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lukranit [14]
3 years ago
10

Mr. Theorem writes the expressions Six to the second power divided by 2m on the board.what is the value of this expression when

the value of m=3
Mathematics
2 answers:
Rudik [331]3 years ago
6 0
The value of the epression is 6 because 6^2 is 36 and since you're substituting m for 3 the value of the denominator is 6 so essentially you are doing the following operation: 6^2/6 --> 36/6=6
Ulleksa [173]3 years ago
4 0
6, if m=3 then it must be 36/6 which is equal to 6
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What is the solution to x- 9> -15?
Elanso [62]

Answer:

x < -6

Step-by-step explanation:

quick explanation: when you transfer (-9) to the other side it becomes +9.

Therefore, -15+9= -6

4 0
3 years ago
If 3x – 10 = 17, what is the value of 6x + 20 ?
IceJOKER [234]
3x-10=17\ \ \ \ /+10\\\\3x-10+10=17+10\\\\3x=27\ \ \ \ /:3\\\\3x:3=27:3\\\\x=9\\\\\\\\6x+20=6\cdot9+20=54+20=74



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3 0
3 years ago
Read 2 more answers
Please help! I’ll mark brainliest
Usimov [2.4K]

Answer: 19 ≥ 3z + 1 ≥ - 5

8 0
3 years ago
Please please help guys
dexar [7]

Answer:

234.85

Step-by-step explanation:

This was easy!!!

(The gardener comes 4 times a month... so divide 85.40 by 4 and you will get 21.35. Each time he comes, he gets 21.35 and by the end of the month, he has 85.40. It says he already came 11 times. So 4 + 4 = 8 + another 4 that equals 12. He came only two months and a few weeks. This means multiply 85.40 by 2 so you would get 170.8. It said he came 11 times so substract 11 from 8 and you would get 3. So multiply 21.35 by 3 and get a total of 64.05. Add 170.8 and 64.05 to get 234.85!!! The answer is 234.85 or C!!!!

5 0
3 years ago
Read 2 more answers
a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

|s_n -s| < \epsilon for all n, and that implies that the series converges to a value, s.

And this complete the proof.

<em>Part c</em>

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where s_n represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most b_{n+1} by the following theorem:

<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any  two consecutive partial sums sn and sn+1.

\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

And this complete the proof.

5 0
4 years ago
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