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2 years ago

15

The circumference of a circle is 11 inches. What is the area, in square inches, of the circle? Express your answer in terms of p

i

2 answers:

kakasveta [241]2 years ago

7 0

Answer

A = 30.25pi. PLS GIVE BRAINLIEST

Step-by-step explanation:

so if circumference is 11, and formula for circumference is 2pi*r, we can work backwards to find the radius.

Radius: 11pi/2 = 5.5pi = 5.5 is radius

Area = pi * r^2

so Area = pi*5.5^2

so A = 30.25pi left in terms of pi

forsale [732]2 years ago

5 0

Answer: 11/4pi

Step-by-step explanation:

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(4x-5)+2(3x+1)<br> Can you help me solve this plz for me

Korvikt [17]

Answer:

10x - 3

Step-by-step explanation:

First, distribute 2 to all terms within the second parenthesis:

2(3x + 1) = (2 * 3x) + (2 * 1) = 6x + 2

4x - 5 + 6x + 2

Combine like terms (terms with the same amount of variables).

4x + 6x + 2 - 5

(4x + 6x) + (2 - 5)

10x + (-3)

10x - 3

10x - 3 is your answer.

~

8 0

3 years ago

Read 2 more answers

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 $R_1\rightarrow R_1-R_3$

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

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

3 years ago

Solve?????????????????

o-na [289]

Answer:

20/98

Step-by-step explanation:

im for sure

3 0

2 years ago

Find the values of x and y.

Harrizon [31]

4x+5x = 90
9x = 90
x = 10

10y + 10 + 5x = 180
10y + 10 +5(10) = 180
10y + 60 = 180
10y = 120
y = 12

answer
<span>d) x = 10, y = 12</span>

5 0

3 years ago

Read 2 more answers

Write the linear equation that gives the rule for this table.

Maksim231197 [3]

to get the equation of any straight line all we need is two points, well, let's just grab them from the table.

hmmm let's use (-49 , -39) and hmmm say (67 , 77)

$(\stackrel{x_1}{-49}~,~\stackrel{y_1}{-39})\qquad (\stackrel{x_2}{67}~,~\stackrel{y_2}{77}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{77}-\stackrel{y1}{(-39)}}}{\underset{run} {\underset{x_2}{67}-\underset{x_1}{(-49)}}}\implies \cfrac{77+39}{67+49}\implies \cfrac{116}{116}\implies 1$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-39)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-49)}) \\\\\\ y+39=1(x+49)\implies y+39=x+49\implies y=x+10$

8 0

2 years ago