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3 years ago

12

Please help me solve inequalities and show work. Will get brainiest!!

1 answer:

Dahasolnce [82]3 years ago

8 0

The answer depends on whether you're looking for the x or the y. If y, the answers are the ones on the left, and the x on the right.

1) -4x + 3y ≤ 15 -4x + 3y <span>≤ 15
</span><span> +4x +4x -3y -3y

3y </span>≤ 4x + 15 -4x <span>≤ -3y + 15
</span><span> --- --------- ---- ------------
3 3 -4 -4

y </span>≤ 4/3x + 5 x <span>≥ 3/4y + 11

2) -6x - 4y </span>< 12 -6x - 4y < 12<span>
+6x +6x +4y +4y

-4y </span>< 6x +12 -6x < 4y + 12<span>
---- ---------- ---- -----------
-4 -4 -6 -6

y </span>> -3/2 - 3 x > -2/3y - 2

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Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

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