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1 month ago

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Which equation is the inverse of (x minus 4) squared minus two-thirds = 6y minus 12?

2 answers:

Inessa [12.5K]1 month ago

7 0

Answer:

The correct choice is option B.

Step-by-step explanation:

The provided function is

$(x - 4)^{2} - \frac{2}{3} = 6y - 12$

We can rearrange the function to obtain

$(x - 4)^{2} - \frac{2}{3} = 6y - 12$

⇒ 3(x - 4)² - 2 = 18y - 36

⇒ 3(x - 4)² = 18y - 34

⇒ $(x - 4)^{2} = 6y - \frac{34}{3}$

⇒ $x - 4 = \pm \sqrt{6y - \frac{34}{3}}$

⇒ $x = 4 \pm \sqrt{6y - \frac{34}{3}}$

Thus, the inverse function becomes $y = 4 \pm \sqrt{6x - \frac{34}{3}}$

Therefore, the accurate answer is option B. (Response)

PIT_PIT [12.4K]1 month ago

7 0

Answer:

EDGE 2020 IS B:)

Step-by-step explanation:

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What is the domain of the square root function graphed below ?

AnnZ [12381]

The graph indicates that x never goes below 0. This means the point (-1,0) is not included in the graph. Therefore, D is the only valid option.

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3 months ago

Read 2 more answers

If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in

Inessa [12570]

Answer:

The highest achievable volume for the box is 2000000 cubic meters.

Step-by-step explanation:

Below is an outline of the volume ( $V$ ), measured in cubic centimeters, and surface area ( $A_{s}$ ), measured in square centimeters, for a box featuring a square base:

$A_{s} = l^{2}+h\cdot l$ (1)

$V = l^{2}\cdot h$ (2)

Where:

$l$ - The length of the base's side, in centimeters.

$h$ - The height of the box, in centimeters.

Using (2), we isolate $h$ in the formula:

$h = \frac{V}{l^{2}}$

Then, we substitute into (1) and simplify the outcome:

$A_{s} = l^{2}+ \frac{V}{l}$

$A_{s}\cdot l = l^{3}+V$

$V = A_{s}\cdot l -l^{3}$ (3)

Next, we calculate the first and second derivatives of this expression:

$V' = A_{s}-3\cdot l^{2}$ (4)

$V'' = -6\cdot l$ (5)

If $V' = 0$ and $A_{s} = 30000\,cm^{2}$ , then we find that the critical value for the base's side length is:

$30000-3\cdot l^{2} = 0$

$3\cdot l^{2} = 30000$

$l = 100\,cm$

Subsequently, we assess this outcome using the second derivative's expression:

$V'' = -600$

According to Second Derivative Test, this critical value signifies an absolute maximum. Consequently, the largest volume obtainable for the box is:

$V = 30000\cdot l - l^{3}$

$V = 2000000\,cm^{3}$

The highest achievable volume for the box is 2000000 cubic meters.

4 0

2 months ago