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Popular >

30<= sqrt(980)cos(θ)3.19

Solución

30 \leq 980 cos (θ) 3.19

Solución

- arccos (\frac{300 5}{2233}) + 2 πn \leq θ \leq arccos (\frac{300 5}{2233}) + 2 πn

Notación de intervalos

[- arccos (\frac{300 5}{2233}) + 2 πn, arccos (\frac{300 5}{2233}) + 2 πn]

Notación decimal

- 1.26567 \dots + 2 πn \leq θ \leq 1.26567 \dots + 2 πn

Pasos de solución

30 \leq 980 cos (θ) \cdot 3.19

Intercambiar lados

980 cos (θ) \cdot 3.19 \geq 30

Multiplicar ambos lados por

100

980 cos (θ) \cdot 3.19 \geq 30

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

980 cos (θ) \cdot 3.19 \cdot 100 \geq 30 \cdot 100

Simplificar

44665 cos (θ) \geq 3000

44665 cos (θ) \geq 3000

Dividir ambos lados entre

44665

44665 cos (θ) \geq 3000

Dividir ambos lados entre

44665

\frac{4466 5 cos ( θ )}{4466 5} \geq \frac{3000}{4466 5}

Simplificar

\frac{4466 5 cos ( θ )}{4466 5} \geq \frac{3000}{4466 5}

Simplificar

\frac{4466 5 cos ( θ )}{4466 5} : cos (θ)

\frac{4466 5 cos ( θ )}{4466 5}

Dividir:

\frac{4466}{4466} = 1

= \frac{5 cos ( θ )}{5}

Eliminar los terminos comunes:

5

= cos (θ)

Simplificar

\frac{3000}{4466 5} : \frac{300 5}{2233}

\frac{3000}{4466 5}

Eliminar los terminos comunes:

2

= \frac{1500}{2233 5}

Factorizar

1500 : 5^{3} \cdot 2^{2} \cdot 3

Factorizar

1500 = 5^{3} \cdot 2^{2} \cdot 3

= \frac{5 ^{3} \cdot 2 ^{2} \cdot 3}{2233 5}

Cancelar

\frac{2 ^{2} \cdot 3 \cdot 5 ^{3}}{2233 5} : \frac{2 ^{2} \cdot 3 \cdot 5 ^{\frac{5}{2}}}{2233}

\frac{2 ^{2} \cdot 3 \cdot 5 ^{3}}{2233 5}

Aplicar las leyes de los exponentes:

5 = 5^{\frac{1}{2}}

= \frac{5 ^{3} \cdot 2 ^{2} \cdot 3}{2233 \cdot 5 ^{\frac{1}{2}}}

Aplicar las leyes de los exponentes:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{5 ^{3}}{5 ^{\frac{1}{2}}} = 5^{3 - \frac{1}{2}}

= \frac{2 ^{2} \cdot 3 \cdot 5 ^{- \frac{1}{2} + 3}}{2233}

Restar:

3 - \frac{1}{2} = \frac{5}{2}

= \frac{2 ^{2} \cdot 3 \cdot 5 ^{\frac{5}{2}}}{2233}

= \frac{2 ^{2} \cdot 3 \cdot 5 ^{\frac{5}{2}}}{2233}

5^{\frac{5}{2}} = 5^{2} 5

5^{\frac{5}{2}}

5^{\frac{5}{2}} = 5^{2 + \frac{1}{2}}

= 5^{2 + \frac{1}{2}}

Aplicar las leyes de los exponentes:

x^{a + b} = x^{a} x^{b}

= 5^{2} \cdot 5^{\frac{1}{2}}

Simplificar

= 5^{2} 5

= \frac{2 ^{2} \cdot 3 \cdot 5 ^{2} 5}{2233}

2^{2} \cdot 3 \cdot 5^{2} 5 = 3005

2^{2} \cdot 3 \cdot 5^{2} 5

2^{2} = 4

= 5^{2} \cdot 4 \cdot 35

5^{2} = 25

= 4 \cdot 3 \cdot 255

Multiplicar los numeros:

4 \cdot 3 \cdot 25 = 300

= 3005

= \frac{300 5}{2233}

cos (θ) \geq \frac{300 5}{2233}

cos (θ) \geq \frac{300 5}{2233}

cos (θ) \geq \frac{300 5}{2233}

Para

cos (x) \geq a

, si

- 1 < a < 1

entonces

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{300 5}{2233}) + 2 πn \leq θ \leq arccos (\frac{300 5}{2233}) + 2 πn

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