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babunello [35]
2 years ago
5

What is the sum of the infinite geometric series? Sigma-Summation Underscript n = 1 Overscript 4 EndScripts (negative 144) (one-

half) Superscript n minus 1
Mathematics
2 answers:
11Alexandr11 [23.1K]2 years ago
8 0

Step-by-step explanation:

∑⁴ₙ₌₁ -144 (½)ⁿ⁻¹

This is a finite geometric series with n = 4, a₁ = -144, and r = ½.

S = a₁ (1 − rⁿ) / (1 − r)

S = -144 (1 − (½)⁴) / (1 − ½)

S = -270

If you wanted to find the infinite sum (n = ∞):

S = a₁ / (1 − r)

S = -144 / (1 − ½)

S = -288

VLD [36.1K]2 years ago
4 0

Answer:

The answer is C (40)

Step-by-step explanation:

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Answer:

The sports car, because it has less mass and therefore less inertia

Step-by-step explanation:

When an object has less inertia it is easier to be put into and out of motion, and a sports car would obviously weigh less than a van.

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Express 3/4 in sixty-fourths
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It would be 48/64 this is done by finding how many times 4 fits in 64 (64/4) then multiplying 3 by that answer
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HELP 64 POINTS
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Answer:  {30, 39, 33, 15, 15, 18, 20, 33, 19, 41, 45, 18}

Step-by-step explanation:

1. The key indicates you how to read the values of the stem-and-leaf plot. Then, 1|5 means 15.

2. Then, the first digit is written in the colum Stem and the second digit in the column Leaf.

3. Based on this information, you can write the data given in the stem-and-leaf plot, as following:

15, 15, 18, 18, 19 (The first digit is 1 and the second digit are: 5,5,8,8 and 9)

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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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