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Black_prince [1.1K]
3 years ago
10

Write an equation that models the situation and find its solution.

Mathematics
2 answers:
atroni [7]3 years ago
8 0
Lindsay has 6 friends. Divide the amount they spent total by the six friends.

36 / 6 = 6
vaieri [72.5K]3 years ago
7 0

Answer:they each contributed 6 $

Step-by-step explanation:

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Someone plz answer this I need help will give you brainiest
Olenka [21]

Answer:

80

Step-by-step explanation:

7 0
2 years ago
What's the answer to this question
Cerrena [4.2K]
10.1 x 8 - 6 = 80.8 - 6 = 74.8
u just had to substitute the variables
3 0
3 years ago
HELP MEEEEEE PLSSSSS
frutty [35]

Answer:

64

Step-by-step explanation:

If you measure the cubes inside for the length, width and height, you get:

Length - 5 cubes

Width -5 cubes

Height - 3 cubes

The volume of the prism is 5 * 5 * 3 = 75. This means the prism can hold a total of 75 cubes

The amount of cubes already in there is 11

so 75 - 11= 64

5 0
2 years ago
Please help asap 20 pts
iVinArrow [24]

in 3 seconds

16 (3^2)

= 16 (9)

= 144 ft

Answer

D. 144 ft

7 0
3 years ago
Read 2 more answers
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
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