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3 years ago

Which of the following sets represents a function?

2 answers:

8 0

The answer would be B because to have a function none of the x's can be the same, and we know that it's (x,y) so all of the first numbers are x's
In the first answer choice the x's would be 0,1,3,5,&5 but the 5 is repeated so we toss that answer choice
In answer choice B, the x's are 3,4,&5, so that's a maybe
In C, the x's are 1,5,1,&5 so we know that's not correct because the 5's and the 1's are both repeated
Last but not least, in D, the x's are 5,7,&5, the 5's are repeated so that leaves us with our answer of B. {(0,1), (1,2), (3,4), (5,6), (5,7)}

gogolik [260]3 years ago

5 0

It would be D
Your welcome ;)

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What is the quotient? URGENT!!

jeyben [28]

Answer:

The answer is A.

Step-by-step explanation:

You have to multiply by converting the second fraction into upside down :

$\frac{4x + 1}{6x} \div \frac{x}{3x - 1}$

$= \frac{4x + 1}{6x} \times \frac{3x - 1}{x}$

$= \frac{(4x + 1)(3x - 1)}{x(6x)}$

$= \frac{12 {x}^{2} - 4x + 3x - 1}{6 {x}^{2} }$

$= \frac{12 {x}^{2} - x - 1 }{6 {x}^{2} }$

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3 years ago

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7 0

3 years ago

Read 2 more answers

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi

julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45) there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$ , due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a unique solution defined in an open interval around $x_0$ .

1. (-4,6) there is no a solution to the equation through this point because $y_0=6$

2. (2,−6) there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45) there is a solution to the equation through this point because

$y_0>6$

6 0

3 years ago

Hey who ever can help me, I give 25 points

Nadusha1986 [10]

4x+2y=18

-2(2x+3y=15)=-4x-6y=-30

4x+2y=18

-4x-6y=-30

-4y=-12

y=3

2x+3(3)=15

2x+9=15

2x=6

x=3

(3,3)

5 0

3 years ago

Which expression is equivalent to (73)5 • 74 ?

Alex777 [14]

The correct answer is D. 73 • 5 • 74

Explanation:

In mathematics, multiplication can be represented by using the symbol × or the symbol ·. Moreover, multiplication can be also represented just by leaving the numbers or values side by side without any symbol. For example, the expression 5x + y implies 5 should be multiplied by x. In this context, the expression (73)5 • 74 indicates 5 should be multiplied by 74, and the result of this should be multiplied by 73 as the lack of symbol indicates multiplication. Therefore, this is equivalent to 73 • 5 • 74.

7 0

3 years ago