If it exists, solve for the inverse function of each of the following:
1. f(x) = 25x - 18
6. gala? +84 - 7
7. 10) = (b + 6) (6-2)
3. A(7)=-=-
4. f(x)=x
9. h(c) = V2c +2
+30
10. f(x) =
5. f(a) = a +8
ox-1
2. 9(x) = -1
2x+17
8. () - 2*​

Answers

Answer 1

Answer:

The solution is too long. So, I included them in the explanation

Step-by-step explanation:

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

Required: Determine the inverse

1:

[tex]f(x) = 25x - 18[/tex]

Replace f(x) with y

[tex]y = 25x - 18[/tex]

Swap y & x

[tex]x = 25y - 18[/tex]

[tex]x + 18 = 25y - 18 + 18[/tex]

[tex]x + 18 = 25y[/tex]

Divide through by 25

[tex]\frac{x + 18}{25} = y[/tex]

[tex]y = \frac{x + 18}{25}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{x + 18}{25}[/tex]

2. [tex]g(x) = \frac{12x - 1}{7}[/tex]

Replace g(x) with y

[tex]y = \frac{12x - 1}{7}[/tex]

Swap y & x

[tex]x = \frac{12y - 1}{7}[/tex]

[tex]7x = 12y - 1[/tex]

Add 1 to both sides

[tex]7x +1 = 12y - 1 + 1[/tex]

[tex]7x +1 = 12y[/tex]

Make y the subject

[tex]y = \frac{7x + 1}{12}[/tex]

[tex]g'(x) = \frac{7x + 1}{12}[/tex]

3: [tex]h(x) = -\frac{9x}{4} - \frac{1}{3}[/tex]

Replace h(x) with y

[tex]y = -\frac{9x}{4} - \frac{1}{3}[/tex]

Swap y & x

[tex]x = -\frac{9y}{4} - \frac{1}{3}[/tex]

Add [tex]\frac{1}{3}[/tex] to both sides

[tex]x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}[/tex]

[tex]x + \frac{1}{3}= -\frac{9y}{4}[/tex]

Multiply through by -4

[tex]-4(x + \frac{1}{3})= -4(-\frac{9y}{4})[/tex]

[tex]-4x - \frac{4}{3}= 9y[/tex]

Divide through by 9

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

[tex]-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y[/tex]

[tex]\frac{-4x}{9} - \frac{4}{27}= y[/tex]

[tex]y = \frac{-4x}{9} - \frac{4}{27}[/tex]

[tex]h'(x) = \frac{-4x}{9} - \frac{4}{27}[/tex]

4:

[tex]f(x) = x^9[/tex]

Replace f(x) with y

[tex]y = x^9[/tex]

Swap y with x

[tex]x = y^9[/tex]

Take 9th root

[tex]x^{\frac{1}{9}} = y[/tex]

[tex]y = x^{\frac{1}{9}}[/tex]

Replace y with f'(x)

[tex]f'(x) = x^{\frac{1}{9}}[/tex]

5:

[tex]f(a) = a^3 + 8[/tex]

Replace f(a) with y

[tex]y = a^3 + 8[/tex]

Swap a with y

[tex]a = y^3 + 8[/tex]

Subtract 8

[tex]a - 8 = y^3 + 8 - 8[/tex]

[tex]a - 8 = y^3[/tex]

Take cube root

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

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

Replace y with f'(a)

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

6:

[tex]g(a) = a^2 + 8a- 7[/tex]

Replace g(a) with y

[tex]y = a^2 + 8a - 7[/tex]

Swap positions of y and a

[tex]a = y^2 + 8y - 7[/tex]

[tex]y^2 + 8y - 7 - a = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 8[/tex]; [tex]c = -7 - a[/tex]

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

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

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

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 }[/tex]

Factorize

[tex]y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }[/tex]

[tex]y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }[/tex]

[tex]y = -4 \±\sqrt{(23 + a)}[/tex]

[tex]g'(a) = -4 \±\sqrt{(23 + a)}[/tex]

7:

[tex]f(b) = (b + 6)(b - 2)[/tex]

Replace f(b) with y

[tex]y = (b + 6)(b - 2)[/tex]

Swap y and b

[tex]b = (y + 6)(y - 2)[/tex]

Open Brackets

[tex]b = y^2 + 6y - 2y - 12[/tex]

[tex]b = y^2 + 4y - 12[/tex]

[tex]y^2 + 4y - 12 - b = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 4[/tex]; [tex]c = -12 - b[/tex]

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

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

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

Factorize:

[tex]y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{16+b}}{2}[/tex]

[tex]y = -2\±\sqrt{16+b}[/tex]

Replace y with f'(b)

[tex]f'(b) = -2\±\sqrt{16+b}[/tex]

8:

[tex]h(x) = \frac{2x+17}{3x+1}[/tex]

Replace h(x) with y

[tex]y = \frac{2x+17}{3x+1}[/tex]

Swap x and y

[tex]x = \frac{2y+17}{3y+1}[/tex]

Cross Multiply

[tex](3y + 1)x = 2y + 17[/tex]

[tex]3yx + x = 2y + 17[/tex]

Subtract x from both sides:

[tex]3yx + x -x= 2y + 17-x[/tex]

[tex]3yx = 2y + 17-x[/tex]

Subtract 2y from both sides

[tex]3yx-2y =17-x[/tex]

Factorize:

[tex]y(3x-2) =17-x[/tex]

Make y the subject

[tex]y = \frac{17 - x}{3x - 2}[/tex]

Replace y with h'(x)

[tex]h'(x) = \frac{17 - x}{3x - 2}[/tex]

9:

[tex]h(c) = \sqrt{2c + 2}[/tex]

Replace h(c) with y

[tex]y = \sqrt{2c + 2}[/tex]

Swap positions of y and c

[tex]c = \sqrt{2y + 2}[/tex]

Square both sides

[tex]c^2 = 2y + 2[/tex]

Subtract 2 from both sides

[tex]c^2 - 2= 2y[/tex]

Make y the subject

[tex]y = \frac{c^2 - 2}{2}[/tex]

[tex]h'(c) = \frac{c^2 - 2}{2}[/tex]

10:

[tex]f(x) = \frac{x + 10}{9x - 1}[/tex]

Replace f(x) with y

[tex]y = \frac{x + 10}{9x - 1}[/tex]

Swap positions of x and y

[tex]x = \frac{y + 10}{9y - 1}[/tex]

Cross Multiply

[tex]x(9y - 1) = y + 10[/tex]

[tex]9xy - x = y + 10[/tex]

Subtract y from both sides

[tex]9xy - y - x = y - y+ 10[/tex]

[tex]9xy - y - x = 10[/tex]

Add x to both sides

[tex]9xy - y - x + x= 10 + x[/tex]

[tex]9xy - y = 10 + x[/tex]

Factorize

[tex]y(9x - 1) = 10 + x[/tex]

Make y the subject

[tex]y = \frac{10 + x}{9x - 1}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{10 + x}{9x -1}[/tex]


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Answers

Answer:

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Step-by-step explanation:

Given that:

Number of calories in a [tex]\frac{3}{4}^{th}[/tex] cup serving = 90 calories

To find:

Unit rate for calories per cup = ?

Number of calories in 2 cups of cereal = ?

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Answers

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Answers

Answer:

Step-by-step explanation:

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Mr. Lee is a hotdog and drink vendor at a baseball game. Hot dogs are 3.50 and drinks are 1.50. If Mr. Lee sold enough to make 450 dollars on Monday. He remembers that on Monday he sold three times as many drinks as he did hot dogs. How many hot dogs and how many drinks did Mr. Lee sell on Monday?

Answers

Answer:

33 hotdogs and 225 drinks on Monday

Step-by-step explanation:

We would use the ratio of 1:3 hotdogs to drinks

Split the ratio in 450 and we get 112.5:337.5

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Answers

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Step-by-step explanation:

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Twice as many students are in the chess club as in the fencing club.The number of students in the fencing club is one fifth the number of students in the cooking club.There are 42 more students in the cooking club than in the chess club.How many students are in the chess club?

Answers

Answer:

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Step-by-step explanation:

Let number of students in cooking club = x

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Number of students in chess club = 2(1/5)x

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

To find the volume of something, you multiply.

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HOPE THIS HELPS YOU!!!!    :)

Answer:

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Step-by-step explanation:

Multiply the diameters all together to get your answers.

5.5 x 3.2 x 7.25

Multiply

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Multiply some more

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You can also just multiply like this:

5.5 x 3.2 x 7.25 = 127.6

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

233 seconds

Step-by-step explanation:

When h = 0 the jumper ends his/her jump. Therefore let h = 0 so that we can solve for t; forming a quadratic equation

0 = 3t^2 - 700t + 200

I am going to solve  using the quadratic formula, but there are other approaches--

[tex]t_{1,\:2}=\frac{-\left(-700\right)\pm \sqrt{\left(-700\right)^2-4\cdot \:3\cdot \:200}}{2\cdot \:3},\\\\\sqrt{\left(-700\right)^2-4\cdot \:3\cdot \:200} = 20\sqrt{1219},\\\\t_{1,\:2}=\frac{-\left(-700\right)\pm \:20\sqrt{1219}}{2\cdot \:3}[/tex]

[tex]\:t_2=\frac{-\left(-700\right)-20\sqrt{1219}}{2\cdot \:3}[/tex]

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Good morning can anybody help me out with this assignment please and thanks! \
please this if for an grade

Answers

Good Morning hope this helped.

Answer:

1: The cost of twelve bottles is $24 cause 12x2=24.

2: The ratio is 24/12 or 2/1

3: $2 every time

The graph of the quadratic function shown on the left is
y = –0.7(x + 3)(x – 4).

The roots, or zeros, of the function are
.



Determine the solutions to the related equation
0 = –0.7(x + 3)(x – 4).

The solutions to the equation are x =
.

Answers

Answer: The roots are x = -3 and x = 4

Explanation:

We'll use the zero product property. This is the idea where if A*B = 0, then either A = 0 or B = 0.

So if

-0.7(x+3)(x-4) = 0

then either

x+3 = 0 or x-4 = 0

Both of those equations solve to

x = -3 or x = 4

The roots are the locations of the x intercepts, which is where the graph crosses or touches the x axis.

Answer:

1)   3- and 4

2)  3- and 4

Step-by-step explanation:

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