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Andre45 [30]
3 years ago
5

Which of the following is not an equation of a simple, even polynomial function? (Select all to apply.)

Mathematics
2 answers:
Ne4ueva [31]3 years ago
4 0

Wouldn't it be y = | x |? I am not 100% positive but I'm pretty sure that you need to have a least one value. Again not 100% sure

Marina CMI [18]3 years ago
4 0

Answer:

y=x^3

Step-by-step explanation:

A function is an even polynomial function when f(-x) is equal to f(x)

Lets check each option, replace x with -x and see whether we get same y equation

y=|x|, Replace x with -x

y=|-x|=x, for any x value the value of y is positive always. So it is an even function

y=x^2

y=(-x)^2=x^2 same as the given equation . So it is an even function

y=x^3

y=(-x)^3=-x^3 It is not same as the given equation . So it is not an even function

y=-x^2

y=-(-x)^2=-x^2 same as the given equation . So it is an even function

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How many solutions will the following system of linear equations have? –2x + 4y = –7 y= -1/2x + 5 A. 0 B. 1 C. 2 D. infinite
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Write a ratio, in the simplest form and a percent for the shaded area (there's a 5x4 box and 6 cubes are shaded)
Alexxandr [17]

Answer:

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Step-by-step explanation:

Given

Shaded = 6\ cubes

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Required

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Box = 5\ x\ 4

Express as multiplication

Box = 5\ *\ 4

Box = 20

This implies that there are 20 cubes in the box.

The percentage of the shaded cubes is:

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5 0
2 years ago
Write the equation, in standard form, of a polynomial of the least degree with integral
Hitman42 [59]

Answer:

f

(

x

)

=

3

x

3

−

5

x

2

−

47

x

−

15

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of

−

3

,

−

1

3

,

5

, the factors are

(

x

+

3

)

(

x

+

1

3

)

(

x

−

5

)

Let's take a look at the factor

(

x

+

1

3

)

. Using the factor in this form will not result in integer coefficients because

1

3

is not an integer.

Move the

3

in front of the x and leave the

1

in place:

(

3

x

+

1

)

.

When set equal to zero and solved, both

(

x

+

1

3

)

=

0

and

(

3

x

+

1

)

=

0

result in

x

=

−

1

3

.

f

(

x

)

=

(

x

+

3

)

(

3

x

+

1

)

(

x

−

5

)

Multiply the first two factors.

f

(

x

)

=

(

3

x

2

+

10

x

+

3

)

(

x

−

5

)

Multiply/distribute again.

f

(

x

)

=

3

x

3

+

10

x

2

+

3

x

−

15

x

2

−

50

x

−

15

Combine like terms.

f

(

x

)

=

3

x

3

−

5

x

2

−

47

x

−

15

5 0
3 years ago
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