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2 years ago

Find the value of x in the following equation: 9x + 8.0 = -12.7

Mathematics

1 answer:

andriy [413]2 years ago

7 0

Answer: x= − 2.3 and the total equation looks like this: 9(-2.3) + 8.0 = -12.7

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In rectangle DATE, the diagonals DT and AE intersect as S. If LaTeX: AE=40 A E = 40 and LaTeX: ST=x+5 S T = x + 5 , find the val

svetoff [14.1K]

For this case we have the following diagonals:
AE = 40
ST = x + 5
The intersection of the diagonals occurs when:
AE = ST
Therefore, matching we have:
40 = x + 5
Clearing x we have:
x = 40-5
x = 35
Answer:
The value of x is:
x = 35

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3 years ago

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Help me quick plz thanks

rodikova [14]

add 12-3 = 9

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3 years ago

HW-02 Problem No.2.1 / 10 pas 5x1 - x2 = 1 ( 3x2 - 2x1 = -3 Solve the system of linear equations by modifying it to REF and to R

IrinaVladis [17]

Answer:

$REF= \left[\begin{array}{ccc}5&-1&1\\0&\frac{-7}{5} &\frac{-18}{5}\end{array}\right]$

$RREF=\left[\begin{array}{ccc}1&0&\frac{5}{7} \\0&1&\frac{18}{7}\end{array}\right]$

Step-by-step explanation:

The augmented matrix of the system is: $\left[\begin{array}{ccc}5&-1&1\\3&-2&-3\end{array}\right]$

First we find the stepped form of A (REF):

1. We subtract 3/5 from row 1 to row 2 (R2- $\frac{3}{5}$ R1) and get the matrix

$\left[\begin{array}{ccc}5&-1&1\\0&\frac{-7}{5} &\frac{-18}{5}\end{array}\right]$

Note that this matrix is in echelon form.

Now we find the reduced row echelon form of the augmented matrix (RREF)

2. From the previous matrix, we multiply the first row by 1/5 and the second row by -5/7 and obtain the matrix:

$\left[\begin{array}{ccc}1&\frac{-1}{5} &\frac{1}{5} \\0&1&\frac{18}{7}\end{array}\right]$

3. From the previous matrix, to row 1 we add 1/5 of row 2 (R1 + $\frac{1}{5}$ R2) and we obtain the matrix

$\left[\begin{array}{ccc}1&0&\frac{5}{7} \\0&1&\frac{18}{7}\end{array}\right]$

which is the reduced row echelon form of the augmented matrix.

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3 years ago