Basic Math Examples

$| {({(\frac{2}{5})}^{2} \cdot {(\frac{20}{7})}^{2})}^{- 1} |$

Step 1

Simplify the expression.

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Step 1.1

Apply the product rule to $\frac{2}{5}$ .

$| {(\frac{2^{2}}{5^{2}} \cdot {(\frac{20}{7})}^{2})}^{- 1} |$

Step 1.2

Raise $2$ to the power of $2$ .

$| {(\frac{4}{5^{2}} \cdot {(\frac{20}{7})}^{2})}^{- 1} |$

Step 1.3

Raise $5$ to the power of $2$ .

$| {(\frac{4}{25} \cdot {(\frac{20}{7})}^{2})}^{- 1} |$

Step 1.4

Apply the product rule to $\frac{20}{7}$ .

$| {(\frac{4}{25} \cdot \frac{20^{2}}{7^{2}})}^{- 1} |$

Step 2

Combine.

$| {(\frac{4 \cdot 20^{2}}{25 \cdot 7^{2}})}^{- 1} |$

Step 3

Simplify the expression.

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Step 3.1

Raise $20$ to the power of $2$ .

$| {(\frac{4 \cdot 400}{25 \cdot 7^{2}})}^{- 1} |$

Step 3.2

Raise $7$ to the power of $2$ .

$| {(\frac{4 \cdot 400}{25 \cdot 49})}^{- 1} |$

Step 3.3

Multiply $4$ by $400$ .

$| {(\frac{1600}{25 \cdot 49})}^{- 1} |$

Step 3.4

Multiply $25$ by $49$ .

$| {(\frac{1600}{1225})}^{- 1} |$

Step 4

Cancel the common factor of $1600$ and $1225$ .

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Step 4.1

Factor $25$ out of $1600$ .

$| {(\frac{25 (64)}{1225})}^{- 1} |$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $25$ out of $1225$ .

$| {(\frac{25 \cdot 64}{25 \cdot 49})}^{- 1} |$

Step 4.2.2

Cancel the common factor.

$| {(\frac{25 \cdot 64}{25 \cdot 49})}^{- 1} |$

Step 4.2.3

Rewrite the expression.

$| {(\frac{64}{49})}^{- 1} |$

Step 5

Change the sign of the exponent by rewriting the base as its reciprocal.

$| \frac{49}{64} |$

Step 6

$\frac{49}{64}$ is approximately $0.765625$ which is positive so remove the absolute value

$\frac{49}{64}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{49}{64}$

Decimal Form:

$0.765625$