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3 years ago

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Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A? yes, be

cause all the elements of set A are in set R yes, because all the elements of set R are in set A no, because each element in set A is not represented in set R no, because each element in set R is not represented in set A

2 answers:

mr_godi [17]3 years ago

8 0

Answer:

yes, becoz all the elements of set R is in set A

Sergio039 [100]3 years ago

5 0

Answer:

2 option

Step-by-step explanation:

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The graph of this function is shifted downwards and the axis of symmetry remains x=1. Which function below the equation of the n

Natali5045456 [20]

Answer:

$y=-x^{2}+2x$

$y=-x^{2}+2x-4$

$y=-x^{2}+2x-3$

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

so

$x=h$

If a> 0 then the parabola open upward (vertex is a minimum)

If a< 0 then the parabola open downward (vertex is a maximum)

In this problem we have

$y=-x^{2} +2x+3$

The vertex is the point $(1,4)$ ------> observing the graph

The axis of symmetry is $x=1$

If the graph of this function is shifted downwards and the axis of symmetry remains x=1

then

The x-coordinate of the vertex of the new graph must be equal to 1

The y-coordinate of the vertex of the new graph must be less than 4

The parabola of the new graph open downward

therefore

<u>Verify each case</u>

case a) $y=-x^{2}+2x$

Convert to vertex form

$y=-(x^{2}-2x)$

$y-1=-(x^{2}-2x+1)$

$y-1=-(x-1)^{2}$

$y=-(x-1)^{2}+1$

The vertex is (1,1)

therefore

The function could be the equation of the new graph

case b) $y=-x^{2}-2x+3$

Convert to vertex form

$y-3=-(x^{2}+2x)$

$y-3-1=-(x^{2}+2x+1)$

$y-4=-(x+1)^{2}$

$y=-(x+1)^{2}+4$

The vertex is (-1,4)

therefore

The function cannot be the equation of the new graph

case c) $y=-x^{2}+2x-4$

Convert to vertex form

$y+4=-(x^{2}-2x)$

$y+4-1=-(x^{2}-2x+1)$

$y+3=-(x-1)^{2}$

$y=-(x-1)^{2}-3$

The vertex is (1,-3)

therefore

The function could be the equation of the new graph

case d) $y=-x^{2}+2x+4$

Convert to vertex form

$y-4=-(x^{2}-2x)$

$y-4-1=-(x^{2}-2x+1)$

$y-5=-(x-1)^{2}$

$y=-(x-1)^{2}+5$

The vertex is (1,5)

therefore

The function cannot be the equation of the new graph

case e) $y=-x^{2}+2x-3$

Convert to vertex form

$y+3=-(x^{2}-2x)$

$y+3-1=-(x^{2}-2x+1)$

$y+2=-(x-1)^{2}$

$y=-(x-1)^{2}-2$

The vertex is (1,-2)

therefore

The function could be the equation of the new graph

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3 years ago