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mylen [45]
3 years ago
7

Is the following relation a function?

Mathematics
2 answers:
IRINA_888 [86]3 years ago
8 0

The answer is B (no)

Eduardwww [97]3 years ago
6 0
The answer is No.
Because the 0 in x oval is pointing to BOTH 1 and 3 in y oval.
And that must not happen in order to relation be a function.
The following conditions must be satisfied for relation to be a function:
1. condition: EVERY element in first oval has to be connected to some element in second oval.
2. condition: Elements in first oval MUST NOT have more than one connection with elements in second oval.
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I usually have an answer to the question
Andru [333]

Answer:is this the question?

Step-by-step explanation:

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3 years ago
In November 2010 the average price of 5 gallons of regular unleaded gasoline in the United states was $14.46. what was the price
Lapatulllka [165]

Propecia:

5 gallons ====== $ 14.46.

16 gallons ======= x dollars

X=16*14.46: 5=46.27  

Answer: $ 46.27.

3 0
3 years ago
How do you write a quadratic equation in standard form?
Lemur [1.5K]

Answer:

  • Powers of the variable descending left to right
  • right side of the equal sign is 0

Step-by-step explanation:

For some constants a, b, and c, the standard form* is ...

  ax^2 + bx + c = 0

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It is nice if the leading coefficient (a) is positive, but that is not required.

The main ideas are that ...

  • Powers of the variable are descending
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* This is the <em>standard form</em> for a quadratic. For other kinds of equations, when the expression is equal to zero, this would be called "general form."

3 0
3 years ago
after sharing a pizza you and two friends divide the cost by three. You each owe $2.66. Explain how to round this to the nearest
Stolb23 [73]

Answer:

Hi so you have the answer          

Say you have 1.23 if the last digit is under 5 it would be 1.20                

Now say you have 3.47 if the last digit is above 5 it would be 3.50    

If it is something like 4.45  the 5 is the last digit so it would be 4.50

Step-by-step explanation:

um hope this helped :)

7 0
2 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
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