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3 years ago

Which rule is an example of rigid transformation?

Mathematics

1 answer:

GalinKa [24]3 years ago

6 0

We know that

Rigid transformation:

A rigid transformation (also called an isometry) is a transformation of the plane that preserves length.

Reflections, translations, rotations, and combinations of these three transformations are "rigid transformations"

so, it's length must be preserved

now, we will check each option

option-A:

we have (x,3y)

y-value changes but x-value will remain same

It changes length

so, this is not rigid transformation

option-B:

we have (3x,y)

x-value changes but y-value will remain same

It changes length

so, this is not rigid transformation

option-C:

(2x, y+2)

It changes length of x-value

but it is only shifting y-value

so, it changes length

so, this is not rigid transformation

option-D:

Both shifts values

but it's length will always be same

so, this is rigid transformation..............Answer

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AfilCa [17]

<h3>4 Answers: A, B, C, D</h3>

=======================================================

Explanation:

f(x) is continuous when x >= 1. The only discontinuity for f(x) is when x = 0, but 0 is not part of this interval.

f(x) is positive for any valid x value in the domain since x^6 is always positive. In general, x^n is positive for all x when n is any even number.

f(x) is decreasing. You can see this through a table of values or through a graph. For anything in the form 1/(x^k), it will be a decreasing function because x^k gets larger, so 1/(x^k) gets smaller, when x goes to infinity.

--------------------

The conditions to use the integral test have been met. So we have to see if $\displaystyle \int_1^{\infty}f(x)dx$ converges or not.

Let's integrate and find out

$\displaystyle \int \frac{1}{x^6} dx = \int x^{-6} dx\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{1+(-6)}x^{-6+1}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{-5}x^{-5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5}*\frac{1}{x^5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5x^5}+C\\\\$

So we have

$\displaystyle g(x) = \int f(x) dx\\\\\\\displaystyle g(x) = \int \frac{1}{x^6} dx\\\\\\\displaystyle g(x) = -\frac{1}{5x^5}+C\\\\\\$

Meaning that,

$\displaystyle \int_{a}^{b} f(x) dx = g(b)-g(a)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = \left(-\frac{1}{5b^5}+C\right)-\left(-\frac{1}{5a^5}+C\right)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = -\frac{1}{5b^5}+\frac{1}{5a^5}\\\\\\$

If we plug in a = 1 and apply the limit as b approaches positive infinity, then the expression $-\frac{1}{5b^5}+\frac{1}{5a^5}$ will turn into $\frac{1}{5}$