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2 years ago

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When Ethan began a weight loss program, he weighed 284 pounds. He was able to lose exactly 3 pounds every month. How much would

weigh 41 months after starting the weight loss program? Round to the nearest tenth (if necessary)

1 answer:

OlgaM077 [116]2 years ago

8 0

Answer:

161 pounds

Step-by-step explanation:

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In order to either condense or expand logarithms using their properties, what must be?

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Answer:

Answer C

Step-by-step explanation:

The rules of logarithmic functions require the two log functions to have the same base. If they do not have the same base, you cannot apply any of the rules.

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taurus [48]

8.8 squared + 9.5 squared = square root 12.94 rounded 12.1

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3 years ago

The length of a rectangle is twice its width. if the perimeter is 72 meters, find the length and width of the rectangle

madam [21]

Length(l)= 2w
width(w)= w

Perimeter(P)= 2w+2l= 72 (simplify expression: divide each side by 2 )
P= w+l= 36 (plug in "2w" for "l")
P= w+(2w)= 36
P= 3w= 36 (divide each side by 3 to find the width)
w= 12 units

find length:
l=2w
l= 2(12)
l= 24 units

Answer:
The length of this rectangle is 24 units and the width is 12 units.

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3 years ago

(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:

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

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2 years ago

I need help answering

Sever21 [200]

Rupees 937.36

think it's right

4 0

2 years ago