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GaryK [48]
3 years ago
12

What is 6x6x6x6x6 in exponential form

Mathematics
2 answers:
melisa1 [442]3 years ago
6 0

Answer:

6x6x6x6x6 in exponential form is, 6^5

Step-by-step explanation:

Using the exponent rules:

a^2 = a\cdot a

a^3 = a \cdot a \cdot a

a^4 = a \cdot a \cdot a \cdot a

a^5 = a\cdot a \cdot a \cdot a \cdot a and so on....

As per the statement:

6x6x6x6x6

To write this in exponential form.

By exponent rules:

6 repeats five times.

6 \times 6 \times 6 \times 6 \times 6  = 6^5

Therefore,  6x6x6x6x6 in exponential form is, 6^5

REY [17]3 years ago
5 0
\underbrace{6\cdot6\cdot6\cdot6\cdot6}_{5}=6^5
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The price of a new car is $35,100 which includes a sales tax of 8%. What is the original cost for the car before tax?
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Fred is making a bouquet of carnationsand roses. The carnations cost $5.25 in all. The roses are $1.68 each. If Fred spent $18.6
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A student club holds a meeting. The predicate M(x) denotes whether person x came to the meeting on time. The predicate O(x) refe
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Answer:

a) \exists \, x \in C : O(x) = 0

b) \{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}

c) \{ x \in C: M(x) = 1 \} = C

d) \{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C

e) \exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1

f) \exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0

Step-by-step explanation:

  • M(x) = 1 if the person x came to the meeting, and 0 otherwise.
  • O(x) = 1 if the person is an officer of the club and 0 otherwise.
  • D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

\exists \, x \in C : O(x) = 0

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

\{ x \in C: M(x) = 1 \} = C

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0

I hope that works for you!

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