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dexar [7]
4 years ago
7

Write the following expression as a function of a positive acute angle. Use "+" to indicate a positive value and "-" to indicate

a negative value.
cos(-240°) = cos
Mathematics
1 answer:
Mama L [17]4 years ago
4 0

cos(-240°) = cos 120°; the "adjacent sides" and the "hypotenuses" are the same in both cases.  The supplement of 120° is 60°, and 60° is a positive acute angle.  Since cos(-240°) is negative, we need to write

cos(-240°) = -cos(60°).  

Check:  cos(60°)= 1/2, so - cos(60°) = -1/2, which is the value of cos(-240°).

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g100num [7]
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6 0
4 years ago
Find the value of x give reasons to justify your solutions b ∈ ac, d ∈
Novay_Z [31]

Answer:

x = 27°

Step-by-step explanation:

180° - 72° = 108°

108° / 4 = 27°

72° + (27° * 4) = 180°

Hope it helps!

6 0
3 years ago
Show all work to identify the asymptotes and zero of the function f(x) = 6x / x^2 - 36
eduard

Answer:

Zero of the function f(x) is at x = 0

Vertical Asymptotes at x = ±6

Horizontal Asymptotes at y = 0

Step-by-step explanation:

<h3>Vertical Asymptotes </h3>

For a given function f(x):

Vertical Asymptotes are obtained at those values of x, where the function f(x) tends to infinity, I.e.,

<em>When</em><em> </em><em>x</em><em> </em><em>approaches</em><em> </em><em>some</em><em> </em><em>constant</em><em> </em><em>value</em><em> </em><em>b</em><em>u</em><em>t</em><em> </em><em>th</em><em>e</em><em> </em><em>curve</em><em> </em><em>moves</em><em> </em><em>towards</em><em> </em><em>infinity</em><em>.</em><em> </em>

  • If f(x) is a fraction, it'll tend to infinity when it's denominator becomes zero.

Vertical Asymptotes of the given function can be obtained by walking thru the following steps:

<u>Step I</u>

(Factorise the numerator and denominator)

\mathsf{ f(x) = \frac{6x}{ {x}^{2} - 36 } }

<em>x</em><em>²</em><em> </em><em>-</em><em> </em><em>36</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>facto</em><em>rised</em><em> </em><em>into</em><em> </em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>6</em><em>)</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>6</em><em>)</em>

<em>and</em><em>,</em><em> </em><em>ofcourse</em><em>,</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>write</em><em> </em><em>6</em><em>x</em><em> </em><em>as</em><em> </em><em>6</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>0</em><em>)</em><em> </em>

\mathsf{ f(x) = \frac{6(x - 0)}{ (x + 6)(x - 6) } }

<u>Step</u><u> </u><u>II</u>

(Reduce the fraction to its simplest form by canceling out the common factors)

<em>There aren't any common factors in the numerator and denominator in this case.</em>

<u>Step</u><u> </u><u>III</u>

(Look for the values of x which cause the denominator to be zero)

<em>If</em><em> </em><em>we</em><em> </em><em>put</em><em> </em>x = 6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em>

Also,

<em>If</em><em> </em><em>we</em><em> </em><em>substitute</em><em> </em><em>x</em><em> </em><em>with</em><em> </em> -6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em><em>.</em><em> </em>

The two values of x indicate the two Vertical Asymptotes of the function f(x).

Therefore,

<u>Vertical</u><u> </u><u>Asymptotes</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>function</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u><u>are</u><u>:</u>

\boxed{ \mathsf{x =  \pm6}}

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3 /><h3>Horizontal Asymptotes:</h3>

Horizontal Asymptotes are obtained When x tends to infinity and y approaches some constant value.

I'll be using the concept of limits for this.

\mathsf{y = \frac{6x}{ {x}^{2} - 36 }  }

<em>dividing</em><em> </em><em>and</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>x</em><em>²</em><em> </em><em>(</em><em>Yep</em><em>!</em><em> </em><em>so</em><em> </em><em>if</em><em> </em><em>x</em><em> </em><em>becomes</em><em> </em><em>infinity</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em> </em><em>and</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em>²</em><em> </em><em>all</em><em> </em><em>such</em><em> </em><em>terms</em><em> </em><em>become</em><em> </em><em>0</em><em>,</em><em> </em><em>'</em><em>cause</em><em> </em><em>1</em><em>/</em><em> </em><em>∞</em><em> </em><em>is</em><em> </em><em>0</em><em>)</em><em> </em>

\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6x}{ {x}^{2} } }{  \frac{ {x}^{2} - 36 }{ {x}^{2} }  } ) }

\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6}{ x } }{  1-  \frac{36 }{ {x}^{2} }  } ) }

Substitute x with ∞, you get zero/ 1

\implies  \boxed{\mathsf{y = 0}}

So, the horizontal Asymptote of the function is y = 0, that is the x axis

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>Zeroes of a function:</h3>

The values of x that reduces f(x) to zero are called the zeroes of f(x).

Here, only x = 0 acts as the zero of the function.

[NOTE:

  • For finding <u>Vertical Asymptotes</u><u>,</u>Equate the denominator to 0. And
  • For finding <u>Zeroes</u><u>,</u> Equate the numerator to 0]

__________________

[That's what it's graph looks like. ]

3 0
3 years ago
Explain the difference between perimeter and area. What do they measure? What types of units are they measured in? NEED ANSWER S
kirill115 [55]
First the difference between the perimeter and area is that area is the amount of space inside a shape, and perimeter is the distance around the shape
^thats what they measure
And since perimeter is a length or distance, the unit of measure is inches, feet, yards, or miles. The area is usually found by doing l x w
Hope this helped
5 0
4 years ago
-4 (5a-4)=2 (6-10a)+4
Anestetic [448]

-20a+8=12-20a+4

8=8

This is the answer.

4 0
4 years ago
Read 2 more answers
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