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Vinvika [58]
3 years ago
11

In trigonometry sec/cosec?is what​

Mathematics
1 answer:
Stels [109]3 years ago
5 0

Answer:

tan x

Step-by-step explanation:

Using the trigonometric identities

secx = \frac{1}{cosx} , cosecx = \frac{1}{sinx} , then

\frac{secx}{cosecx}

= \frac{\frac{1}{cosx} }{\frac{1}{sinx} }

= \frac{1}{cosx} × sinx

= \frac{sinx}{cosx}

= tan x

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The difference of a number and 6 is the same as 5 times the sum of the number and 2. What is the number?
S_A_V [24]

Step-by-step explanation:

Lets consider the unknown number as x

according to the question,

6-x= 5(x+2)

6-x= 5x+10

-x-5x=10-6

-6x=4

x=4/-6= 2/-3

x= -2/3

<em>hope this helps </em>

<em>please mark me as brainliest.</em>

6 0
3 years ago
Evan swims 321 laps in a month. He swims for 22 days in the month, How
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Answer:

Step-by-step explanation:

321

8 0
3 years ago
Based on the question
nlexa [21]

From the given table, we have that the lateral limits of f(x) as x -> 3 are different, hence the limit of f(x) does not exist at x = 3.

<h3>What is a limit?</h3>

A limit is given by the value of function f(x) as x tends to a value. For the limit to exist, the lateral limits have to be the same, as follows:

\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)

In this problem, we have that:

  • To the left of x = 3, that is, for values that are less than x = 3, f(x) - > -3.
  • To the right of x = 3, that is, for values that are greater than x = 3, f(x) -> 4.

Hence the lateral limits are given as follows:

  • \lim_{x \rightarrow 3^-} f(x) = -3
  • \lim_{x \rightarrow 3^+} f(x) = 4

Since the lateral limits are different, the limit does not exist.

More can be learned about lateral limits at brainly.com/question/26270080

#SPJ1

8 0
2 years ago
Need help ASAP it due today
Harlamova29_29 [7]

Answer:

3/4

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Plz help
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Answer:

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3 years ago
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