Please someone answer this.

Mathematics

1 answer:

klio [65]2 years ago

7 0

The triangle ABC may be rotated along a fixed point, translated to a different position and reflected across a mirror line.

<h3>Types of transformation</h3>

The three types of transformation are;

Rotation
Reflection
Translation

Rotation of shapes entails moving them around a fixed point. The shape remains the same, but its position in the space changes.

Translating a shape means moving it entirely to a different position, without changing it in any way.

Reflecting a shape is to flip it across a mirror line. Each point in the shape is moved to the other side of the mirror line but has same distance away from the line. The reflected image faces the opposite direction to the original object.

Thus, the triangle ABC may be rotated along a fixed point, translated to a different position and reflected across a mirror line.

Learn more about transformation here:

brainly.com/question/4289712

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How do i find the antiderivative of sin^2(x, cos^2(x and tan^2(x? i know how to antidifferentiate sinx, cosx, and tanx, but thes

Lena [83]

For $\sin^2x$ and $\cos^2x$ , it's helpful to remember the half-angle identities:

$\sin^2x=\dfrac{1-\cos2x}2$
$\cos^2x=\dfrac{1+\cos2x}2$

So

$\displaystyle\int\sin^2x\,\mathrm dx=\frac12\int(1-\cos2x)\,\mathrm dx=\frac x2-\frac{\sin2x}4=C$
$\displaystyle\int\cos^2x\,\mathrm dx=\frac12\int(1+\cos2x)\,\mathrm dx=\frac x2+\frac{\sin2x}4=C$

For $\tan^2x$ , the Pythagorean identity suffices:

$\sin^2x+\cos^2x=1\implies \tan^2x+1=\sec^2x$

which means

$\displaystyle\int\tan^2x\,\mathrm dx=\int(\sec^2x-1)\,\mathrm dx=\tan x-x+C$

since $\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x$ .