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k0ka [10]
3 years ago
13

The figure has been decomposed into three rectangles as shown. A figure is decomposed into 3 rectangles. The rectangles are 4 in

ches by 2 inches, 6.5 inches by 2 inches, and 4 inches by 1.5 inches. Find the area of each piece of the composite figure. The area of the top rectangle is A = 4 in.(2 in.) = 8 in2. The area of the middle rectangle is A = 2 in.( in.) = in2. The area of the bottom rectangle is A = 4 in.( in.) = in2. The total area of the figure is in2.
Mathematics
2 answers:
kompoz [17]3 years ago
5 0

Answer:

27 in³

Step-by-step explanation:

4 × 2 = 8 in²

6.5 × 2 = 13 in²

4 × 1.5 = 6 in²

Total area = 8 + 13 + 6

27 in²

Orlov [11]3 years ago
3 0

Answer:

A= 4in.(2in.) = 8in.2

A=2in.(2in.)= 6in.2

A=4in.(1.5in.)= 6in.2

The total area of the figure is 20in.2

Step-by-step explanation:

I did the work

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A=63°
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Answer:

B. 7.35 inches

Step-by-step explanation:

In the triangle:

  • A=63°
  • c = 7.75 inch
  • B = 47°

Now we know that:

\angle A+\angle B+\angle C=180^\circ$ (Sum of angles in a \triangle)\\63^\circ+47^\circ+\angle C=180^\circ\\\angle C=180^\circ-(63^\circ+47^\circ)\\\angle C=70^\circ

Using the Law of Sines

\dfrac{a}{\sin A} =\dfrac{c}{\sin C}\\\\\dfrac{a}{\sin 63^\circ} =\dfrac{7.75}{\sin 70^\circ} \\\\a=\dfrac{7.75}{\sin 70^\circ} \times \sin 63^\circ\\\\a=7.35$ inches (to the nearest hundredth of an inch)

6 0
4 years ago
Read 2 more answers
Which students line segments have a midpoint of (-1,4)? Select all that apply A. Student 1 B. Student 2 C. Student 3 D. Student
SashulF [63]

Answer:

C. Student 3

E. Student 5

Step-by-step explanation:

we know that

The formula to calculate the midpoint between two points is equal to

M(\frac{x1+x2}{2},\frac{y1+y2}{2})

<u><em>Verify the midpoint of each student</em></u>

student 1

we have the endpoints

(-9,0) and (11,-8)

substitute in the formula

M(\frac{-9+11}{2},\frac{0-8}{2})

M(1,-4)

so

The midpoint is not (-1,4)

student 2

we have the endpoints

(-6,-1) and (4,-7)

substitute in the formula

M(\frac{-6+4}{2},\frac{-1-7}{2})

M(-1,-4)

so

The midpoint is not (-1,4)

student 3

we have the endpoints

(-5,2) and (3,6)

substitute in the formula

M(\frac{-5+3}{2},\frac{2+6}{2})

M(-1,4)

so

<u>The midpoint is equal to (-1,4)</u>

student 4

we have the endpoints

(-3,10) and (5,-2)

substitute in the formula

M(\frac{-3+5}{2},\frac{10-2}{2})

M(1,4)

so

The midpoint is not (-1,4)

student 5

we have the endpoints

(0,-3) and (-2,11)

substitute in the formula

M(\frac{0-2}{2},\frac{-3+11}{2})

M(-1,4)

so

<u>The midpoint is equal to (-1,4)</u>

therefore

Student 3 and student 5

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