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ella [17]
3 years ago
13

Find the Area of the figure below, composed of a rectangle and a semicircle. Round to

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

129.1 square units

Step-by-step explanation:

First, let's find the area of the rectangle. We can use the formula below.

a=lw

where l is the length and w is the width.

The length of the rectangle is 13 and the width is 8.

l=13\\w=8

Substitute the values into the formula.

a=13*8

a=104

Next, let's find the area of the semicircle. We can use the formula below.

a=\frac{1}{2} *\pi r^2

First, find the radius. The radius is half of the diameter, and the diameter is 8.

⇒ r= d/2

⇒ r= 8/2

⇒ r= 4

a=\frac{1}{2} *\pi (4)^2

Evaluate the exponent.

⇒ 4² = 4*4 = 16

a=\frac{1}{2} *\pi *16

Multiply pi and 16.

a=\frac{1}{2} *50.2654825

Multiply.

a=25.1327412

Finally, find the area of the figure.

Add the area of the rectangle (104) and the area of the semicircle (25.1327412)

104+25.1327412

129.132741

Round to the nearest tenth. The 3 in the hundredth place tells us to leave the 1 in the tenth place.

129.1

The area of the figure is about 129.1 (square units).

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A2 = [1 2 3; 4 5 6; 7 8 9; 3 2 4; 6 5 4; 9 8 7]
34kurt

Answer:

Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.

We need to find the echelon form of the matrix augmented matrix of the system A2x=b2

B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]

We apply row operations:

1.

  • To row 2 we subtract row 1, 4 times.
  • To row 3 we subtract row 1, 7 times.
  • To row 4 we subtract row 1, 3 times.
  • To row 5 we subtract row 1, 6 times.
  • To row 6 we subtract row 1, 9 times.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]

2.

  • We subtract row two twice to row three of the previous matrix.
  • we subtract 4/3 from row two to row 4.
  • we subtract 7/3 from row two to row 5.
  • we subtract 10/3 from row two to row 6.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]

3.

we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]

Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set

\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}

is a basis for Row (A2)

Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.

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Step-by-step explanation:

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Step-by-step explanation:

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