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3 years ago

The function f(t)=5tan2t does not have an amplitude and has a period of \pi yes or no?

Mathematics

2 answers:

Bess [88]3 years ago

8 0

It's multiplied by 5, so the amplitude is 5.

The period is pi/2 since it's tan(2t)

Fittoniya [83]3 years ago

4 0

Answer:

Step-by-step explanation:

The function is a tangent, therefore, there is no amplitude. The range is all reals, but the domain is where it gets funky.

there is an asymptote at pi/4, and the period is located at every multiple of pi/2. Because pi/2 is not equivilant to pi, this portion of the question is false.

This is a conditional question, so both have to be true. Therefore, the answer is no.

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creativ13 [48]

Answer:

${f}^{ - 1} (x)= \frac{x}{2} - 2$

Step-by-step explanation:

The given function is

$f(x) = 2x + 4$ Let

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Interchange x and y.

$x = 2y + 4$

Solve for y,

$x - 4 = 2y$

Divide both sides by 2

$y = \frac{x}{2} - 2$

Therefore;

${f}^{ - 1} (x)= \frac{x}{2} - 2$

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Lady bird [3.3K]

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