Powers: Using the laws of exponents

Previously mentioned in the order of operations that exponents come before multiplication and division which come before addition and subtraction (and parentheses always come first),

Let's calculate therefore first the value of the expression inside the parentheses in the first term from the left (by calculating the values of the terms in the root inside the parentheses first) , let's handle the expression inside the parentheses separately first:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9}$Let's recall the definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$And we'll convert the roots in the first term to powers:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9} =8^{\frac{1}{2}}\cdot2^{\frac{1}{2}}-\sqrt{9}$

Next, we'll note that the two multiplication terms in the first term from the left have the same power, so let's recall the law of exponents for powers applied to parentheses, but in the opposite direction:

$x^n\cdot y^n=(x\cdot y)^n$

The literal interpretation of this law in the direction given here is that we can write a multiplication of two terms with equal exponents as a multiplication of the bases in parentheses raised to that same power, let's apply this to the expression we got in the last step:

$8^{\frac{1}{2}}\cdot2^{\frac{1}{2}}-\sqrt{9} =(8\cdot2)^{\frac{1}{2}}-\sqrt{9}=16^{\frac{1}{2}}-\sqrt{9}$

Where in the first stage we applied the above law of exponents and in the second stage we calculated the result of the multiplication in parentheses,

Let's return to writing roots using the definition of root as a power, but in the opposite direction:

$a^{\frac{1}{n}} = \sqrt[n]{a}$

And we'll convert back the power of one-half to a square root (while remembering that a square root is actually a root of order 2), then we'll simplify the expression by calculating the values of the roots (without a calculator!):

$16^{\frac{1}{2}}-\sqrt{9} =\sqrt{16}-\sqrt{9}=4-3=1$

Where in the last stage we calculated the result of the subtraction operation,

Let's summarize then the development stages for the expression in parentheses in the original problem, we got that:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9} = (8\cdot2)^{\frac{1}{2}}-\sqrt{9}= \sqrt{16}-\sqrt{9}=4-3=1$

Let's return now to the expression in the original problem and substitute this result:

$(\sqrt{8}\cdot\sqrt{2}-\sqrt{9})^2:2^2+3^2 \\ \downarrow\\ 1^2:2^2+3^2$

We'll continue and after we calculate the values of the numbers in powers, we'll perform the division operation and then the addition operation:

$1^2:2^2+3^2=1:4+9=\frac{1}{4}+9=9\frac{1}{4}$

Where in the second stage we converted the division operation to fraction notation,

Therefore the correct answer is answer C.

To solve the given power expression, we need to apply the formula for powers of a fraction. The expression we are given is:
$\left(\frac{2}{3}\right)^3$

Let's break down the steps:

• When we raise a fraction to a power, we apply the exponent to both the numerator and the denominator separately. This means raising both 2 and 3 to the power of 3.
• Thus, we calculate:
  $2^3 = 8$ and $3^3 = 27$.
• Therefore, $\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$.

So, the result of the expression $\left(\frac{2}{3}\right)^3$ is $\frac{8}{27}$.

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 $\frac{25}{36}$