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2 years ago

Solve for a and x in ax^2=48

Mathematics

1 answer:

Gelneren [198K]2 years ago

6 0

Answer:

$ax^2 = 48\\a = \frac{48}{x^2}$

$ax^2 = 48\\x^2 = \frac{48}{a} \\x = \frac{\sqrt{48} }{\sqrt{a} } \\x = \frac{4\sqrt{3} }{\sqrt{a}} \\x = \frac{4\sqrt{3a} }{a}$

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Please tell me the steps to this problem. i will give brainliest

gayaneshka [121]

Option C: -3 is the average rate of change between $x=-1$ and $x=1$

Explanation:

The formula to determine the average rate of change is given by

Average rate of change = $\frac{y_2-y_1}{x_2-x_1}$

We need to find the average rate of change between $x=-1$ and $x=1$

Thus, from the table, we have,

$x_1=-1$ , $y_1=5$

$x_2=1$ , $y_2=-1$

Thus, substituting these values in the formula, we get,

Average rate of change = $\frac{-1-5}{1+1}$

$=\frac{-6}{2}$

$=-3$

Thus, the average rate of change between $x=-1$ and $x=1$ is -3.

Hence, Option C is the correct answer.

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2 years ago

Reduce to simpleast form. -5/8 - (-4/3)=

vfiekz [6]

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3 years ago

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2 years ago

Solve for a and x in ax^2=48​

Solve for a and x in ax^2=48