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sleet_krkn [62]
4 years ago
11

Six friends ate in a restaurant together and split the total cost, c, equally. Each person paid less than $8. Which statements r

epresent the scenario? Check all that apply.
The total cost of the food, c, can be represented by the inequality StartFraction c Over 6 EndFraction less-than 8.

The total cost of the food could be $48
.
The total cost of the food could be $36.

When graphed, the number line would be shaded to the left of the maximum value.

The total cost of the food must be greater than $60.

100 BRAINLY POINTS
Mathematics
2 answers:
dolphi86 [110]4 years ago
4 0

Answer:

A, C, and D are the correct answers.

Step-by-step explanation:

A) The total cost of the food, c, can be represented by the inequality c/6 < 8

C) The total cost of the food could be $36.

D) When graphed, the number line would be shaded to the left of the maximum value.

It's A because the 6 friends SPLIT the total cost which is c, and each person paid LESS than $8 dollars, not $8 and above. So it will look like, "total cost/6 people is less than 8" aka "c/6 < 8"

It's C because if each person paid less than $8 dollars, the total is anything under 48. (6 x 8 = 48, and 6 x 6 = 36, and 6 is less than 8.)

It's D because anything LESS than the max value could be the value to c/6. So anything to the left of 8, is less than 8, and anything to the right of 8, is greater than 8, and we want what is "< 8" less than 8 so it has to be to the left.

Natali [406]4 years ago
3 0

Answer:

The total cost could be $36

Step-by-step explanation:

36/6 = 6

6 is less than 8

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Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Ple
pshichka [43]

Answer:

a)Reflexive, not symmetric, transitive

b)Reflexive, not symmetric, transitive

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Step-by-step explanation:

a)

R=\left \{ (a,b)\epsilon  \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}

The relation R is reflexive for

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it is transitive

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b)  

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}

R is reflexive because m=1.m for every integer m

R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2

R is transitive:  if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp

c)

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}

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R is symmetric: if a different to b, then b different to a

R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1

d)

R=\left \{ A,B\mid A\subseteq B \right \}

R is reflexive for every set A is a subset of itself

R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}

R is transitive: if A is subset of B and B is subset of C, then A is subset of C

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