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2 years ago

Find an angle in each quadrant with a common reference angle with 216°, from 0°=0<360°

Mathematics

1 answer:

Naddik [55]2 years ago

7 0

Answer:

Quadrant 1: 36°

Quadrant 2: 144°

Quadrant 3 (Given): 216°

Quadrant 4: 324°

Step-by-step explanation:

A 216° angle in standard position has its terminal side in quadrant 3 since 216° is between 180° and 270°.

216° - 180° = 36°

The reference angle for 216° is 36°

Quadrant 1: 36°

Quadrant 2: 180° - 36° = 144°

Quadrant 3 (Given): 216°

Quadrant 4: 360° - 36° = 324°

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Nimfa-mama [501]

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3 years ago

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Answer:

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Step-by-step explanation:

Using the Secant and Tangent Intersection Theorem we can say;

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3 years ago

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garri49 [273]

Answer:

$3^{75}$ .

Step-by-step explanation:

We have been an division problem: $\frac{45^{45}*15^{15}}{75^{30}}$ .

We will simplify our division problem using rules of exponents.

Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

$45^{45}=(9*5)^{45}=9^{45}*5^{45}$

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Substituting these values in our division problem we will get,

$\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}$

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Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

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$\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}$

$\frac{3^{105}*5^{60}}{3^{30}*5^{60}}$

$\frac{3^{105}}{3^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{105}}{3^{30}}=3^{105-30}$

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Therefore, our resulting quotient will be $3^{75}$ .

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