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Sedaia [141]
3 years ago
5

Help asap thank youu

Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

Yes

Step-by-step explanation

Linear means no exponents i believe i hope this is right and i help if its not im soooooo sorry :)

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Alina [70]

Area = length * width

Area = 12 * 14.5

Area = 174 square feet

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4 years ago
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PLEASE HELP ME WITH MY MATH
Alina [70]

Answer:

Step-by-step explanation:

1)

y is 3 times more than x.

X=0 Y=0  3x0=0

X=1  Y=3

X=2 Y=6

X=3 Y=9

X=4 Y=12

Then just plot it as co-ordinates (x,y)

2)

Y is 1/4 of x. Divide by 4

X=0 Y=0

X=4 Y=1

X=8 Y=2

X=12 Y=3

X=16 Y=4

Then just plot it as co-ordinates (x,y)

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3 years ago
It’s brainiest time ! One thing to do!
balandron [24]

Answer:

116.4545454545455

Step-by-step explanation:

5 0
3 years ago
Which of the following is a solution to the equation y = 3x - 1 ?
Korvikt [17]
One way of determining the answer is by substituting the values to the equation. It is done as follows:

A. (4,1)
<span>y = 3x - 1
</span>1 = 3(4) - 1 = 11 -------> not equal

B. (2,5)
<span>y = 3x - 1
</span>5 = 3(2) - 1 = 5 --------> equal

C. (4,3)
<span>y = 3x - 1
</span>3 = 3(4) -1 --------------> not equal

D. (0, -3)
<span>y = 3x - 1
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8 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}

3 0
3 years ago
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