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yuradex [85]
3 years ago
15

Suppose u=<2,−1> and v=<1,−8> are two vectors that form the sides of a parallelogram. Then the lengths of the two di

agonals of the parallelogram are?
Mathematics
1 answer:
tigry1 [53]3 years ago
8 0
| u | = √(2² + (-1²)) = √5
| v | = √ ( 1² + (-8)² = √65
cos (u,v) = ( u * v ) / (| u | * | v |) =
 (2 * 1 + ( -1 ) * ( - 8 )) / √5 √ 65 = (2 + 8) / √5 √65 = 10 / (√5 √ 65 )
The length of a larger diagonal:
d 1² = | u |² + 2 |u| |v| + | v |² = 5 + (2 √5 √65 * 10 / √5 √65 )+65
d 1² = 70 + 20 = 90
d 1 = √ 90 = 3√10
d 2² = 70 - 20 = 50
d 2 = √50 = 5√2
Answer:
The lengths of the diagonals are: 3√10 and 5√2 .
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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}

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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}

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Can someone please help me with this problem
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Answer:

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∠1 = 35

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4. ∠5 = 90. It is given

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