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Anna007 [38]
3 years ago
14

Someone please help me with these Geometry questions.

Mathematics
1 answer:
AnnZ [28]3 years ago
6 0
<h3>Answer: 112.5 square units</h3>

Work Shown:

A = area of the triangle

A = base*height/2

A = 9*7/2 = 63/2 = 31.5 ... see note1 below

B = area of rectangle

B = base*height

B = 9*3 = 27

C = area of parallelogram

C = base*height .... see note2 below

C = 9*6 = 54

If you're confused where I got the 6 from, check out the attached image below.

D = total area of the composite figure

D = A+B+C

D = 31.5 + 27 + 54

D = 112.5

-------------------------

note1: The vertical component of the triangle is 9 units because this is part of the rectangle. The rectangle has opposite sides that are the same length. The parallelogram's vertical sides are 9 units long as shown by the single tickmark.

note2: a rectangle is essentially a special case of a parallelogram. In other words, a rectangle is a parallelogram with 4 right angles. So this is why the formula "area = base*height" shows up identical for both rectangles and parallelograms. Keep in mind that the height is always perpendicular to the base. You will not use the segments marked with double tickmarks.

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A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1
kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = \int\limits^\infty_0 {f(t)e^{-st} } \, dt

          =  \int\limits^\infty_0[2t +1] e^{-st} dt

          = 2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt

         = 2 L(t) + L(1)

L(1) = \int\limits^\infty_0e^{-st} dt

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = 2\int\limits^\infty_0 te^{-st}

        =  2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

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8 0
11 months ago
Given a standard form linear equation 33 - 4y = 12​
Kisachek [45]

Answer:

5 1/4

Step-by-step explanation:

4 0
2 years ago
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Solve the system of equations: 3x + 6y=6 and 2y - 3x = 6.
weeeeeb [17]

Answer:

x = -1, y = 1.5

Step-by-step explanation:

3x + 6y = 6 -----> x + 2y = 2

-3x + 2y = 6 -----> -3x + 2y = 6

------------------

4x = -4

x = -1, y = 1.5

7 0
1 year ago
Equations with
andriy [413]
The anwser is 9
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3 0
3 years ago
PLEASE HELP ME!!!!!!!!!
vaieri [72.5K]
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