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egoroff_w [7]
1 year ago
13

Which choice shows a function with a domain of {-4, -2, 2, 4)?

Mathematics
2 answers:
Luden [163]1 year ago
7 0

Answer:

  C  {(-4, 2), (-2, 1), (2, 0), (4, 5)}

Step-by-step explanation:

The domain is the set of x-values for which the function is defined.

<h3>Choices</h3>

A) The relation is <em>only defined for x=2</em>. It is not a function, because x=2 maps to an infinite number of y-values.

B) The domain is an infinite number of x-values. The listed values are included in the domain, but they are <em>not the entire domain</em>.

C) The list of first-values in the ordered pairs is the domain of the function it matches the given domain.

D) The list of first-values in the ordered pairs is <em>not the same as the given domain</em>.

elixir [45]1 year ago
7 0
Answer: C(-4,2), (-2,1), (2,0), (4,5)
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For what values of b the relation R:{(b^2, 5), (5b, 6)} is not a function?
xenn [34]

Answer:

B=5,0.

Step-by-step explanation:

It's not that hard. It's just that it might throw you off a bit with the R and the other things. Functions if you don't remember are where the coordinates don't have the same x and y. So since the y1 and y2 are different then to make it not a function, you have to make x1 and x2. To make it that b would have to be = to 5 and 0.

6 0
3 years ago
Brent is saving for a computer that is $840. He will also need to pay a 6% sales tax. What is the total cost of the computer aft
Kobotan [32]

Answer:

$890.4

Step-by-step explanation:

840 + .06(840)

840 + 50.4

890.4

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3 years ago
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Which is not an equation of the line going through (3, -6) and (1, 2)?
777dan777 [17]

Answer:

B, C

Step-by-step explanation:

The equation of a line can be given by y -y₁= m(x -x₁), where m is the slope. This is also known as the point-slope form.

\boxed{ slope = \frac{y _{1} - y_2 }{x_1 - x_2} }

Slope of the line

=  \frac{2 - ( - 6)}{1 - 3}

=  \frac{2 + 6}{ - 2}

= \frac{8}{ - 2}

= -4

<em>Substitute m= -4 into the equation:</em>

y -y₁= -4(x -x₁)

<em>Substitute a pair of coordinates into (x₁, y₁):</em>

Let's start by substituting (1, 2).

y -2= -4(x -1)

This gives us the same equation as D, making D an incorrect option. Note that the question asks for which is not the correct equation.

Let's change the above into the slope-intercept form, where by y is the subject of formula.

<em>Start by expanding the right-hand side:</em>

y -2= -4x +1

<em>+</em><em>2</em><em> on both sides:</em>

y= -4x +3

This equation is not the same as C. C is thus the correct option.

Let's check for options A and B.

The equation in option B is not the correct equation either as they have substituted (2, 1) instead of (1, 2) into (x₁, y₁). Thus, option B is also correct.

y -y₁= -4(x -x₁)

Substitute (3, -6) into (x₁, y₁):

y -(-6)= -4(x -3)

y +6= -4(x -3)

This is the same as option A, making option A incorrect too.

4 0
2 years ago
Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''
Nonamiya [84]

Answer:

a-e^x,e^{-x}

b-x^{-2}

c-x^3

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

D^2-1=0

(D-1)(D+1)=0

D=1,-1

Then , the solution of given differential equation

y=e^x,y=e^{-x}

2.7x^2y''+14xy'-14 y=0

Y=y=x^3

y=3x^2

y''=6x

Substitute in the given differential equation

42x^3+42x^3-14x^3\neq 0

Hence, x^3 is not a solution of given differential equation

e^x,e^{-x} are also not a solution of given differential equation.

y=x^{-2}

y'=-2x^{-3}

y''=6x^{-4}

Substitute the values in the differential equation

7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}

=42x^{-2}-28x^{-2}-14x^{-2}=0

Hence, x^{-2} is a solution of given differential equation.

c.7x^2y''-42y=0

y=x^3

y'=3x^2

y''=6x

Substitute the values in the differential equation

42x^3-42x^3=0

Hence, x^3 is a solution of given differential equation.

a-e^x,e^{-x}

b-x^{-2}

c-x^3

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