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3 years ago

Does anybody know advanced trig??? jk this is for kaitlynn have a good one u guys :)

Mathematics

2 answers:

DerKrebs [107]3 years ago

8 0

Answer:

Well even if it was a joke, I don't know it but I can tell you a few things about it

Step-by-step explanation:

1. It covers the inverse trigonometric functions

2. It's solving equations involving trig concepts

3. 3 identities for trig are sine, cosine, and tangent

grandymaker [24]3 years ago

5 0

Answer:

yeno

Step-by-step explanation:

Advance trigonometry covers the inverse trigonometric functions, solving equations involving trig concepts, and additional identities, including those of double and half angles Topics include: The Inverse Sine Function.

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Find the midpoint of the segment with the following endpoints.<br> (7, -2) (1, -10)

dem82 [27]

Answer:

Step-by-step explanation:

The formula of a midpoint of the segment AB:

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We have A(7, -2) and B(1, -10). Substitute:

$M_{AB}\left(\dfrac{7+1}{2};\ \dfrac{-2+(-10)}{2}\right)\\\\M_{AB}\left(\dfrac{8}{2};\ \dfrac{-12}{2}\right)\\\\M_{AB}(4;\ -6)$

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3 years ago

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Assume there are 200 mendaliens living on an island in my back yard. if my island has an area of 20 hectares, what is the popula

Anit [1.1K]

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4 years ago

Convert the rectangular coordinates (4√3,−4) to polar form. Let r>0 and 0≤θ<2π.

spayn [35]

Answer:

(8,150)

Step-by-step explanation:

Polar form is in,

(r,theta).

To find r, apply pythagorean theorem using both coordinates to find r.

$(4 \sqrt{3} ) {}^{2} + ( - 4 {}^{2} ) = {r}^{2}$

$48 + 16 = {r}^{2}$

$r = 8$

To find theta, since we know the opposite and adjacent side, apply the tangent function.

$\tan(x) = \frac{ - 4}{4 \sqrt{3} }$

Apply inverse tan function

$\tan {}^{ - 1} ( \frac{ - 4}{4 \sqrt{3} } ) = - 30$

Since the degrees have to be in between 0 and 360. Add 180 to -30.

$- 30 + 180 = 150$

So the answer is

(8,150).

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3 years ago

Please help! I will give brainliest!!!

Romashka-Z-Leto [24]

Answer:

what i the answerrrr

Step-by-step explanation:

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Answer:

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2 years ago

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