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3 years ago

Wezi has sixty-eight $1-bills. He takes the $1-bills to the bank to exchange them for $5-bills. How many $5-bills does Wezi get?

2 answers:

4 0

Find the total value
68 pieces of $1 = $68

Find the number of $5 notes he can exchange
68 ÷ 5 = 13 Remainder 3
So he got 13 $5 notes and the remaining 3 in one dollar notes.

m_a_m_a [10]3 years ago

3 0

Thirteen five dollar bills

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Elan Coil [88]

Answer:

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Step-by-step explanation:

7 - 3 = 4 in y

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Simplify

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3 years ago

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

What is 3/3 simplified

REY [17]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

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:

The initial value problem is given as:

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

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

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