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katen-ka-za [31]
2 years ago
10

Find the value of all trigonometric functions of 135°

Mathematics
1 answer:
Vera_Pavlovna [14]2 years ago
5 0

Answer:

sin (- 135°)= – sin 135°= – sin (1 × 90°+ 45°) = – cos 45° = – 1√2

cos (- 135°)= cos 135°= cos (1 × 90°+ 45°) = – sin 45°= – 1√2

tan (- 135°) = – tan 135° = – tan ( 1 × 90° + 45°) = – (- cot 45°) = 1

csc (- 135°)= – csc 135°= – csc (1 × 90°+ 45°)= – sec 45° = – √2

sec (- 135°)= sec 135°= sec (1 × 90°+ 45°)= – csc 45°= – √2

cot (- 135°) = – cot 135° = – cot ( 1 × 90° + 45°) = – (-tan 45°) = 1

Step-by-step explanation:

hope this helps

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There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,
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<h2 /><h2>Here we go ~ </h2>

According to given information there are :

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\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{total \: red \: balls}{total \: balls}

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<h3>2. what is the probability that the ball chosen is blue ?</h3>

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{total \: blue \: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{25}

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{total \: yellow\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{6}{5 + 4 + 6 + 10}

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<h3>4. what is the probability that the ball chosen is green ?</h3>

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{total \: green\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{25}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{2}{5}

<h3>5. what is the probability that the ball chosen is not green ?</h3>

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\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}

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\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{3}{5}

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