All categories

3 years ago

Is the function even,odd, or neither

Mathematics

2 answers:

alukav5142 [94]3 years ago

8 0

Answer:

even I think

Step-by-step explanation:

slega [8]3 years ago

3 0

Answer:

Neither

Step-by-step explanation:

If f (–x) = f (x), then the function is even. if f (–x) = –f (x), (all the signs are flipped) the it is odd.

To apply this to the function here, simply replace the x in $(x-2)^{2}$ with -x.

$(x-2)^{2}$ = $x^{2}$ - 4x + 4, and $(-x - 2)^{2}$ = $x^{2}$ +4x + 4.

Since $x^{2}$ +4x + 4 doesnt equal $x^{2}$ - 4x + 4 (even) or $-x^{2}$ +4x -4 (odd), the function is neither even nor odd.

You might be interested in

En un festival para niños se cobraron $50 por niño y $100 por adulto, si el total de dinero recaudado por boleto pagado fue de $

myrzilka [38]

Was this meant for Spanish or French ?

7 0

3 years ago

For this question, f(x) = 4kx2 + (4k + 2)x + 1, where k is a real constant.

snow_tiger [21]

Answer:

fghhjhhfgkus+($+#($!?$)

6 0

3 years ago

Which are whole numbers ? 0,4,-2,2/3,-1.8,16,3.2,-5/4

Rasek [7]

Answer:

0,4,16

Step-by-step explanation:

i hope that helps

7 0

3 years ago

Read 2 more answers

Find the length of CD shown in red below. Show all work.

PIT_PIT [208]

By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.

<h3>How to calculate the length of an arc</h3>

The figure presents a circle, the arc of a circle (s), in inches, is equal to the product of the <em>central</em> angle (θ), in radians, and the radius (r), in inches. Please notice that a complete circle has a central angle of 360°.

If we know that θ = 52π/180 and r = 6 inches, then the length of the arc CD is:

s = [(360π/180) - (52π/180)] · (6 in)

s ≈ 32.254 in

By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.

<h3>Remark</h3>

The statement has typing mistakes, correct form is shown below:

<em>Find the length of the arc EF shown in red below. Show all the work.</em>

To learn more on arcs: brainly.com/question/16765779

#SPJ1

4 0

2 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 4 in the manner described. (Ent

QveST [7]

Answer:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Step-by-step explanation:

$(x-h)^2+(y-k)^2=r^2$ has parametric equations:

$(x-h)=r\cos(t) \text{ and } (y-k)=r\sin(t)$ .

Let's solve these for x and y respectively.

$x-h=r\cos(t)$ can be solved for x by adding h on both sides:

$x=r\cos(t)+h$ .

$y-k=r \sin(t)$ can be solve for y by adding k on both sides:

$y=r\sin(t)+k$ .

We can verify this works by plugging these back in for x and y respectively.

Let's do that:

$(r\cos(t)+h-h)^2+(r\sin(t)+k-k)^2$

$(r\cos(t))^2+(r\sin(t))^2$

$r^2\cos^2(t)+r^2\sin^2(t)$

$r^2(\cos^2(t)+\sin^2(t))$

$r^2(1)$ By a Pythagorean Identity.

$r^2$ which is what we had on the right hand side.

We have confirmed our parametric equations are correct.

Now here your h=0 while your k=1 and r=2.

So we are going to play with these parametric equations:

$x=2\cos(t)$ and $y=2\sin(t)+1$

We want to travel clockwise so we need to put -t and instead of t.

If we were going counterclockwise it would be just the t.

$x=2\cos(-t)$ and $y=2\sin(-t)+1$

Now cosine is even function while sine is an odd function so you could simplify this and say:

$x=2\cos(t)$ and $y=-2\sin(t)+1$ .

We want to find $\theta$ such that

$2\cos(t-\theta_1)=2 \text{ while } -2\sin(t-\theta_2)+1=1$ when t=0.

Let's start with the first equation:

$2\cos(t-\theta_1)=2$

Divide both sides by 2:

$\cos(t-\theta_1)=1$

We wanted to find $\theta_1$ for when $t=0$

$\cos(-\theta_1)=1$

Cosine is an even function:

$\cos(\theta_1)=1$

This happens when $\theta_1=2n\pi$ where n is an integer.

Let's do the second equation:

$-2\sin(t-\theta_2)+1=1$

Subtract 2 on both sides:

$-2\sin(t-\theta_2)=0$

Divide both sides by -2:

$\sin(t-\theta_2)=0$

Recall we are trying to find what $\theta_2$ is when t=0:

$\sin(0-\theta_2)=0$

$\sin(-\theta_2)=0$

Recall sine is an odd function:

$-\sin(\theta_2)=0$

Divide both sides by -1:

$\sin(\theta_2)=0$

$\theta_2=n\pi$

So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means $\theta_1=2n\pi=2(0)\pi=0$ .

We also don't have to shift the sine parametric equation either since at n=0, we have $\theta_2=n\pi=0(\pi)=0$ .

So let's see what our equations look like now:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Let's verify these still work in our original equation:

$x^2+(y-1)^2$

$(2\cos(t))^2+(-2\sin(t))^2$

$2^2\cos^2(t)+(-2)^2\sin^2(t)$

$4\cos^2(t)+4\sin^2(t)$

$4(\cos^2(t)+\sin^2(t))$

$4(1)$

$4$

It still works.

Now let's see if we are being moving around the circle once around for values of t between $0$ and $2\pi$ .

This first table will be the first half of the rotation.

t 0 pi/4 pi/2 3pi/4 pi

x 2 sqrt(2) 0 -sqrt(2) -2

y 1 -sqrt(2)+1 -1 -sqrt(2)+1 1

Ok this is the fist half of the rotation. Are we moving clockwise from (2,1)?

If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).

Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1). We have now made half a rotation around the circle whose center is (0,1) and radius is 2.

Let's look at the other half of the circle:

t pi 5pi/4 3pi/2 7pi/4 2pi

x -2 -sqrt(2) 0 sqrt(2) 2

y 1 sqrt(2)+1 3 sqrt(2)+1 1

So now for the talk half going clockwise we should see the x's increase since we are moving right for them. The y's increase after the half rotation but decrease after the 3/4th rotation.

We also stopped where we ended at the point (2,1).

3 0

3 years ago