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Ber [7]
3 years ago
5

Pleasee help !!!!!!!

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0

Answer: x = -1 and x= -5

Step-by-step explanation:

Try plugging in the solutions for x (remember that whatever number comes out of the absolute value bars is ALWAYS POSITIVE)

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Match the medication with its food recommendation
Ann [662]

Answer:

we need more information then that....

Step-by-step explanation:

6 0
2 years ago
Help me please I will give brainliest
ExtremeBDS [4]
The correct answer would be B because it is is written in correct simplified format
8 0
3 years ago
If (x+3)divided by 3 = 9 what is x
defon

Answer:

x = 24

Step-by-step explanation:

Given that:

(x+3)/3 = 9

multiplying both sides by 3

(x+3) = 9*3

(x+3)= 27

Subtracting 3 from both sides

x= 27-3

x=24

I hope it will help you!

3 0
3 years ago
Read 2 more answers
Someone please help me with this and explain it please.
Yuki888 [10]
<h3>Answer: Choice D) \sqrt{34}, 6, 2\pi, \sqrt{42}</h3>

===========================================

Explanation:

Use a calculator to find the decimal form of each of the following below (values are approximate; rounded to 2 decimal places)

  • 2pi = 6.28
  • sqrt(42) = 6.48
  • sqrt(34) = 5.83

We see that the order of the decimal values is: 5.83, 6, 6.28, 6.48

So the order of the terms, from smallest to largest, is \sqrt{34}, 6, 2\pi, \sqrt{42}

5 0
3 years ago
(12x + y + z = 26
mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}26\\17\\23\end{array}\right].

Step 3:

The linear system B is represented as

\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}23\\17\\26\end{array}\right].

Step 4:

The linear system C is represented as

\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\11\\12\end{array}\right].

Step 5:

The linear system D is represented as

\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}12\\11\\4\end{array}\right].

Step 6:

Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.

4 0
2 years ago
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