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2 years ago

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Sister’s recipe needs 6 eggs to make 5 pancakes. If you want to make 20 pancakes for your family using her recipe,how many eggs

will you need?

1 answer:

uranmaximum [27]2 years ago

5 0

Answer:

Sister’s recipe needs 24 Eggs to make 20 pancakes for your family.

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Please answer the question above.

Crank

Answer:

Waffle house :3

Step-by-step explanation:

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2 years ago

A square box has an area of 90,000 square inches. What's the length of each side of the box?

Sever21 [200]

Answer: A 300 inches

Step-by-step explanation:

Area (a) = 90,000

Side = √a

Side = √90000

Side = 300

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3 years ago

Read 2 more answers

Ken went to the store for his mother. he spent $2.37 on groceries. in addition, his mother said he could spend 35¢ on candy. wha

Viefleur [7K]

$2.37 (the amount he spent on groceries) + <span>35¢ (or $0.35 he could spend on candy) = $2.72
Now you subtract this number from the five dollar bill to get the change.
5 - 2.72 = $2.28
His change was $2.28.
</span>

5 0

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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3 years ago

Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to the

blsea [12.9K]

Hello,
Please, see the attached files.
Thanks.

7 0

3 years ago

Read 2 more answers