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ElenaW [278]
1 year ago
10

In which of the following intervals does the trigonometric inequality sec(x) < cot(x) always hold true.

Mathematics
1 answer:
Ray Of Light [21]1 year ago
8 0

The inequality for sec (x) < cot (x) is;  π/2 < x < π

<h3>How to express trigonometric inequality?</h3>

We are given that;

We want to find the intervals that the trigonometric inequality sec (x) < cot (x) always hold true.

This can also be expressed as;

1/cos (x) < 1/tan (x)

Now, this can happen only in the quadrant where tan (x) is negative and cos x is positive which is in fourth quadrant where;

π/2 < x < π

The inequality for sec (x) < cot (x) is;  π/2 < x < π

Read more about Trigonometric Inequality at; brainly.com/question/12094532

#SPJ1

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masya89 [10]

25x=65

We divide 25 to cancel it out.

65/25=2.6

I would assume 3, because 2.6 is 3 when rounded up.

---

hope it helps

3 0
3 years ago
What is the solution to the equation?
Gelneren [198K]

Answer:

Hello,

answer B

2

Step-by-step explanation:

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3 0
3 years ago
Aya = a1 + (n - 1)d
Alenkasestr [34]

Answer:

The 26th term of an arithmetic sequence is:

a_{26}=67

Hence, option A is true.

Step-by-step explanation:

Given

  • a₁ = -33
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An arithmetic sequence has a constant difference 'd' and is defined by

a_n=a_1+\left(n-1\right)d

substituting a₁ = -33 and d = 4 in the nth term of the sequence

a_n=-33+\left(n-1\right)4

\:a_n=-33+4n-4

a_n=4n-37

Thus, the nth term of the sequence is:

a_n=4n-37

now substituting n = 26 in the nth term to determine the 26th term of the sequence

a_n=4n-37

a_{26}=4\left(26\right)-37

a_{26}=104-37

a_{26}=67

Therefore, the 26th term of an arithmetic sequence is:

a_{26}=67

Hence, option A is true.

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2 years ago
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Norma-Jean [14]
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5 0
3 years ago
I need help on this math problem ​
Mila [183]

Answer:

x = 2

y = —3

Step-by-step explanation:

3x — 9 = —2x + 1

x = 2

y = —3

8 0
3 years ago
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