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Hatshy [7]
3 years ago
6

Fencing How much fencing is required to enclose a circular garden whose radius is 28 ​m? Use 3.14 for . About nothing m of fenci

ng is required to enclose the circular garden. ​(Type an integer or decimal rounded to the nearest hundredth as​ needed.)
Mathematics
1 answer:
Andru [333]3 years ago
8 0

Answer:

175.84 m

Step-by-step explanation:

Given that:

A circular garden which has a radius of 28 m.

\pi = 3.14

To find:

The fencing required to enclose the circular garden ?

Solution:

Here, we are given a circle and its radius.

We are required to find the perimeter of the circle in order to find the fencing required to enclose the circular garden..

Formula for finding the perimeter of a circle is:

Perimeter = 2\pi r

where r is the radius of the circle.

Using the given values in the formula:

Perimeter = 2\times 3.14 \times 28 = <em>175.84 m</em>

Therefore, 175.84 m of fencing is required.

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Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

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The 14th term of an arithmetical progression is 15 and it's 9th term is 35 find the 15th term​
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the 15th term is 11

Step-by-step explanation:

common difference is= -4

first term is 67

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Perimeter of the triangle below ?
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Answer:

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Step-by-step explanation:

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combine like terms

5x^2+8x^2-18x^2= 31x^2

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31x^2-13x

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There are 60 ways to arrange the letters in the word PRIOR.

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25 POINTS please help me
prisoha [69]

If you have a graphing calculator or even a website you can graph the equation down then look at the x axis where 3 and 4 is. After you do that look at the y axis where you will find what 3 and 4 equals to then you will subtract those numbers and that should be the answer

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