Assuming all image points have the same symmetry about y=2, each point (x, y) has image (x, 4-y).
A' = (1, 4-1) = (1, 3)
B' = (-3, 4-0) = (-3, 4)
C' = (3, 4-4) = (3, 0)
D' = (-2, 4-(-1)) = (-2, 5)
Answer: Choice B
Choice A is not the answer because if you expand everything out, you end up with a 7th degree polynomial. The leading terms are x, x^3 and x^3 for each factor, which overall becomes x*x^3*x^3 = x^7. The graph shown is an even function so there is no way that any of the terms have odd exponents. Furthermore, the end behavior suggests that the function has a leading term with an even exponent.
Choice C is not the answer because of similar reasons as choice A. This time the polynomial is a 9th degree polynomial, which is also odd.
Choice D is not the answer because the (x-a)^5 term means that one of the roots cuts through the x axis instead of merely touching it at one point like what is shown in the image
The only thing that is left is choice B. It is an even function and the leading term has an even exponent. Each root has multiplicity that is even so that means each root touches the x axis instead of crossing over.
3x² + 8x + 4
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x + )(x + )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2.
Our options are:
(3x + 1)(x + 4)
(3x + 4)(x + 1)
(3x + 2)(3x +2)
To figure out which one to use, I'm just going to FOIL them all.
(3x + 1)(x + 4) = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4
The factored form is:
(3x + 2)(x + 2)
To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:
(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots.
3x + 2 = 0
3x = -2
x = -2/3
x + 2 = 0
x = -2
The coordinates of a point is the location of the point in a plane.
<em>The coordinates of the centers of holes are: (48.5, 28) and (28, 48.5)</em>
Given
I've added an attachment as an illustration
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<u>Considering </u>
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To solve for x1, we make use of cosine ratio.
So, we have:
Make x the subject
To solve for y1, we make use of sine ratio.
So, we have:
Make y the subject
So, we have:
<u>Considering </u>
<u />
To solve for x2, we make use of cosine ratio.
So, we have:
Make x the subject
To solve for y1, we make use of sine ratio.
So, we have:
Make y the subject
So, we have:
<em>Hence, the coordinates of the centers of the holes are: (48.5, 28) and (28, 48.5)</em>
Read more about coordinate geometry at:
brainly.com/question/8121530
