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almond37 [142]
2 years ago
6

Is 8/15 0.53 with the 3 repeating or no?

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
8 0

Answer:

Yes.

Step-by-step explanation:

I don't really know how to explain this other than use a calculator.

Hope that this helps!

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The graph shows the cost per pound for bananas
kenny6666 [7]

Answer:

wheres the graph

Step-by-step explanation:

it's not theree

6 0
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
Garland put 2b + 3 dollars in the bank in the first week. The following week he doubled the first week’s savings and put that am
KengaRu [80]

Answer:

1st week deposit= 2b+3

2nd week deposit=2(2b+3)

3rd week deposit=4(2b+3)

Total deposit=2b+3+2(2b+3)+6(2b+3)

=9(2b+3)

=18b+27

18b+27=477

18b. =477-27=450

18b. =450

;

b=450/18=25

Therefore 2b+3=2(25)+3 =50+3=53

PLEASE GIVE BRAINLIEST

6 0
2 years ago
Can anyone help with this question please!
Snowcat [4.5K]

Answer:

6 in

Step-by-step explanation:

A= bh

18 as our area= 3 as our base and (x) as the height

18/3 = 6

so 6 is the height

Let's make sure that the answer is right:

A= bh

A= 3(6)

A= 18

6 0
3 years ago
What substitution Should be used to re-write 26(x^3+1)^2-22(x^3+1)-3=0 as a quadratic function?
mel-nik [20]

Answer:

x^3+1

Step-by-step explanation:

26(x^3+1)^2-22(x^3+1)-3=0

Comparing to

A(u)^2+B(u)+C =0

Where A,B,C are constants

You should see that we need to substitute the x^3+1 with u.

7 0
3 years ago
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