Powers and Roots: Using fractions

Let's start by solving the given expression $(\frac{1}{4})^2+\frac{1}{16}$.

We will first evaluate the power: $(\frac{1}{4})^2$.

• When you square a fraction, you square both the numerator and the denominator. Therefore,
• Solve $\left(\frac{1}{4}\right)^2 = \frac{1^2}{4^2} = \frac{1}{16}$.

Now that we have the squared term, the expression simplifies to $\frac{1}{16} + \frac{1}{16}$.

To add these fractions, we simply add the numerators since they have the same denominator:

• $\frac{1}{16} + \frac{1}{16} = \frac{1+1}{16} = \frac{2}{16}$.

Simplify the fraction $\frac{2}{16}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

• $\frac{2}{16} = \frac{2 \div 2}{16 \div 2} = \frac{1}{8}$.

Therefore, the answer to the expression is $\frac{1}{8}$.

To solve $(\frac{3}{4})^2$, you need to square both the numerator and the denominator separately:

1. Square the numerator: $3^2 = 9$

2. Square the denominator: $4^2 = 16$

3. Combine the results to get $\frac{9}{16}$

To solve $(\frac{5}{6})^2$, you need to square both the numerator and the denominator separately:

1. Square the numerator: $5^2 = 25$

2. Square the denominator: $6^2 = 36$

3. Combine the results to get $\frac{25}{36}$

According to the order of operations, we should first solve the expression inside of the parentheses:

$(\sqrt{380.25}-\frac{1}{2})=(19.5-\frac{1}{2})=(19)$

In the next step, we will proceed to solve the exponentiation, and finally the subtraction:

$(19)^2-11=(19\times19)-11=361-11=350$

Let's solve the problem step-by-step using the order of operations, specifically addressing powers and fractions.

Given the expression: $(\frac{1}{4})^2+\frac{1}{16}$

• First, evaluate the power: $(\frac{1}{4})^2$
• Squaring a fraction means squaring both the numerator and the denominator:
• $(\frac{1}{4})^2 = \frac{1^2}{4^2} = \frac{1}{16}$
• Next, add the fractions: $\frac{1}{16} + \frac{1}{16}$
• Since the denominators are the same, simply add the numerators:
• $\frac{1+1}{16} = \frac{2}{16} = \frac{1}{8}$

Therefore, the value of the expression is $\frac{1}{8}$.

To solve the expression $(\frac{1}{2})^2+(\frac{1}{3})^2+\frac{1}{4}$, we will follow the order of operations and break it down step by step:

Step 1: Calculate the squares of the fractions:

- For$(\frac{1}{2})^2$:

$(\frac{1}{2})^2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

- For $(\frac{1}{3})^2$:

$(\frac{1}{3})^2 = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Step 2: Add the fractions together:

- Adding $\frac{1}{4}$, $\frac{1}{9}$, and $\frac{1}{4}$:

The least common denominator (LCD) of $4$ and $9$ is $36$.

Rewrite each fraction with the LCD:

• $\frac{1}{4} = \frac{9}{36}$

• $\frac{1}{9} = \frac{4}{36}$

• $\frac{1}{4} = \frac{9}{36}$

Now add them together:

$\frac{9}{36} + \frac{4}{36} + \frac{9}{36} = \frac{22}{36}$

Step 3: Simplify the result:

$\frac{22}{36}$ can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is $2$:

• $\frac{22 \div 2}{36 \div 2} = \frac{11}{18}$

Thus, the simplified result of the expression $(\frac{1}{2})^2+(\frac{1}{3})^2+\frac{1}{4}$ is $\frac{11}{18}$, which matches the provided correct answer.

In the numerator we solve the square root exercise:

$\sqrt{49}=7$

In the denominator we solve the exercise within parentheses:

$(2+2\times3)=$

$2+6=8$

The exercise we now have is:

$\frac{21:7+2}{8-8}=$

We solve the exercise in the numerator of fractions from left to right:

$21:7=3$

$3+2=5$

We obtain the exercise:

$\frac{5}{8-8}=\frac{5}{0}$

Since it is impossible for the denominator of the fraction to be 0, it is impossible to solve the exercise.