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Yuri [45]
1 year ago
15

Use trigonometric identities, algebraic methods, and inverse trigonometric functions, as necessary, to solve the following trigo

nometric equation on the interval [0, 27).Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."Scos(x) - 5 = 0

Mathematics
1 answer:
uranmaximum [27]1 year ago
6 0

we have the function

5\cos x-5=0

Rewrite

\begin{gathered} 5\cos x=5 \\ \cos x=\frac{5}{5} \\ \cos x=1 \end{gathered}

The domain for x is the given interval [0,2pi) ----> 2pi is not included

the values of x are

x=0 radians

<h2>The solution is x=0</h2>
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The number of laughs (denoted by L) can be defined as a function of the number of jokes (denoted by J), the amount of knowledge
Blababa [14]

Answer:

L = f(J, K, F)

L = (JK²)/F

So, as the formula appears, the appropriate units will be option A.

Unit for number of laughs = (jokes*knowledge²) / familiarity

Just like the formula suggests.

Step-by-step explanation:

Complete Question

The number of laughs (denoted by L) can be defined as a function of the number of jokes (denoted by J), the amount of knowledge about the joke material (denoted by K) and the familiarity with the jokes (denoted by F) using this formula: L = (JK²)/F

Select an appropriate unit for number of laughs:

A) (jokes*knowledge²) / familiarity

B) familiarity / (jokes*knowledge²)

C) familiarity² / (jokes*knowledge)

D) (jokes*knowledge) / familiarity²

Solution

L = f(J, K, F)

where

L = number of laughs

J = number of jokes

K = amount of knowledge about the joke material

F = familiarity with the jokes

Analysing how the dependent variable depends on each of the independent variables.

- Number of jokes

The number of laughs will increase with the number of jokes and vice versa. It can be stated that there is a direct variation between the number of laughs and number of jokes.

- Amount of knowledge about the joke material

The more one understands the joke material, the funnier the joke. In fact, the joke can be termed funnier when one understands the joke material deeply. The direct variation of number of laughs to knowledge about joke material isn't just enough, the number of laughs varying directly as the square of the knowledge of joke material seems more fitting.

- Familiarity with the joke

The more familiar one is with a joke, the less funny it is. Hence, it's an inverse relationship between number of laughs and the familiarity with the joke.

L = (JK²)/F

So, as the formula appears, the appropriate units will be

(jokes*knowledge²) / familiarity

Just like the formula suggests.

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
Which expression is equivalent to ^3 square root 64a^6 b^7 c^9
sergij07 [2.7K]

Answer:

4a^{2} b^{2} c^{3} (\sqrt[3]{b})

Step-by-step explanation:

The given expression is :

\sqrt[3]{(64}a^{6}b^{7} c^{9} )

Writing 64 ,a,b,c as cubes we have:

= \sqrt[3]{(}4^{3}( a^{2})^3( b^{2})^3.b( c^{3} )^3)

Using radical rule we have :

=4a^{2} b^{2} c^{3} (\sqrt[3]{b}).

The second option is the right answer.



7 0
3 years ago
Read 2 more answers
Which linear inequality is represented by the graph
Anarel [89]

Answer:

the second choice

Step-by-step explanation:

The last two choices could not be correct, because it would need a dashed line.  The line under the greater than or less than symbol mean  that the line is included in the answer and it will be a solid line like you see in the picture.  For both of the top two answers the slope that they y intercept is correct.  You just have to decide is the symbol will be greater than or equal to or less than and equal to.  

If we put 0 in x for in the form y = mx + b, what would y be?

y = 1/3 x - 1

y = 1/3(0) - 1

y = -1

The ordered pair is (0,1).  That is the part that is shaded so that shows us that the second answer is correct.

4 0
2 years ago
Please help need answer ​
gtnhenbr [62]
Your answer is The first one
6 0
3 years ago
Hat is the 4th equivalent fraction to 1/11​
quester [9]

Answer:

4/44 i think

Step-by-step explanation:

5 0
3 years ago
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