What is the inverse of the function f(x)=-6x-7

A sequence is defined by the function f ( n ) = f ( n − 1 ) + 5 , where n represents the number of the term for n > 1 , and f

Shkiper50 [21]

Step-by-step explanation:

if f(1)=-4, then adding 5 would give you 1, again would give you 6, again would give 11. Therefore, your first four terms are -4,1,6,11.

6 0

2 years ago

Which transformations could have been used to create A'B'C'? Choose all that apply.

Eduardwww [97]

The correct transformation is a rotation of 180° around the origin followed by a translation of 3 units up and 1 unit to the left.

<h3>Which transformation is used to get A'B'C'?</h3>

To analyze this we can only follow one of the vertices of the triangle.

Let's follow A.

A starts at (3, 4). If we apply a rotation of 180° about the origin, we end up in the third quadrant in the coordinates:

(-3, -4)

Now if you look at A', you can see that the coordinates are:

A' = (-4, -1)

To go from (-3, -4) to (-4, -1), we move one unit to the left and 3 units up.

Then the complete transformation is:

A rotation of 180° around the origin, followed by a translation of 3 units up and 1 unit to the left.

If you want to learn more about transformations:

brainly.com/question/4289712

#SPJ1

6 0

1 year ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$