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Ilia_Sergeevich [38]
2 years ago
13

A linear function and its inverse are given.

Mathematics
2 answers:
jarptica [38.1K]2 years ago
7 0

Answer:

x: -5,  -3,   0,  1,  4

y:-23, -15, -3, 1, 13        for the function.

x:-23, -15, -3, 1, 13

y: -5,   -3,  0,  1,  4         for the inverse.

Step-by-step explanation:

we know that if we have the function f(x) = y, then the inverse of f(x) (let's call it g(x)) is such that:

g(y) = x.

now we have

y=4x-3

y=(1/4)x+3/4

The only table that works for our first function is:

x: -5,  -3,   0,  1,  4

y:-23, -15, -3, 1, 13

You can see this by replacing the values of x and see if the value of y also coincides.

Then, using the fact that the other table must be for the inverse, we should se a table with the same values, but where the values of x and y are interchanged.

The second table is that one:

x:-23, -15, -3, 1, 13

y: -5,   -3,  0,  1,  4

AlexFokin [52]2 years ago
6 0

Answer: B and D

x:-23, -15, -3, 1, 13

y:-5, -3, 0, 1, 4

x:-5, -3, 0, 1, 4

y:-23, -15, -3, 1, 13

Step-by-step explanation:

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tigry1 [53]
Let x = liters 20% solution to be added
then

.12(12) + .20x = .14(12+x)
.12(12) + .20x = .14(12+x)
1.44 + .20x = 1.68 + .14x
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Hope this helped!!
3 0
2 years ago
What did I do wrong so I don’t do it again on the test
denpristay [2]

Answer:

nothing

Step-by-step explanation:

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8 0
2 years ago
Suppose you can factor x^2 + bx + c as (x + p)(x + q). If c > 0, what could be possible values of p and q?
ad-work [718]

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3 0
2 years ago
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
ddd [48]

Answer:

f'(x) = 4x^2 - 3 for x \le -3

Step-by-step explanation:

See attachment for proper question

Given

f(x) = -\frac{1}{2}\sqrt{x + 3}

For

x \ge -3

Required

Determine the inverse function

f(x) = -\frac{1}{2}\sqrt{x + 3}

Replace f(x) with y

y = -\frac{1}{2}\sqrt{x + 3}

Swap the positions of x and y

x = -\frac{1}{2}\sqrt{y + 3}

Multiply both sides by -2

-2 * x =-2 *  -\frac{1}{2}\sqrt{y + 3}

-2x =\sqrt{y + 3}

Square both sides

(-2x)^2 =(\sqrt{y + 3})^2

4x^2 =y + 3

Make y the subject

y = 4x^2 - 3

The inverse has been solved. So, we need to replace y with f'(x)

f'(x) = 4x^2 - 3

Next, is to determine the interval

x \ge -3

Change inequality to \le

x \le -3

Hence, the inverse function is:

f'(x) = 4x^2 - 3 for x \le -3

7 0
3 years ago
Factor the following expression.<br> x2 + 4x - 5
sashaice [31]

Answer:(x+5)(x-1)

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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