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Viktor [21]
2 years ago
13

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

Mathematics
2 answers:
disa [49]2 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]2 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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I need help with the last 2 I need the next 3 in the sequence and the nth term ​
kkurt [141]

Answer:

f. the sequence goes by the half of the previous number.

= 16, 8, 4, 2, 1, 1/2, 1/4, 1/8...

g. the sequence goes by adding the consecutive odd number added to the previous number.

first number= 3.

second number= 3+3

= 6

third number= 6+5

= 11

fourth number= 11+7

= 18

fifth number= 18+9

= 27

sixth number= 27+11

= 38

seventh number= 38+13

= 51, etc.

= 3, 6, 11, 18, 27, 38, 51...

7 0
3 years ago
7(2x+5) = 11x-9-x <br><br> Please HELP
mr_godi [17]

Step-by-step explanation:

7(2x+5)=11x-9-x

14x+35=10x-9

4x+35=-9

4x=-44

x=-11

3 0
2 years ago
Please answer these 2 questions for me! Take your time. These are my last points because no one answers my questions :( I need h
umka2103 [35]

Wil you mark me brainliest?

7 0
3 years ago
alvin and simon raced a distance of 2.9 x 10^2 kilometers. they need to race 7 times that distance in kilometer. how far must th
dimaraw [331]

The distance in kilometers in scientific notation 2.03 \times 10^3 km

<h3>Scale factors</h3>

Scale factors are values or constants multiplied with a variable

If Alvin and Simon raced a distance of 2.9 \times 10^2 kilometers and they need to race 7 times that distance in a kilometer, the distance traveled will be:

  • Distance travelled = 7(2.9 \times 10^2)
  • Distance travelled = 2.03 \times 10^3 km

Hence the distance in kilometers in scientific notation 2.03 \times 10^3 km

Learn more on distance here: brainly.com/question/23848540

8 0
2 years ago
The distance from Kenya's house to her job is 18 miles. One morning she left her house at 7:15 a.m. She traveled at the rate of
Snezhnost [94]

20 minutes = 1/3 of an hour

1/3 = 0.333

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18-15 = 3

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0.483 x 60 = 28.98 so approximately 29 minutes total driving time

7:15 + 29 minutes is 7:44 am she arrived at work

8 0
3 years ago
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