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serious [3.7K]
3 years ago
10

Tyson and his friends go ice skating. Tyson must pay an initial fee of $18.30 plus $4.80 per hour. Write an equation for the cos

t c Tyson must pay after skating h hours. Solve the equation to find the cost to go ice skating for four hours. Write your answer as an ordered pair.
A, c = 4.80h – 18.30 (4, 0.9)

B. c = 4.80 + 18.30h (4, 78.00)

C. c = 18.30 + 4.80h (4, 37.50)

D. c = 18.30 + 4.80 + h (4, 27.10)
Mathematics
2 answers:
Helga [31]3 years ago
6 0
The answer should be C.
Alex_Xolod [135]3 years ago
4 0
C (cost) = 18.30 + 4.8 h

if skate for 4 hours then
18.30 + 4.8(4)
= 18.30 + 19.2
= 37.5

skate 4 hours, cost $37.50

answer

<span>C. c = 18.30 + 4.80h (4, 37.50)</span>
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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}

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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
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Goryan [66]

Answer:

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10.4g = 8 pods

Total mass / Total pods = Individual pod weight

10.4g / 8 pods = 1.30g per pod

Lets check our work to confirm:

1.30g x 8 pods = 10.4g

5 0
3 years ago
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