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Viktor [21]
3 years ago
13

What is the range of y=sec^-1(x)?

Mathematics
2 answers:
disa [49]3 years ago
7 0

Answer:

Range is [0, π/2) ∪ [π/2, π)

Step-by-step explanation:

For the given function y=sec^{-1}x if we draw a graph we find that it's discontinuous at x = -1 and 1.

Therefore domain of the function will be = (-∞, -1]∪[1, ∞)

and the range of the function will be [0, π/2) ∪ (π/2, π].

Bingel [31]3 years ago
3 0
Assuming that the x is not part of the ^-1 the range would be 

[0,π/2) U (π/2,<span>π]</span>
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24$

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$80 - $32 - $24 = $24

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Solve the right triangle, ΔABC, for all the missing side and angles to the nearest tenth given sides a = 10.6 and b = 19.2.
Aliun [14]

Answer:

Part 1) c=21.9\ units

Part 2) B=61.1\°

Part 3) A=28.9\°

Part 4) C=90\°

Step-by-step explanation:

step 1

Find the length side c

Applying the Pythagoras Theorem

c^{2}=a^{2}+b^{2}

substitute the given values

c^{2}=10.6^{2}+19.2^{2}

c^{2}=481

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step 2

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we know that

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substitute the given values

tan(B)=19.2/10.6

B=arctan(19.2/10.6)=61.1\°

step 3

Find the measure of angle A

we know that

The measure of interior angles in a triangle must be equal to 180 degrees

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∠A+∠B+∠C=180°

Remember that in a right triangle the measure of angle C is 90 degrees

we have

B=61.1\°

C=90\°

substitute

A+61.1\°+90\°=180\°

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3 years ago
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