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const2013 [10]
2 years ago
14

Pretest: Unit 1

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
3 0

A vertical line that the graph of a function approaches but never intersects. The correct option is B.

<h3>When do we get vertical asymptote for a function?</h3>

Suppose that we have the function f(x) such that it is continuous for all input values < a or > a and have got the values of f(x) going  to infinity or -ve infinity (from either side of   x = a) as x goes near a, and is not defined at   x = a, then at that point, there can be constructed a vertical line  x = a and it will be called as vertical asymptote for f(x) at   x = a

A vertical asymptote can be described as a vertical line that the graph of a function approaches but never intersects.

Hence, the correct option is B.

Learn more about Vertical Asymptotes:

brainly.com/question/2513623

#SPJ1

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Answer:

78

Step-by-step explanation:

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X squared = −169. A= undefined B= 13 C= ±13 D= −13
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Step-by-step explanation:

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3 years ago
Sin5A/SinA-cos5A/cosA=4cos2A​
irina1246 [14]

Answer:

See Explanation

Step-by-step explanation:

\frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ LHS = \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}   \\  \\  =  \frac{ \sin5A \:\cos A -  \cos5A \:  \sin A}{\sin A \:\cos A }  \\  \\  =  \frac{ \sin(5A -A )}{\sin A \:\cos A}  \\  \\ =  \frac{ \sin 4A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =4\cos 2A \\  \\  = RHS \\  \\ thus \\  \\  \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ hence \: proved

5 0
3 years ago
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