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ziro4ka [17]
2 years ago
14

For each bracelet Julie makes, SHES uses 5 green beads, 6 red beads, and 7 white beads, how many bracelets can she make?

Mathematics
2 answers:
Masteriza [31]2 years ago
6 0

Answer:

d; 151 bracelets

Step-by-step explanation:

to find this out we must divide each number of beads by their matching color amounts

1012 / 5 is 202.4 or 202

998 / 6 is 164.6... or 164

1057 / 7 is 151

151 is the least number out of all of these and in this case you can only make 151 bracelets or else you would run out of white

adell [148]2 years ago
4 0
I believe it’s 519 bracelets sorry if I’m wrong
have a nice day!
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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
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a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

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y' +y = 7+\delta (t-3) \\ \\ y(0)=0

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Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

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