Algebra Examples

$(3, \frac{π}{2})$ , $(8, \frac{4 π}{3})$

Step 1

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 2

Substitute the actual values of the points into the distance formula.

$\sqrt{{(8 - 3)}^{2} + {(\frac{4 π}{3} - \frac{π}{2})}^{2}}$

Step 3

Simplify.

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Step 3.1

Subtract $3$ from $8$ .

$\sqrt{5^{2} + {(\frac{4 π}{3} - \frac{π}{2})}^{2}}$

Step 3.2

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

$\sqrt{25 + {(\frac{4 π}{3} - \frac{π}{2})}^{2}}$

Step 3.3

To write $\frac{4 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\sqrt{25 + {(\frac{4 π}{3} \cdot \frac{2}{2} - \frac{π}{2})}^{2}}$

Step 3.4

To write $- \frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\sqrt{25 + {(\frac{4 π}{3} \cdot \frac{2}{2} - \frac{π}{2} \cdot \frac{3}{3})}^{2}}$

Step 3.5

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.5.1

Multiply $\frac{4 π}{3}$ by $\frac{2}{2}$ .

$\sqrt{25 + {(\frac{4 π \cdot 2}{3 \cdot 2} - \frac{π}{2} \cdot \frac{3}{3})}^{2}}$

Step 3.5.2

Multiply $3$ by $2$ .

$\sqrt{25 + {(\frac{4 π \cdot 2}{6} - \frac{π}{2} \cdot \frac{3}{3})}^{2}}$

Step 3.5.3

Multiply $\frac{π}{2}$ by $\frac{3}{3}$ .

$\sqrt{25 + {(\frac{4 π \cdot 2}{6} - \frac{π \cdot 3}{2 \cdot 3})}^{2}}$

Step 3.5.4

Multiply $2$ by $3$ .

$\sqrt{25 + {(\frac{4 π \cdot 2}{6} - \frac{π \cdot 3}{6})}^{2}}$

Step 3.6

Combine the numerators over the common denominator.

$\sqrt{25 + {(\frac{4 π \cdot 2 - π \cdot 3}{6})}^{2}}$

Step 3.7

Simplify the numerator.

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Step 3.7.1

Multiply $2$ by $4$ .

$\sqrt{25 + {(\frac{8 π - π \cdot 3}{6})}^{2}}$

Step 3.7.2

Multiply $3$ by $- 1$ .

$\sqrt{25 + {(\frac{8 π - 3 π}{6})}^{2}}$

Step 3.7.3

Subtract $3 π$ from $8 π$ .

$\sqrt{25 + {(\frac{5 π}{6})}^{2}}$

Step 3.8

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.8.1

Apply the product rule to $\frac{5 π}{6}$ .

$\sqrt{25 + \frac{{(5 π)}^{2}}{6^{2}}}$

Step 3.8.2

Apply the product rule to $5 π$ .

$\sqrt{25 + \frac{5^{2} π^{2}}{6^{2}}}$

Step 3.9

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

$\sqrt{25 + \frac{25 π^{2}}{6^{2}}}$

Step 3.10

Raise $6$ to the power of $2$ .

$\sqrt{25 + \frac{25 π^{2}}{36}}$

Step 3.11

To write $25$ as a fraction with a common denominator, multiply by $\frac{36}{36}$ .

$\sqrt{25 \cdot \frac{36}{36} + \frac{25 π^{2}}{36}}$

Step 3.12

Combine $25$ and $\frac{36}{36}$ .

$\sqrt{\frac{25 \cdot 36}{36} + \frac{25 π^{2}}{36}}$

Step 3.13

Combine the numerators over the common denominator.

$\sqrt{\frac{25 \cdot 36 + 25 π^{2}}{36}}$

Step 3.14

Factor $25$ out of $25 \cdot 36 + 25 π^{2}$ .

$\sqrt{\frac{25 (36 + π^{2})}{36}}$

Step 3.15

Rewrite $\sqrt{\frac{25 (36 + π^{2})}{36}}$ as $\frac{\sqrt{25 (36 + π^{2})}}{\sqrt{36}}$ .

$\frac{\sqrt{25 (36 + π^{2})}}{\sqrt{36}}$

Step 3.16

Simplify the numerator.

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Step 3.16.1

Rewrite $25 (36 + π^{2})$ as $5^{2} (6^{2} + π^{2})$ .

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Step 3.16.1.1

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{5^{2} (36 + π^{2})}}{\sqrt{36}}$

Step 3.16.1.2

Rewrite $36$ as $6^{2}$ .

$\frac{\sqrt{5^{2} (6^{2} + π^{2})}}{\sqrt{36}}$

Step 3.16.2

Pull terms out from under the radical.

$\frac{5 \sqrt{6^{2} + π^{2}}}{\sqrt{36}}$

Step 3.16.3

Raise $6$ to the power of $2$ .

$\frac{5 \sqrt{36 + π^{2}}}{\sqrt{36}}$

Step 3.17

Simplify the denominator.

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Step 3.17.1

Rewrite $36$ as $6^{2}$ .

$\frac{5 \sqrt{36 + π^{2}}}{\sqrt{6^{2}}}$

Step 3.17.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{5 \sqrt{36 + π^{2}}}{6}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{5 \sqrt{36 + π^{2}}}{6}$

Decimal Form:

$5.64392522 \dots$

Step 5