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3 years ago

The mirror image of the point (2, 5) in the x – axis is?

Mathematics

2 answers:

zubka84 [21]3 years ago

5 0

Answer:

(2,-5).

Step-by-step explanation:

You see, when you reflect over something, one variable has to change. Our line of reflection is the x-axis, or y=0.

Here is a rule to help with these kinds of problems.

If you reflect over the x-axis, the x value is still the same, but the y value changes to be the inverse or opposite value. The opposite of positive is negative, so it will be -5.

For reflected over the y-axis, the y value stays the same, but the x value changes to be the inverse of the original y value.

Alekssandra [29.7K]3 years ago

4 0

Answer:

(2,-5)

Step-by-step explanation:

Reflecting (2,5) across x-axis will invert it's y-coordinate and will make it negative.

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Step-by-step explanation:

$\frac{ {x}^{2} - 8x + 16 }{ {x}^{2} - 7x + 12}$

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$= > \frac{ {x}^{2} - 4x - 4x + 16}{ {x}^{2} - 3x - 4x + 12}$

Lets factorise the numerator and denominator.

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2 years ago

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

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Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$