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eduard
2 years ago
10

Find the exact value.

Mathematics
1 answer:
levacccp [35]2 years ago
4 0

\text{Given that,}\\\\\cot A = -\dfrac{9}{\sqrt{40}}\\\\\\\implies \tan A = -\dfrac{\sqrt{40}}9\\\\\\\implies \tan^2 A = \dfrac{40}{81}\\\\\\\implies \sec^2 A -1 = \dfrac{40}{81}\\\\\implies \sec^2 A = 1 + \dfrac{40}{81}\\\\\implies \sec^2 A = \dfrac{121}{81}\\\\\implies \sec A = \pm \sqrt{\dfrac{121}{81}}= \pm\dfrac{11}{9}\\\\\\\text{Since the angle A is in quadrant II, the value of sec A will be negative.}\\\\\\\text{Hence,}~ \sec A=-\dfrac{11}{9}

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This is the answer :
100%-20%=80%
\frac{80}{100}  =  \frac{x}{200}  \\ x =  \frac{200 \times 80}{100}  =  \frac{16000}{100}  = 160
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Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N inter
JulijaS [17]

Answer:

A circle is shown. Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N intersects the circle at point M. The length of P N is 32, the length of Q N is x, the length of L M is 22, and the length of M N is 14.

In the diagram, the length of the external portion of the secant segment PN is <u>X</u>

The length of the entire secant segment LN is <u>36</u>.

The value of x is <u>15.74</u>

Step-by-step explanation:

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Jona_Fl16

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Read 2 more answers
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kupik [55]

Answer:

i think its a trick question

Step-by-step explanation:

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2 years ago
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2 years ago
Law of radicals
andrey2020 [161]

Answer:

1)  \sqrt{x^7}=x^{\frac{7}{2}

2)  \sqrt[3]{y^5}=y^{\frac{5}{3}

3) \sqrt[5]{a^{12}}=a^{\frac{12}{5} }

4) \sqrt[4]{z^{9}}=z^\frac{9}{4}

Step-by-step explanation:

1) \sqrt{x^7}

We know that \sqrt{x}=x^{\frac{1}{2}

So, \sqrt{x^7}=x^{\frac{7}{2}

2) \sqrt[3]{y^5}

We know that \sqrt[3]{x}=x^{\frac{1}{3}

So, \sqrt[3]{y^5}=y^{\frac{5}{3}

3) \sqrt[5]{a^{12}}

We know that \sqrt[5]{x}=x^{\frac{1}{5}

So, \sqrt[5]{a^{12}}=a^{\frac{12}{5} }

4) \sqrt[4]{z^{9}}

We know that \sqrt[4]{x}=x^{\frac{1}{4}

So, \sqrt[4]{z^{9}}=z^\frac{9}{4}

6 0
3 years ago
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