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2 years ago

Does (5,-8),(3,5),(3,-3),(2,8) represent a function

Mathematics

1 answer:

Rzqust [24]2 years ago

8 0

Answer:

Step-by-step explanation:

So to determine whether this is a function, we first need to know what a function is. A function is defined as for each input, there is only 1 output.

This output doesn't have to be unique and other inputs can output this output, but it only matters that each input outputs only one output.

So let's look at what you provided and specifically these two points:

(3, 5)

(3, -3)

This input 3, outputs 5 and -3, which is not a function since this input outputted 2 different outputs. One thing to note is had you provided the two points: (3, 5), (3, 5) instead then it would still be a function because the input outputted the same output of 5 both times.

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I need help fast thanks

BabaBlast [244]

Answer:

I wanna say its all real numbers

Step-by-step explanation:

3 0

3 years ago

So the people who are good at math Im going to ask 17 questions that invlove math so if u want 20 points in each question then g

PilotLPTM [1.2K]

Answer: ok will do

Step by step explanation

3 0

2 years ago

What is the quotient of the following division problem? x 1 startlongdivisionsymbol x squared 3 x 2 endlongdivisionsymbol. minus

Monica [59]

The quotient of the provided division in which binomial is divided by a quadratic equation is x+2.

<h3>What is the quotient?</h3>

Quotient is the resultant number which is obtained by dividing a number with another. Let a number <em>a</em> is divided by number b. Then the quotient of these two number will be,

$q=\dfrac{a}{b}$

Here, (a, b) are the real numbers.

The given division expression is,

$\dfrac{x^2+3x+2}{x+1}$

Let the quotient of this division problem is f(x). Thus,

$f(x)=\dfrac{x^2+3x+2}{x+1}$

Factor the numerator expression as,

$f(x)=\dfrac{x^2+2x+x+2}{x+1}\\f(x)=\dfrac{x(x+2)+1(x+2)}{x+1}\\f(x)=\dfrac{(x+2)(x+1)}{x+1}\\f(x)={x+2}$

Thus, the quotient of the provided division in which binomial is divided by a quadratic equation is x+2.

Learn more about the quotient here;

brainly.com/question/673545

5 0

2 years ago

Alan mixes 1 1/3 cups of milk with a can of condensed soup. He makes a total of 2 5/8 cups of soup. How many cups of condensed s

Alborosie

Subtract 2_5/8 minus 1_1/3:
First, get the fractions to have a common denominator by multiplying the denominators together: 8*3 = 24
Rewrite both fractions with this new denominator:
5/8 needs to be multiplied by 3 on top and bottom to make it have a denominator of 24:
5/8 * 3/3 = 15/24
1/3 needs to be multiplied by 8 on top and bottom to make it have a denominator of 24:
1/3 * 8/8 = 8/24
Now that the fractions have been rewritten with the same denominator, subtract the mixed numbers:
2_15/24 - 1_8/24
Whole numbers:
2 - 1 = 1
Fractions:
15/24 - 8/24 = 7/24.
The can of condensed soup contains 1_7/24 cup

3 0

3 years ago

Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

erastova [34]

Answer:

The probability is $\frac{56!}{64!}$

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, $64 \choose 8 .$

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function $f : A \rightarrow A ,$ with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}$

7 0

3 years ago