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3 years ago

Select the ordered pairs that are solutions to the inequality

Mathematics

1 answer:

sveticcg [70]3 years ago

5 0

Answer:

I would say equal to twelve

Step-by-step explanation:

Hope this helps.

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Please helppp (had to have 20 words haha)

s2008m [1.1K]

Answer:

w==77

Step-by-step explanation:

The sum of the 2 opposite interior angles equals the exterior angle.

w-36 +w-20=w+21

combine like terms: 2w-56=w+21

subtract w from both sides: w-56=21

add 56 to both sides: <u>w=77</u>

6 0

3 years ago

Analyze the conditional statement below and complete the instructions that follow.

morpeh [17]

Answer:

If you did not break it, Then you did not by it.

Step-by-step explanation:

8 0

3 years ago

(4x-1) = 6<br> 2<br> How to solve rational equations

9966 [12]

Answer:

educated guess

Step-by-step explanation:

cos ( 390 ) sin ( 330 ) − 4 cot ( − 45 )

8 0

2 years ago

Below are the steps to solve an equation:

otez555 [7]

Answer:

x-5=-3

Step-by-step explanation:

|x-5|+2=5

|x-5|=5-2

|x-5|=3

x-5=±3

so x-5=-3 is a correct step.

4 0

2 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 $R_1\rightarrow R_1-R_3$

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

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