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3 years ago

Use the given information to determine the exact trigonometric value.

Mathematics

1 answer:

hoa [83]3 years ago

6 0

Answer:

Step-by-step explanation:

Use the main trigonometric equality:

$\cos ^2\theta+\sin^2\theta=1.$

Since $\sin\theta=-\dfrac{1}{5},$ you have that

$\cos^2\theta=1-\left(-\dfrac{1}{5}\right)^2=1-\dfrac{1}{25}=\dfrac{24}{25}.$

You know that $\pi$ thus $\cos \theta$

and

$\cos\theta=-\sqrt{\dfrac{24}{25}}=-\dfrac{2\sqrt{6}}{5}.$

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3 years ago

Read 2 more answers

(3x-5)/(x-5)>=0 <br> How do I get the answers for this problem?

Diano4ka-milaya [45]

$\frac{3x-5}{x-5}$ > 0

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: $\frac{3x-5}{x-5}$ = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = $\frac{5}{3}$
Sixth, from the values of x above, we have these 3 intervals to test:
x < $\frac{5}{3}$
$\frac{5}{3}$ < x < 5
x > 5
Seventh, pick a test point for each interval.

1. For the interval x < $\frac{5}{3}$ :
Let's pick x - 0. Then, $\frac{3x0-5}{0-5}$ > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval $\frac{5}{3}$ < x < 5:
Let's pick x = 2. Then, $\frac{3x2-5}{2-5}$ > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, $\frac{3x6-5}{6-5}$ > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
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Answer: x < $\frac{5}{3}$ and x > 5

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Answer:

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Step-by-step explanation:

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