Find the Secant Given the Point (-2/7,(3 square root of 5)/7)

$(- \frac{2}{7}, \frac{3 \sqrt{5}}{7})$

To find the $\sec (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(- \frac{2}{7}, \frac{3 \sqrt{5}}{7})$ , draw the triangle between the three points $(0, 0)$ , $(- \frac{2}{7}, 0)$ , and $(- \frac{2}{7}, \frac{3 \sqrt{5}}{7})$ .

Opposite : $\frac{3 \sqrt{5}}{7}$

Adjacent : $- \frac{2}{7}$

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

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

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Apply the product rule to $- \frac{2}{7}$ .

$\sqrt{{(- 1)}^{2} {(\frac{2}{7})}^{2} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

Apply the product rule to $\frac{2}{7}$ .

$\sqrt{{(- 1)}^{2} \frac{2^{2}}{7^{2}} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

Raise $- 1$ to the power of $2$ .

$\sqrt{1 \frac{2^{2}}{7^{2}} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

Multiply $\frac{2^{2}}{7^{2}}$ by $1$ .

$\sqrt{\frac{2^{2}}{7^{2}} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

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

$\sqrt{\frac{4}{7^{2}} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

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

$\sqrt{\frac{4}{49} + {(\frac{3 \sqrt{5}}{7})}^{2}}$

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

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Apply the product rule to $\frac{3 \sqrt{5}}{7}$ .

$\sqrt{\frac{4}{49} + \frac{{(3 \sqrt{5})}^{2}}{7^{2}}}$

Apply the product rule to $3 \sqrt{5}$ .

$\sqrt{\frac{4}{49} + \frac{3^{2} {\sqrt{5}}^{2}}{7^{2}}}$

Simplify the numerator.

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Raise $3$ to the power of $2$ .

$\sqrt{\frac{4}{49} + \frac{9 {\sqrt{5}}^{2}}{7^{2}}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\sqrt{\frac{4}{49} + \frac{9 {(5^{\frac{1}{2}})}^{2}}{7^{2}}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5^{\frac{1}{2} \cdot 2}}{7^{2}}}$

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5^{\frac{2}{2}}}{7^{2}}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5^{\frac{2}{2}}}{7^{2}}}$

Rewrite the expression.

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5^{1}}{7^{2}}}$

Evaluate the exponent.

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5}{7^{2}}}$

Simplify the expression.

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

$\sqrt{\frac{4}{49} + \frac{9 \cdot 5}{49}}$

Multiply $9$ by $5$ .

$\sqrt{\frac{4}{49} + \frac{45}{49}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{4 + 45}{49}}$

Add $4$ and $45$ .

$\sqrt{\frac{49}{49}}$

Divide $49$ by $49$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

$\sec (θ) = \frac{Hypotenuse}{Adjacent}$ therefore $\sec (θ) = \frac{1}{- \frac{2}{7}}$ .

$\frac{1}{- \frac{2}{7}}$

Simplify $\sec (θ)$ .

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Cancel the common factor of $1$ and $- 1$ .

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Rewrite $1$ as $- 1 (- 1)$ .

$\sec (θ) = \frac{- 1 \cdot - 1}{- \frac{2}{7}}$

Move the negative in front of the fraction.

$\sec (θ) = - \frac{1}{\frac{2}{7}}$

Multiply the numerator by the reciprocal of the denominator.

$\sec (θ) = - (1 (\frac{7}{2}))$

Multiply $\frac{7}{2}$ by $1$ .

$\sec (θ) = - \frac{7}{2}$

Approximate the result.

$\sec (θ) = - \frac{7}{2} \approx - 3.5$

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