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3 years ago

Determine if the given functions are even, odd or neither. f(x) = -2x2 - 7 and g(x) = 2x2 + 3

Mathematics

1 answer:

Anettt [7]3 years ago

3 0

Answer:

It's odd.

Explanation:

A function is odd when f(−x)=−f(x)

f(−x)=6(−x)3−2(−x)

=−6x3+2x

=−(6x3−2x)

=−f(x)

Thus, the function is odd.

If a function is odd, then it is symmetric over the x-axis.

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Answer:

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Step-by-step explanation:

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3 years ago

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marishachu [46]

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Step-by-step explanation:

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3 years ago

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3 years ago

The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

4 0

2 years ago