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maks197457 [2]
3 years ago
13

Write a rule for the nth term of the arithmetic sequence: a11= 50, d = 7

Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0

Answer:

Required rule for n^{th} is a_{n}=7n-27.

Step-by-step explanation:

Given that,

a_{11} = 50,\  \  d=7

From the question: we have to write the n^{th} term of Arithmetic sequence.

Arithmetic Sequence or Arithmetic progression (A.P) : It is a sequence which possess that difference between of two successive sequence is always constant.

                a_{1} ,a_{2},a_{3},a_{4}.....................a_{n-1},a_{n}

                                        where, a_{1} is the first term of A.P

                                                     d is the common difference.

                                                     a_{n} is the last term or general term.

The above sequence to be in A.P then their common difference should be equal.

          d=a_{2}-a_{1} =a_{3}-a_{2}=a_{4} -a_{3} ..........................a_{n}-a_{n-1}

Now, Formula of General Term is a_{n}=a+(n-1)d

So,                                                    a_{11}= a+(11-1)d\\        a_{11} = a+10d

 Substituting the value of   a_{11} = 50,\  \  d=7 we get,

                                                         50=a+10\times7\\50=a+70\\a=-20

Then General term (a_{n}) of given data is

                                                         a_{n}=-20+(n-1)7\\a_{n}=-20+7n-7\\a_{n}=7n-27

Therefore, Required rule for n^{th} is a_{n}=7n-27.

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Answer:

Step-by-step explanation:

From the given information,

Suppose

X represents the Desktop computer

Y represents the DVD Player

Z represents the Two Cars

Given that:

n(X)=275

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n(Z)=405

n(XUY)=145

n(YUZ)=195

n(XUZ)=110

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n(X ∩ Y ∩ Z) = 1000-265

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n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)

145 = 275+455 - n(X ∩ Y)

n(X ∩ Y) = 585

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195 = 455+405-n(Y ∩ Z)

n(Y ∩ Z) = 665

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n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735

n(X ∪ Y ∪ Z) = 50

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n(X ∪ Y ∪ C') = 145-50

n(X ∪ Y ∪ C') = 95

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3 years ago
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Answer:

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Step-by-step explanation:

7 0
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In a certain town, the wind speed, x, in km/h on a certain day is described by two statements: If 2 times the wind speed is incr
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<span>1) if 2 times the wind speed is increased by 2, the wind speed is still less than 46 km/h.

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2) Twice the wind speed minus 27 is greater than 11 km/h.

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Part A: Create a compound inequality to represent the wind speed range. (3 points)


from 2x + 2 < 46

=> 2x < 44

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=> 2x > 11 + 27

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The set of inequalities is

2x + 2 <46
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The solution is x < 22 and x > 19, which is:

19 < x < 22 <----- answer

Part B: Can the wind speed in this town be 20 km/h? Justify your answer by solving the inequalities in Part A. (3 points)

Yes, the wind speed can be 20 km/h, because the solution of the inequality is the range (19,22).

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x ≥ 23 - 4 => x ≥ 19

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for x = 0, it will take the "x if x ≥ 0" and just have y = 0 which is what |0| does
4 0
3 years ago
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