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3 years ago

3. In the figure shown, a circle with center O and radius of length 3 units is

Mathematics

1 answer:

Levart [38]3 years ago

7 0

Answer:

Step-by-step explanation:

To find the area of a circle, take its radius (3) and square it. In this case, 3 times 3 = 9. Then, you multiply the answer by pi, which rounds to 3.14, but is more like 3.14159265......etc, etc. There is a rhyme for that: May I have a large container of coffee beans? Where the length of each word is a number of pi. Anyway, most people would write 9pi, but this is asking for the exact answer, so you can do 3.14 times 9, which equals roughly 28.27. Hope this helped!

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Can you please help me

Levart [38]

thanks for your help I really think that it's three

5 0

2 years ago

3n+4 first the 7 terms

kiruha [24]

Answer:

4, 7, 10, 13, 16, 19, 22

Step-by-step explanation:

Not sure if there is more to this question but here is what I assume:

We want to find the first 7 terms of this equation given 3n+4

Assuming we are starting at 0 we plug 0 in for n.

3(0) + 4 → 4

So our first answer is 4

We continue by pugging in the next numbers after that to get to the first 7 terms

3(1) + 4 → 7

3(2) + 4 → 10

3(3) + 4 → 13

3(4) + 4 → 16

3(5) + 4 → 19

3(6) + 4 → 22

5 0

3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$