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vlada-n [284]
3 years ago
6

Find the area of the shape, show work pls ill give brainiest

Mathematics
1 answer:
andre [41]3 years ago
8 0

Answer:

6.2

Step-by-step explanation:

Although there's multiple ways to solve this problem, my method will be to simply find the area for the full triangle (the empty + orange triangles) and subtract the area of the smaller, empty triangle.

Now, you that area for a triangle is 1/2*base*height.

To find the measurements for the full triangle, you must add up the bases for the two smaller triangles:

Base_{total}  = Base_{empty} + Base_{orange}  \\Base_{total} = 1.9ft + 3.1 ft\\Base_{total} = 5 ft

Height is the same for both triangles so Height total = 4 ft.

Now the total area can be calculated:

Area total= 1/2* base_total * height_total

Area total = 1/2 * 5ft * 4 ft  

Area total = 20 / 2 = 10 ft squared

Lastly, subtract the area of the empty triangle from the total triangle to find the orange triangle.

Area Empty Triangle = 1/2 * base_empty * height_empty

Area Empty Triangle = 1/2 * 1.9ft * 4 ft = 7.6 ft / 2 = 3.8 ft squared

Area total - Area empty = 10ft^2 - 3.8ft^2 = 6.2 ft squared

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First check the characteristic solution:

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has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

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y_p = (At^3+Bt^2)e^{-2t}

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{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

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y_p = Ct + D

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Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

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