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3 years ago

The table shows sets of ordered pairs that form a relation.

Mathematics

1 answer:

tiny-mole [99]3 years ago

4 0

A relation is a relationship between sets of values. In math, the relation is between the x-values and y-values of ordered pairs. The set of all x-values is called the domain of the relation, and the set of all y-values is called the range of the relation.

A relation from a set X (domain) to a set Y (range) is called a function if each element of X is related to exactly one element in Y.

A. A relation {(2,-3), (3,-3), (1,4), (9,2)} is a function, because each x has exactly one element y related to.

B. A relation {(8,-3), (3,8), (9,4), (-2,2)} is a function, because each x has exactly one element y related to.

C. A relation {(1,-1), (4,8), (9,-1), (3,2)} is a function, because each x has exactly one element y related to.

D. A relation {(1,-1), (-1,5), (1,-6), (-3,2)} is not a function, because x=1 has two possible outcomes y: y=-1 or y=-6.

Answer: A, B and C are functions and D is not a function

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-5b + 8 < 3<br> What’s the correct answer

sp2606 [1]

Answer:

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Step-by-step explanation:

-5b+8<3

-5b<3-8

-5b<-5

5b<5

b<5/5

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maks197457 [2]

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3 years ago

What are two different ways in which you can use prime factorization to find the prime factors of a number

LekaFEV [45]

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3 years ago

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi

Drupady [299]

Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. $x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. $c = \dfrac{3}{8} \ g/L$

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx = (5·c - 5·c/3)×dt = 20/3·c = $6\frac{2}{3} \cdot c \cdot dt$

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. The amount of salt, x after t minutes is given by the relation

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

$c = \dfrac{3}{8} \ g/L$

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3 years ago