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maxonik [38]
3 years ago
15

The trigonometric ratios sine and secant are reciprocals of each other true or false?

Mathematics
2 answers:
kodGreya [7K]3 years ago
7 0

True: The trigonometric ratios sine and secant are reciprocals of each other

jek_recluse [69]3 years ago
4 0

That's FALSE.


It's sine and cosecant that are reciprocals, as are cosine and secant, as well as tangent and cotangent.


The "co" or complementary functions are their corresponding non-complementary function applied to the complementary angle: cos(x) = sin(90-x), that kind of thing.


The non-co functions sine, secant and tangent all increase with increasing angle in the first quadrant. The co functions all decrease.


Since when sine increases its reciprocal decreases, its reciprocal must be one of the co functions, and of course it's cosecant.



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The length of two sides of a triangle are 10, 15 find the range of possible lengths for the third side
Zepler [3.9K]

Answer:

Range for third side is  

( 5 , 25 )  cm.

Step-by-step explanation:

As two sides of triangle are  10  and  15 ,

the third side would have to be less than the sum of other two sides i.e. less than  25  cm.

On the other hand if it is smaller one than this side plus side of length  10

should be greater than  15  and therefore

this side is greater than  15 − 10 = 5  cm.

Hence range is  

( 5 , 25 )

Hope this answer helps you :)

Have a great day

Mark brainliest

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3 years ago
Q = m-a/x (solve for m)​
DaniilM [7]

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Step-by-step explanation:

Q=m-a/x

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You pick a card at random, put it back, and then pick another card at random. 2 3 4 5 What is the probability of picking a numbe
olga_2 [115]

Answer:

3/4 or a 75% chance

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Then, picking a 4 out of 4 cards is a 1/4 chance; since that is one card out of 4 total possibilities.

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Answer:

1)   x=\dfrac12

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3)  see below

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Step-by-step explanation:

<u>Question 1</u>

15-0.5(4x-2)+4x=17

\implies 15-2x+1+4x=17

\implies 2x+16=17

\implies 2x=1

\implies x=\dfrac12

<u>Question 2</u>

\textsf{rearrange} \ n=m+1 :

\implies -m=-n+1

\textsf{add equations} \ -m=-n+1 \ \textsf{and} \ m=2n+2:\\

\implies0=n+3

\implies n=-3

\textsf{substitute} \ \ n=-3 \ \ \textsf{into} \ \ m=n-1:

\implies m=-3-1

\implies m=-4

<u>Question 3</u>

subtract the second equation from the first

divide both sides by -4

substitute found value for y into first equation

solve for x

<u>Question 4</u>

3j=k

k=3j+1

\textsf{rearrange} \ 3j=k :

\implies -k=-3j

\textsf{add equations}\  -k=-3j \ \ \textsf{and}\ \ k=3j+1:

\implies 0=1

Solution = A

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