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2 years ago

Which is the simplified form of x Superscript negative 12?

Mathematics

2 answers:

chubhunter [2.5K]2 years ago

5 0

Answer:

C. $\frac{1}{x^{12}}$

Step-by-step explanation:

We are given that an expression

$(x)^{-12}$

We have to find the simplified form of given expression.

We know that

$a^{-b}=\frac{1}{a^b}$

Using the property then we get

$x^{-12}=\frac{1}{x^{12}}$

Hence, the simplified form of $x^{-12}$ is given by

$\frac{1}{x^{12}}$

Option C is true.

NikAS [45]2 years ago

5 0

Answer:

its letter C

Step-by-step explanation:

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taurus [48]

Answer:

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1) D. hyperbolic paraboloid

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1 year ago

nikki backyard is in the shape of a rectangle. the length is 27 feet. The width is one-third the length plus 4 feet. write and e

PIT_PIT [208]

Answer:

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Step-by-step explanation:

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1 year ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$