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2 years ago

Please help me. NO LINKS!!

Mathematics

2 answers:

stiks02 [169]2 years ago

6 0

The domain (input values) of the cosine function is all negative and positive angle measures.

Let the function be f(x) = cos(x)

The domain of cos(x) is -∞ < x < ∞

The range is -1 ≤ f(x) ≤ 1

Hence, domain of cos(x) is all (+) and (-) angle measures.

Same goes with sine function as well

For function f(x) = sin(x)

The domain is -∞ < x < ∞ and range -1 ≤ f(x) ≤ 1

However for f(x) = tan(x) the same is not applicable.

NikAS [45]2 years ago

3 0

Answer:

A. all negative and positive angle measures.

Step-by-step explanation:

The domain of a function is the set of input values (x-values).

The cosine function is a <u>continuous function</u>, and therefore has <u>no restrictions</u> throughout its domain.

Therefore, its domain is all real numbers.

<u>Set notation</u>: { x | x ∈ R }

<u>Interval notation</u>: (-∞, ∞)

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Y=4x−9<br> Complete the missing value in the solution to the equation

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Answer:

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The complete question is

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3 years ago

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3 years ago

Read 2 more answers

In each of the following ,make y the subject and hence find the value of y when a=2, b=3 ,and c=4.

zvonat [6]

Part i

$\dfrac{2a-7}{6y} = \dfrac{3b-5}{2c}$

$y = \dfrac{ 2c(2a-7)}{6(3b-5)} = \dfrac{c(2a-7)}{9b-15}$

Substituting,

$y = \dfrac{4(2(2)-7)}{9(3)-15} = \dfrac{ -12}{12} = -1$

Answer: -1

Part ii

I'm not sure that one's typed in correctly but I'll solve it as written.

$3-34y + 2a = \dfrac{3b-5}{2c + 1}$

$34y = 3+2a-\dfrac{3b-5}{2c + 1}$

$y = \frac 1 {34}\left(3+2a-\dfrac{3b-5}{2c + 1} \right)$

We're not asked to simplify it so I wont. Substituting,

$y = \frac 1 {34}\left(3+2(2)-\dfrac{3(3)-5}{2(4) + 1} \right) = \frac 1 {34}(7-4/9) = \dfrac{59}{306}$

Answer: 59/306

3 0

3 years ago

Please help me. NO LINKS!!​

Please help me. NO LINKS!!