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oksian1 [2.3K]
3 years ago
7

If polygon ABCD rotates 70° counterclockwise about point E to give polygon A'B'C'D', which relationship must be true? A. BB' = D

D' B. m∠ABC < m∠A'B'C' C. m∠ABC > m∠A'B'C' D. A'E' = AE
Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

Answer:

A'E' = AE

Step-by-step explanation:

Transformation is the movement of a point from its initial position to a new position. Types of transformation are rotation, reflection, translation and dilation. If an object is transformed then all its points are also transformed.

Rigid transformations are transformations that preserves the shape an size of an object (i.e. the side length and angles). Examples of rigid transformations are rotation, translation and reflection.

Since rotation is a rigid transformation, the side lengths and angles of polygon A'B'C'D' would be the same as that of polygon ABCD. Therefore:

A'E' = AE

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1. Let f(x) = x + 3 and g(x) = 3x + 5. Find f(g(4)) - g(f(4)) 2. Let f(x) = 5x^2 - 5. What is f(f(x))? 3. Let f(g(x)) = 3x + 3 a
ludmilkaskok [199]

Answer:

1. f(g(4)) - g(f(4)) = -6

2. f(f(x)) = 125x^4 - 250x^2 + 120

3. a + b = 0

4. g(4) = -9

Step-by-step explanation:

1.

First we need to find f(4) and g(4):

f(4) = 4 + 3 = 7

g(4) = 3*4 + 5 = 17

Then, we find g(f(4)) = g(7):

g(7) = 3*7 + 5 = 26

And we find f(g(4)) = f(17):

f(17) = 17 + 3 = 20

so f(g(4)) - g(f(4)) = 20 - 26 = -6

2.

To find f(f(x)), we use the value of f(x) for every x in f(x):

f(f(x)) = 5*(f(x))^2 - 5 = 5*(5x^2 - 5)^2 - 5 = 5*(25x^4 - 50x^2 + 25) - 5

f(f(x)) = 125x^4 - 250x^2 + 120

3.

To find f(g(x)), we use the value of g(x) for every x in f(x):

f(g(x)) = g(x) + 6 = ax + b + 6 = 3x + 3

ax + (b+6) = 3x + 3 -> a = 3 and b = -3

a + b = 3 - 3 = 0

4.

If we assume g(x) = ax + b, we have:

g(f(x)) = a*(2x - 3) + b = 2ax - 3a + b = 5 - 4x

2a = -4 -> a = -2

-3a + b = 5

6 + b = 5 -> b = -1

g(x) = -2x - 1

g(4) = -2*4 - 1 = -9

7 0
4 years ago
(Different one)<br><br> Factor 6x^2 - 10x - 4
Alina [70]
We can first factor all the terms by 2:
<span>6x^2 - 10x - 4=2(3x^2-5x-2)
</span>
Then we solve 3x^2-5x-2=0.

Discriminant : 25+4*3*2=49=7^2 hence the roots are (5-7)/6 and (5+7)/6 which are -1/3 and 2

Hence 3x^2-5x-2=(x-2)(3x+1)

Thus 6x^2-10x-4=2(x-2)(3x+1)
6 0
3 years ago
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SSSSS [86.1K]

Answer:

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Step-by-step explanation:

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Answer:

A

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4 years ago
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