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Sunny_sXe [5.5K]
2 years ago
6

Which function represents a reflection of f(x) = 5(0.8)x across the x-axis?

Mathematics
1 answer:
lukranit [14]2 years ago
8 0

The function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

<h3>Reflection of functions and coordinates</h3>

Images that are reflected are mirror images of each other. When a point is reflected across the line y = x, the x-coordinates and y-coordinates change their position. In a similar manner, when a point is reflected across the line y = -x, the coordinates <u>changes position but are negated.</u>

Given the exponential function below

f(x) = 5(0.8)^x

If the function f(x) is reflected over the x-axis, the resulting function will be

-f(x)

This means that we are going to negate the function f(x) as shown;

f(x) = -5(0.8)^x

Hence the function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

Learn more on reflection here: brainly.com/question/1908648

#SPJ1

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Answer:

-16

Step-by-step explanation:

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3 years ago
Write each fraction as a sum or difference 14-6x/19
Oksanka [162]
<span>Given number : 14 - 6x / 19
So we need to write this fraction as a sum or difference.
First, simplify the given number:
=> 14 – 6x / 19
=> <u>14 x 19 – (6x</u>)
            19
=> <u>266 – 6x</u>
         19
=> 266 – 6x = -2 * (3x - 133)
Thus, the final result would be:
=> <u>-2 * (3x - 133) </u> => This will be the fraction sum or difference of the given number<u>
</u>               19

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8 0
3 years ago
Find FG. The diagram is not to scale.
soldier1979 [14.2K]
FG = FE
3n - 4 = n + 8
2n = 12
  n = 6

FG = 3(6) - 4
FG = 18 - 4
FG = 14

answer
14
7 0
3 years ago
Read 2 more answers
Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneou
notsponge [240]

Extraneous solutions are the values that we get when solving equations which aren't really solutions to the equation.

<h3>What are extraneous solutions?</h3>

Your information is incomplete. Therefore, an overview will be given.  An extraneous solution is the root of a transformed equation which is not a root of the original equation since it was excluded from the domain of the original equation.

The reason extraneous solutions exist is simply that some operations produce extra answers, and these operations are a part of the path to solving the problem.

Learn more about equations on:

brainly.com/question/2972832

8 0
2 years ago
Try this hard Math Problem if you dare!!
Dennis_Churaev [7]

Answer:

a. (x - 3)^2 + 16

b. 8(x -7)^2

c. (a^2 - 1)(7x - 6) or (a+1)(a-1)(7x-6)

d. (x^2-4)(x^2+3) or (x-2)(x+2)(x^2+3)

e. (a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

Step-by-step explanation:

a.\ (x + 1)^2 - 8(x - 1) + 16

Expand

(x + 1)(x + 1) - 8(x - 1) + 16

Open brackets

x^2 + x + x + 1 - 8x + 8 + 16

x^2 + 2x + 1 - 8x + 24

Collect Like Terms

x^2 + 2x - 8x+ 1  + 24

x^2 - 6x+ 25

Express 25 as 9 + 16

x^2 - 6x+ 9 + 16

Factorize:

x^2 - 3x - 3x + 9 + 16

x(x -3)-3(x - 3) + 16

(x - 3)(x - 3) + 16

(x - 3)^2 + 16

b.\ 8(x - 3)^2 - 64(x-3) + 128

Expand

8(x - 3)(x - 3) - 64(x-3) + 128

8(x^2 - 6x+ 9) - 64(x-3) + 128

Open Brackets

8x^2 - 48x+ 72 - 64x+192 + 128

Collect Like Terms

8x^2 - 48x - 64x+192 + 128+ 72

8x^2 -112x+392

Factorize

8(x^2 -14x+49)

Expand the expression in bracket

8(x^2 -7x-7x+49)

Factorize:

8(x(x -7)-7(x-7))

8((x -7)(x-7))

8(x -7)^2

c.\ 7a^2x - 6a^2 - 7x + 6

Factorize

a^2(7x - 6) -1( 7x - 6)

(a^2 - 1)(7x - 6)

The answer can be in this form of further expanded as follows:

(a^2 - 1^2)(7x - 6)

Apply difference of two squares

(a+1)(a-1)(7x-6)

d.\ x^4 - x^2 - 12

Express x^4 as x^2

(x^2)^2 - x^2 - 12

Expand

(x^2)^2 +3x^2- 4x^2 - 12

x^2(x^2+3) -4(x^2+3)

(x^2-4)(x^2+3)

The answer can be in this form of further expanded as follows:

(x^2-2^2)(x^2+3)

Apply difference of two squares

(x-2)(x+2)(x^2+3)

e.\ a^{4n} -b^{4n}

Represent as squares

(a^{2n})^2 -(b^{2n})^2

Apply difference of two squares

(a^{2n} -b^{2n})(a^{2n} +b^{2n})

Represent as squares

((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})

Apply difference of two squares

(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

6 0
3 years ago
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