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3 years ago

A house plan Is drawn to a scale 1cm to 2m. What is the length of a window 2.5cm long on the plan?

Mathematics

1 answer:

slamgirl [31]3 years ago

8 0

1cm = 2m

=> 1cm = 200cm

2.5cm = 2.5 × 200cm = 500 cm = 5m

So, the length of window is 500cm or 5m.

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What is the volume of a figure that is 9 inches wide, 3 inches tall and 3 inches long?

motikmotik

Answer:

1/2

Step-by-step explanation:

1/2 of an answer is zero answer

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3 years ago

A farmer sees 56 of his cows out of the barn. He knows that he has 83 cows altogether. Let c represent the number of cows still

irinina [24]

No, the number of cows in the barn is 27. You can get this number by subtracting the 83 cows he has from the 56 that are out of the barn.

83 - 56 = 27

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3 years ago

Read 2 more answers

(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

HELP ME I HAVE TO PASS

raketka [301]

B, I used elimination

5 0

3 years ago

On Monday a team of street sweepers cleaned 2/5 of a city block. Tuesday, the team cleaned 4/5 as

otez555 [7]

Answer:

1 and 1/5

Step-by-step explanation:

$\frac{2}{5}$ + $\frac{4}{5}$ = $\frac{6}{5}$ or $1 \frac {1}{5}$

8 0

2 years ago