Which is true of the infinite solutions of the inequalit x <0?The solutions will contain an infinite

The solutions will contain an infinite amount of negative numbers.

Step-by-step explanation:

Given x < 0

What does this mean?

First, carefully examine the sign in the inequality. It says less (which is different from less and equal)

Then explain it's meaning.

x < 0 just means that the solutions will contain an infinite amount of negative numbers but NOT zero. This defines the correct answer.

Step-by-step explanation:

< = more than
> = less than
So therefore,
D > 5 miles

Hope this helped !

2x + 6y = − 12 ;10x + 32y = −62
Now make sure to solve!
2x + 6y = −12 for x > 2x + 6y + −6y = −12 + −6y (Just add -6y to both sides)
2x = −6y −12
[tex]\frac{2x}{2} = \frac{-6y - 12}{2}[/tex]< Divide both sides by 2
x= − 3y − 6
Now we have to substitute > −3y − 6 for x in 10x + 32y = −62
10x + 32y = −62
10(−3y − 6) + 32y = −62
2y − 60 = −62 (We have to simplify both sides!)
2y − 60 + 60 = −62 + 60 (Now add 60 to both sides!)
2y = −2
[tex]\frac{2y}{2} = \frac{-2}{2}[/tex] (Make sure to divide both sides by 2)
We get y = -1 here.
Substitute −1 for y in x = −3y − 6
 −3y − 6x
(−3)(−1) − 6
x=−3 (Simplify both sides)
We have 2 solutions > x = −3 and y = −1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am not quite sure how to solve the second problem!

1)  Option (3) is correct. The system has one solution.

2)  Option (2) is correct. The solution is (4,0)

Step-by-step explanation:

1) Consider the given system of equation ,

2x +6y = -12          .....(1)

10x + 32y = -62       ....(2)

Here,  [tex]a_1=2,b_1=6,c_1--12[/tex] and [tex]a_2=10,b_2=32,c_2=-62[/tex]

Find the ratio for [tex]\frac{a_1}{a_2},\frac{b_1}{b_2},\frac{c_1}{c_2}[/tex]

Thus, [tex]\frac{a_1}{a_2}=\frac{2}{10}=\frac{1}{5}[/tex]

[tex]\frac{b_1}{b_2}=\frac{6}{32}=\frac{3}{16}[/tex]

and [tex]\frac{c_1}{c_2}=\frac{-12}{-62}=\frac{3}{8}[/tex]

hence, it is clear from above ratios, that [tex]\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq\frac{c_1}{c_2}[/tex]

Thus, the system will have a unique solution.

We will solve the system ,

Multiply equation (1) by 5, we get,

10x + 30y = -60     ...(3)

Subtract equation (3) from (2),

10x + 32y -( 10x + 30y ) = -62-( -60 )

⇒ y = - 1

Put y = 1 in (1), we get,

2x +6(-1) = -12  ⇒ 2x= -12 +6  ⇒ x = -3.

Thus, (-3 , -1) is the solution

Hence the system has one solution.

2) Consider the given system of equation ,

2d + e = 8       ...........(1)

d - e = 4  ..........(2)

Add equation (1) and (2) , we get,

2d + e +(d -e) = 8+ 4

⇒ 2d +d = 12

⇒  3d= 12

⇒  d = 4

Put d = 4 in (1) , we get

2d + e = 8  ⇒ 2(4)+e = 8  ⇒ e = 8 - 8 ⇒ e = 0.

Thus, The solution is (4,0).  

1, 2, 5 and 6 are the answers.

Step-by-step explanation:

2 is correct because the arrow shows that the number line continues till infinity. 5 is correct because any fraction is possible as no restriction of integer value is placed. Any number less than 7 is not included. There are some confusions with 1 as it's not a solid color, So, I assume -7 is not included.

answer:

b

explanation:

Here... I only have one answer.

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

[tex]5x+4y+5z=-1[/tex]

[tex]x+y+2z=1[/tex]

[tex]2x+y-z=-3[/tex]

Firs we find determinant of system of equations

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

[tex]\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}[/tex]

[tex]\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0[/tex]

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 [tex]R_1\rightarrow R_1-4R_2[/tex] and [tex]R_3\rightarrow R_3-2R_2[/tex]

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

Apply[tex]R_2\rightarrow R_2-R_1[/tex]

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

Apply [tex]R_3\rightarrow R_3+R_2[/tex]

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

Apply [tex]R_3\rightarrow- \frac{1}{2}[/tex] and [tex]R_2\rightarrow R_2-R_3[/tex]

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

Apply [tex]R_1\rightarrow R_1-R_3[/tex]

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

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.

It's C) x + 1 < 5; x < 4

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