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sveta [45]
3 years ago
8

what translation rule can be used to describe the result of the composition of (x,y) --> (x - 9,y - 2) and (x,y) --> (x +

1, y - 2) ? PLEASE HELPPPPP!!!!!
Mathematics
2 answers:
N76 [4]3 years ago
5 0
Move x coordinate 9 units left then y coordinate 2 units upwards

Move 1 unit to the right and 2 units down
marshall27 [118]3 years ago
4 0

we know that

<u>the rule of the translation A is</u>

(x,y)-----> (x-9,y-2)

that means

the translation is 9 units to the left and 2 units down

<u>the rule of the translation B is</u>

(x,y)-----> (x+1,y-2)

that means

the translation is 1 unit to the right and 2 units down

therefore

<u>the rule of the composition A+B is</u>

(x,y)-----> (x-9+1,y-2-2)

(x,y)-----> (x-8,y-4)

that means

the translation is 8 units to the left and 4 units down

so

<u>the answer is</u>

(x,y)-----> (x-8,y-4)

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Answer:

Solution given:

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8 0
2 years ago
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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

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

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

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

Integrating, we have:

=> 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:

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

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

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

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

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

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

3 0
3 years ago
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GalinKa [24]

Answer:

it's the 2nd one

Step-by-step explanation:

d = 3 + at^2

-3 | -3

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\text{Hence the answer is A}

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