frac{1}{2} + \frac{1}{4}$ . To add fractions, they must have a common denominator. The least common denominator for $2$ and $4$ is $4$ . We convert $\frac{1}{2}$ to $\frac{2}{4}$ so that both fractions have the same denominator. Now we have $\frac{2}{4} + \frac{1}{4} = \frac{3}{4}$ . Adding the whole number and the fraction gives us $7\frac{3}{4}$ .
First Mixed Numbers: Next, we add the second set of mixed numbers: $2\frac{2}{7} + 3\frac{1}{2}$ . Again, we start by adding the whole numbers: $2 + 3 = 5$ . Now we need to add the fractions $\frac{2}{7}$ and $\frac{1}{2}$ . The least common denominator for $7$ and $2$ is $14$ . We convert $\frac{2}{7}$ to $\frac{4}{14}$ and $\frac{1}{2}$ to $2 + 3 = 5$ $0$ . Adding these gives us $2 + 3 = 5$ $1$ . The sum of the whole number and the fraction is $2 + 3 = 5$ $2$ .
Second Mixed Numbers: Finally, we add the third set of mixed numbers: $4\frac{1}{8} + 6\frac{3}{4}$ . We start with the whole numbers: $4 + 6 = 10$ . Now we add the fractions $\frac{1}{8}$ and $\frac{3}{4}$ . The least common denominator for $8$ and $4$ is $8$ . We convert $\frac{3}{4}$ to $\frac{6}{8}$ so that both fractions have the same denominator. Now we have $\frac{1}{8} + \frac{6}{8} = \frac{7}{8}$ . The sum of the whole number and the fraction is $4 + 6 = 10$ $0$ .