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nevsk [136]
3 years ago
9

Using the 3 functions below, make the desired transformation(s) to the correct function and describe the transformation(s). Writ

e the new equation k(x).
If the functions are: f(x)=2x+1 g(x)=−12x−3 h(x)=−2
then find k(x) if k(x)=g(x)−3
Transformation(s) of g(x) to k(x):

-3:




New equation:

k(x)=

Mathematics
1 answer:
Eduardwww [97]3 years ago
5 0

Answer:

The transformation of g(x) to k(x) consists in a vertical translation. The new equation is k(x) = -12\cdot x-6.

Step-by-step explanation:

Let g(x) = -12\cdot x - 3. We proceed to make the required transformations on g(x), which consists in one vertical translation, 3 units in the -y direction. That is to say:

k(x) = g(x) - 3 (1)

k(x) = (-12\cdot x - 3) - 3

k(x) = -12\cdot x-6

Then, the transformation of g(x) to k(x) consists in a vertical translation. The new equation is k(x) = -12\cdot x-6.

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This diagram of a rectangular city park was drawn using a scale of 1 centimeter to 20 meters
I am Lyosha [343]

Answer:

Scale factor for the drawing from actual park = \frac{1}{20}

Area of the actual park = 1200 cm²

Step-by-step explanation:

Length of the rectangular city park = 5 cm

Width of the rectangular park = 6 cm

Using scale factor 1 cm = 20 meters

Scale factor = \frac{\text{length of the park in drawing}}{\text{Actual length of park}}=\frac{1}{20}

\frac{5}{\text{Actual length of park}}=\frac{1}{20}

Actual length = 5 × 20 = 100 meters

Actual width = 6 × 20 = 120 meters

Area of the rectangular park = length × width

                                               = 100 × 120

                                               = 1200 square meters

Therefore, Scale factor from actual length to the length in drawing = 1 : 20

Area of the rectangular park = 1200 square feet

7 0
3 years ago
Can someone please tell me if any of my answers are wrong!????
ra1l [238]

Couldn’t quite read everything, the writing is very light. But from what I could read, it was correct for the most part. :)

5 0
2 years ago
Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.
guapka [62]

\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}

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2 years ago
Please I need help
Dimas [21]

Answer:

Step-by-step explanation:

The directrix is a vertical line, so the parabola is horizontal. The focus lies to the left of the directrix, so the parabola opens to the left.

For a left-opening parabola:

x = a(y-k)²+h,

a < 0,

vertex (h,k)

focal length p = 1/|4a|

focus (h-p, k)

directrix: x=h+p

Apply your data

focus (1,-4)

directrix x=2

vertex (1.5,-4).

focal length p = 0.5

a = -1/|4p| = -½

x = -½(y-2)²+ ½

8 0
2 years ago
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How do I use this app
Natalka [10]

Answer:

Ask a question and fellow persons will help you .

Step-by-step explanation:

Just type your question in the blue box at the top of course.

4 0
3 years ago
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