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kirill [66]
3 years ago
7

Which formula does NOT describe the sequence 8, 6, 3, –1, –6, … ?

Mathematics
2 answers:
Lapatulllka [165]3 years ago
8 0
<span> To get an explicit formula, we need to find an expression which gives the n-th term without having to compute earlier terms in the sequence. Looking at the numbers, and from the recursive formula, we see that the sequence is built by subtracting n from the previous term. This is similar to the triangular number sequence 1,3,6,10,15,... which has the explicit formula a_n = n(n+1)/2. In our case we are subtracting n from the previous term, so we multiply by -1/2 instead of 1/2. However, we also need to add a constant term to reproduce the numbers of the sequence. We can write a_1 = -1(2)/2 + c = 8. Therefore, c = 9.
So the explict formula is:
a_n = -n(n+1)/2 + 9</span>
lara [203]3 years ago
8 0
Where is the rest of the question
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Would the answer be 12.5
Schach [20]
Yes, it would be 12.5.
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3 years ago
Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
Ipatiy [6.2K]

Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

(C) y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}

(D) y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R},

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

(D) Using C, let be y=me^{3t} we obtain that it must satisfies 3^2m-4m=1\Rightarrow m=1/5 and then the general solution is y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R}, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).  

4 0
3 years ago
In a random sample of 50 West High School students, 14 students have a pet. If 500 students attend West
Korvikt [17]

Answer:

the answer is 140

Step-by-step explanation:

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velikii [3]

Answer: joe

Step-by-step explanation:

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3 years ago
jane works 40 hours per week. due to illness , jane worked 3/5 of her scheduled hours. how many hours did jane work last week?​
Hunter-Best [27]

Answer:

24

Step-by-step explanation:

40/1 * 3/5 = 120/5

120/5 simplified = 24

6 0
2 years ago
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