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Troyanec [42]
2 years ago
11

Three trigonometric functions for a given angle are shown below. cosecant theta = thirteen-twelfths; secant theta = Negative thi

rteen-fifths; cotangent theta = Negative five-twelfths What are the coordinates of point (x, y) on the terminal ray of angle Theta, assuming that the values above were not simplified? (–5, 12) (5, –12) (–12, 5) (12, –5)

Mathematics
1 answer:
Kay [80]2 years ago
8 0

Answer:

(–5, 12) is the correct answer.

Step-by-step explanation:

We are given the following values:

cosec\theta = \dfrac{13}{12}\\sec\theta=-\dfrac{13}{5}\\cot\theta =-\dfrac{5}{12}

Now, we know the following identities:

sin \theta = \dfrac{1}{cosec\theta}\\cos \theta = \dfrac{1}{sec\theta}\\tan \theta = \dfrac{1}{cot\theta}

Now, the values are:

sin\theta = \dfrac{12}{13}\\cos\theta=-\dfrac{5}{13}\\tan\theta =-\dfrac{12}{5}

Sine value is positive and cos, tan values are negative.

It can be clearly observed that \theta is in 2nd quadrant.

2nd quadrant means, the value of x will be negative and y will be positive.

Let us have a look at the value of tan\theta:

tan\theta  = \dfrac{Perpendicular}{Base}\\OR\\tan\theta  = \dfrac{y-coordinate}{x-coordinate} = -\dfrac{12}{5}\\\therefore y = 12,\\x = -5

Please refer to the attached image for clear understanding and detailed explanation.

Hence, the correct answer is coordinate (x,y) is (–5, 12)

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Oksanka [162]
<h3>Answer:</h3>

A) 177.568 thousand.

B) 125.836 thousand.

<h3>Step-by-step explanation:</h3>

In this question, it is asking you to use the equation to find the population of ladybugs in a certain year.

Equation we're going to use:

y = -1.437 x + 197.686

We know that the "x" variable represents the number of years since 2010, so that means our starting year is 2010.

Lets solve the question.

Question A:

We need to find the ladybug population is 2024.

2024 is 14 years after 2010, so our "x" variable will be replaced with 14.

Your equation should look like this:

y = -1.437 (14) + 197.686

Now, we solve.

y = -1.437 (14) + 197.686\\\\\text{Multiply -1.437 and 14}\\\\y=-20.118+197.686\\\\\text{Add}\\\\y=177.568

You should get 177.568

This means that the population of ladybugs in 2024 is 177.568 thousand.

Question B:

We need to find the ladybug population is 2060.

2060 is 50 years after 2010, so the "x" variable would be replaced with 50.

Your equation should look like this:

y = -1.437 (50) + 197.686

Now, we solve.

y = -1.437 (50) + 197.686\\\\\text{Multiply -1.437 and 50}\\\\y=-71.85+197.686\\\\\text{Add}\\\\y=125.836

This means that the population of ladybugs in 2060 would be 125.836 thousand.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
7 0
2 years ago
Find FL if H is the circumcenter of EFG,
Zanzabum

The artistic crop isn't helpful; it cuts off some vertex names.

The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn.  We have right triangle ELH, right angle L, so

EH² = HL² + EL²

EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339

Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.

Answer: 4.25

8 0
2 years ago
Write a real world problem that can be represented by the equation 3/4c=21
Black_prince [1.1K]

Answer: Real world problem is "A student have c toffee he distribute \frac{3}{4}th part of those toffees to his friends. He gave total 21 toffees to his friend".

Explanation:

Let a student have c number of toffees in his bag.

It is given that he distribute \frac{3}{4}th part of those toffees to his friends.

The \frac{3}{4}th part of c toffees is,

\frac{3c}{4}

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It is the same as given equation.

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4 0
2 years ago
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Leto [7]

Answer:

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To write an equation to find the value for c, we need to declare what c is first.

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Now we multiplied c to 2.45 and 1.65 and added them together, because whatever the value of c is will give us the equivalence of the sum of 4.12 + 0.75.

Now to check if the equation is right, let's solve for c.

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Therefore concluding that the value of c is 1.19.

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2 years ago
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kogti [31]
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