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2 years ago

Given that angle 1 and angle 2 are supplementary, m angle 1 = (2x + 6)° and m angle 3 = (3x - 6)º.

1 answer:

7 0

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zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)

5 0

3 years ago

The answer choices are 11 1 2.48 6

Fudgin [204]

Answer:

The ball reached its maximum height of ( $11\ yards$ ) in ( $1\ second$ ).

Step-by-step explanation:

This question is essentially asking one to find the vertex of the parabola formed by the given equation. One could plot the equation, but it would be far more efficient to complete the square. Completing the square of an equation is a process by which a person converts the equation of a parabola from standard form to vertex form.

The first step in completing the square is to group the quadratic and linear term:

$h(t)=-5t^2 + 10t + 6\\\\h(t) = (-5t^2 + 10t) + 6$

Now factor out the coefficient of the quadratic term:

$h(t)=-5(t^2 -2t) + 6$

After doing so, add a constant such that the terms inside the parenthesis form a perfect square, don't forget to balance the equation by adding the inverse of the added constant term:

$h(t) = -5(t^2 -2t) + 6\\\\h(t) = -5(t^2 -2t + 1 -1 ) + 6$

Now take the balancing term out of the parenthesis:

$\\\\h(t)=-5(t^2 -2t + 1) + 6 + ((-1)(-5))$

Simplify:

$h(t) = -5(t^2 -2t + 1) + 6 + 5\\\\h(t) = -5(t-1)^2 + 11$

The x-coordinate of the vertex of the parabola is equal to the additive inverse of the numerical part of the quadratic term. The y-coordinate of the vertex is the constant term outside of the parenthesis. Thus, the vertex of the parabola is:

$(1, 11)$