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topjm [15]
3 years ago
15

"the sum of the digits of a two-digit number is 13. reversing the digits produces a number that is 9 less than the original numb

er. what is the original number?"
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0
Suppose the one's digit is x, the ten's digit is y. the original number is 10y+x, the reversed number is 10x+y

x+y=13
(10x+y)=(10y+x)-9 =>9y-9x=9 =>y-x=1
Add the two equation to solve: 2y=14 =>y=7
x=13-7=6
the original number is 76. 


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Answer with Step-by-step explanation:

Since we have given that

q = 896-20p

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e=-\dfrac{dq}{dp}\times \dfrac{p}{q}\\\\e=-(-20)\times \dfrac{32}{896-20p}\\\\e=\dfrac{20\times 32}{896-20p}\\\\e=\dfrac{640}{896-20p}\\\\e=\dfrac{640}{896-640}\\\\e=\dfrac{640}{256}=2.5

(B) The demand is going down with increase in 15 increase in price at that price level, as we know that there is inverse relationship between price and quantity demanded.

(C)  Also, calculate the price that gives a maximum weekly revenue.

R=pq\\\\R=p(896-20p)\\\\R=896p-20p^2

We first find the first derivative:

R'(p)=896-400p

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Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

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Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

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=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

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Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

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