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mestny [16]
3 years ago
15

Which of the following represents a linear function?

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
D is the only one that is linear. If there is any exponent other than 1 then it is not linear
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Answer:

Step-by-step explanation:

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Help plzzzz second time show work
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3 years ago
What is the approximate length of the radius of a circle with a circumference of 63 inches?
Sergeeva-Olga [200]

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3 years ago
If a sweatshirt is on sale for 35% and now costs $52, <br>what was the original price?​
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3 years ago
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If it exists, solve for the inverse function of each of the following:
nata0808 [166]

Answer:

<em>The solution is too long. So, I included them in the explanation</em>

Step-by-step explanation:

This question has missing details. However, I've corrected each question before solving them

Required: Determine the inverse

1:

f(x) = 25x - 18

Replace f(x) with y

y = 25x - 18

Swap y & x

x = 25y - 18

x + 18 = 25y - 18 + 18

x + 18 = 25y

Divide through by 25

\frac{x + 18}{25} = y

y = \frac{x + 18}{25}

Replace y with f'(x)

f'(x) = \frac{x + 18}{25}

2. g(x) = \frac{12x - 1}{7}

Replace g(x) with y

y = \frac{12x - 1}{7}

Swap y & x

x = \frac{12y - 1}{7}

7x = 12y - 1

Add 1 to both sides

7x +1 = 12y - 1 + 1

7x +1 = 12y

Make y the subject

y = \frac{7x + 1}{12}

g'(x) = \frac{7x + 1}{12}

3: h(x) = -\frac{9x}{4} - \frac{1}{3}

Replace h(x) with y

y = -\frac{9x}{4} - \frac{1}{3}

Swap y & x

x = -\frac{9y}{4} - \frac{1}{3}

Add \frac{1}{3} to both sides

x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}

x + \frac{1}{3}= -\frac{9y}{4}

Multiply through by -4

-4(x + \frac{1}{3})= -4(-\frac{9y}{4})

-4x - \frac{4}{3}= 9y

Divide through by 9

(-4x - \frac{4}{3})/9= y

-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y

\frac{-4x}{9} - \frac{4}{27}= y

y = \frac{-4x}{9} - \frac{4}{27}

h'(x) = \frac{-4x}{9} - \frac{4}{27}

4:

f(x) = x^9

Replace f(x) with y

y = x^9

Swap y with x

x = y^9

Take 9th root

x^{\frac{1}{9}} = y

y = x^{\frac{1}{9}}

Replace y with f'(x)

f'(x) = x^{\frac{1}{9}}

5:

f(a) = a^3 + 8

Replace f(a) with y

y = a^3 + 8

Swap a with y

a = y^3 + 8

Subtract 8

a - 8 = y^3 + 8 - 8

a - 8 = y^3

Take cube root

\sqrt[3]{a-8} = y

y = \sqrt[3]{a-8}

Replace y with f'(a)

f'(a) = \sqrt[3]{a-8}

6:

g(a) = a^2 + 8a- 7

Replace g(a) with y

y = a^2 + 8a - 7

Swap positions of y and a

a = y^2 + 8y - 7

y^2 + 8y - 7 - a = 0

Solve using quadratic formula:

y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

a = 1 ; b = 8; c = -7 - a

y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a} becomes

y = \frac{-8 \±\sqrt{8^2 - 4 * 1 * (-7-a)}}{2 * 1}

y = \frac{-8 \±\sqrt{64 + 28 + 4a}}{2 * 1}

y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}

y = \frac{-8 \±\sqrt{92 + 4a}}{2 }

Factorize

y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }

y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }

y = -4 \±\sqrt{(23 + a)}

g'(a) = -4 \±\sqrt{(23 + a)}

7:

f(b) = (b + 6)(b - 2)

Replace f(b) with y

y  = (b + 6)(b - 2)

Swap y and b

b  = (y + 6)(y - 2)

Open Brackets

b  = y^2 + 6y - 2y - 12

b  = y^2 + 4y - 12

y^2 + 4y - 12 - b = 0

Solve using quadratic formula:

y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

a = 1 ; b = 4; c = -12 - b

y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a} becomes

y = \frac{-4\±\sqrt{4^2 - 4 * 1 * (-12-b)}}{2*1}

y = \frac{-4\±\sqrt{4^2 - 4 *(-12-b)}}{2}

Factorize:

y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}

y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}

y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}

y = \frac{-4\±2\sqrt{16+b}}{2}

y = -2\±\sqrt{16+b}

Replace y with f'(b)

f'(b) = -2\±\sqrt{16+b}

8:

h(x) = \frac{2x+17}{3x+1}

Replace h(x) with y

y  = \frac{2x+17}{3x+1}

Swap x and y

x  = \frac{2y+17}{3y+1}

Cross Multiply

(3y + 1)x = 2y + 17

3yx + x = 2y + 17

Subtract x from both sides:

3yx + x -x= 2y + 17-x

3yx = 2y + 17-x

Subtract 2y from both sides

3yx-2y  =17-x

Factorize:

y(3x-2)  =17-x

Make y the subject

y = \frac{17 - x}{3x - 2}

Replace y with h'(x)

h'(x) = \frac{17 - x}{3x - 2}

9:

h(c) = \sqrt{2c + 2}

Replace h(c) with y

y = \sqrt{2c + 2}

Swap positions of y and c

c = \sqrt{2y + 2}

Square both sides

c^2 = 2y + 2

Subtract 2 from both sides

c^2 - 2= 2y

Make y the subject

y = \frac{c^2 - 2}{2}

h'(c) = \frac{c^2 - 2}{2}

10:

f(x) = \frac{x + 10}{9x - 1}

Replace f(x) with y

y = \frac{x + 10}{9x - 1}

Swap positions of x and y

x = \frac{y + 10}{9y - 1}

Cross Multiply

x(9y - 1) = y + 10

9xy - x = y + 10

Subtract y from both sides

9xy - y - x = y - y+ 10

9xy - y - x =  10

Add x to both sides

9xy - y - x + x=  10 + x

9xy - y =  10 + x

Factorize

y(9x - 1) =  10 + x

Make y the subject

y = \frac{10 + x}{9x - 1}

Replace y with f'(x)

f'(x) = \frac{10 + x}{9x -1}

8 0
2 years ago
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