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kakasveta [241]
3 years ago
5

You know that for any , neither sin nor cos can be greater than 1. how can you explain this using the unit circle definitions of

sine and cosine
Mathematics
1 answer:
Feliz [49]3 years ago
4 0
In the unit circle the hypotenuse of the triangle formed  is equal to radius of circle , which = 1. The point on the circle formed by the intersection of the hypotenuse  is  (cos q , sin q)  where q is the angle between the x axis and the hypotenuse.   As the hypotenuse = 1 the opposite and adjacent sides of the triangle < 1   so sin and cos of q  must both be  <= 1.
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What are the domain and range of y=cot x? Select one choice for domain
Lostsunrise [7]

The domain is \boxed{x \neq n\pi}, where n is an integer.

  • This is because cot x = 1/tan(x), and if tan x = 0, then the fraction is undefined.

The range is <u>all real numbers</u>.

4 0
1 year ago
Given cos ⁡ ( θ ) = 5/6 , then sin ⁡ ( θ ) = a/b .<br><br> Determine the values of a and b.
slega [8]

Answer:

a = √11 and b = 6

Step-by-step explanation:

Refer to attached picture for reference

for an right triangle with angle θ

we are given

cos θ = 5/6 = length of adjacent side / length of hypotenuse

hence

adjacent length = 5 units

hypotenuse length = 6 units

the missing side is the "opposite" length which we can find with the Pythagorean equation. in our case:

hypotenuse ² = adjacent ² + opposite²   (rearrange)

opposite ² = hypotenuse ² - adjacent ²

opposite ² = 6² - 5²

opposite = √ (6²-5²) = √11

sin θ = opposite length / hypotenuse  (substitute values above)

sin θ = √11 / 6

hence a = √11 and b = 6

3 0
3 years ago
Which congruency theorem can be used to prove
irina1246 [14]

Answer:

SAS

Step-by-step explanation:

Given: Two triangles ΔABD and ΔDCA,  

We have, AD=AD (normal)  

∠A=∠A (Given)  

BA=CD (Sides inverse to rise to points are consistently equivalent)  

With the SAS rule of congruency,  

ΔABD≅ΔDCA

Brainliest?

4 0
2 years ago
Someone help ASAP!!!!!!
Alchen [17]

Answer:

Length, l = 11 ft.

Width, w = 9 ft.

Step-by-step explanation:

From the given data, the area of the rectangle = 99 ft².

Area of the rectangle = Length, l X Width, w

Here, Length, l = 7 more than twice the width

⇒       Length, l = 7 + 2w

Therefore, Area, A = 99 = (7 + 2w)w

⇒ 99 = 7w + 2w²

⇒ 2w² + 7w - 99 = 0

Solve the Quadratic equation using the formula: x = $ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ for the quadratic equation ax² + bx + c = 0.

Therefore, w = $ \frac{-7 \pm \sqrt{7^2 -4(2)(-99)}}{2(2)} $

$ = \frac{-7 \pm \sqrt{49 + 8(99)}}{4} $

$ = \frac{-7 \pm + \sqrt{841}}{4} $

Since, $ \sqrt{841} = 29 $ we get:

$ w = \frac{-7 \pm 29}{4} $

This gives two values of 'w', viz., w = $ \frac{-7 - 29}{4} $, $ \frac{-7 + 29}{4} $

$ \implies w = \frac{22}{4}, \frac{-36}{4} $

⇒ w = $ \frac{22}{4} $, -9.

We take the integer values.

If w = -9, then l = 2(-9) + 7

⇒ l = - 18 + 7 = - 11

Therefore, the length, l of the rectangle = - 11 ft.

and the width, w of the rectangle = - 9 ft.

Hence, the answer.

3 0
3 years ago
WILL GIVE BRAINLEST TO WHOEVER ANSWERS THXS &lt;3
ikadub [295]

hmmm i would help but I completely forgot

4 0
2 years ago
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