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3 years ago

I need help to show the work for number 11. I know the answer is option number 3.

Mathematics

1 answer:

EastWind [94]3 years ago

3 0

The given exterior angle and angle Y are supplementary angles and need to equal 180 degrees.

To find Y subtract the outside angle from 180:

Y = 180 - 124 = 56 degrees.

Now the 4 inside angles of the figure need to equal 360 degrees.

To find X Subtract the 3 known angles from 360:

X = 360 - 56 - 65 - 125 = 114 degrees.

The answer is C.

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Awser to this questin

kobusy [5.1K]

180 times 4/100= 7.20- Sales tax

180+7.20= $187.20 is the total cost
Please mark brainliest and give a thanks...

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4 years ago

Read 2 more answers

What is -19 + 37?<br> -56<br> -18<br> 18<br> 56

rodikova [14]

All your doing here is 37-19 adding negative is the same has munising

So 37-19=18

5 0

3 years ago

The table below shows the number of school absences each day.

Vladimir79 [104]

Answer

Step-by-step explanation:

Did it on edge20202

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3 years ago

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(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

Do circumstance of a circle with 17ft

olchik [2.2K]

Answer:

The circumference and area of a circle with a 17 ft diameter are:

C = 53.4 ft

A = 227 ft2

5 0

4 years ago