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3 years ago

Verify that -tan2x = (2tanx) / (sec^2x - 2)

Mathematics

1 answer:

sashaice [31]3 years ago

8 0

Answer:

Below.

Step-by-step explanation:

- tan 2x = -2tanx / (1 - tan^2x)

Using the identity tan^2x = sec^2x - 1 and substituting for tan^2x:

- tan 2x = -2 tanx / (1 - (sec^2x - 1))

= 2 tanx / ( - 1(sec^2x + 2))

= 2 tan x / (sec^2 x - 2)

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2 years ago

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Graph h(x)=0.5(x+2)^2 -4 by following these steps:

Goshia [24]

Answer:

Step 1: Identify a, h, and k.

a=0.5, h= -2, k= -4

Step 3: The axis of symmetry is the line x= -2

Step 4: h(-4)= -2 h(-6)= 4

The graph should have points plotted at (-6, 4), (-4, -2), (-2, -4), (0, -2), and (2, 4)

Step-by-step explanation: The answers were already there, just here to repeat for others who are confused.

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2 years ago

For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

kvv77 [185]

Answer: Option C

$f(x) = x^2;\ k (x) = x ^ 2 -7$

Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

In this case we have

$f (x) = x ^ 2$

We know that this function has its vertex in point (0,0).

Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$ .

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$

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3 years ago

A line is defined by the equation. The line passes through a point whose y-coordinate is 0. What is the x-coordinate of this poi

emmainna [20.7K]

The answer to that would be 0 because the line goes through the orgin witch the points are 0

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3 years ago