Which expression is the simplest form of 5(x-3)-3(2x+4)/9 ?
A. -x-3/9
B. -x-27/9 C.11x-27/9
D.-x-3

Which Expression Is The Simplest Form Of 5(x-3)-3(2x+4)/9 ? A. -x-3/9B. -x-27/9 C.11x-27/9D.-x-3

Answers

Answer 1
Answer : D. -x - 3

Explanation :

5(x - 3) -3(2x + 4) / 9
5x - 15 - 6x - 12 / 9
-1x - 27 / 9
-1x - 3 which is also -x - 3

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Step-by-step explanation:

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Step-by-step explanation:

Answer:

y=(x-1)(x-6)

Step-by-step explanation:

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Answers

Answer:

[tex]42a^{2}[/tex]

Step-by-step explanation:

Split them into sections!

First, [tex]3a[/tex] x [tex]5a[/tex] = [tex]15a^{2}[/tex]

Next, [tex](3a+3a+3a)(3a)[/tex] = [tex]9a[/tex] x [tex]3a[/tex] = [tex]27a^{2}[/tex]

Add [tex]15a^{2}[/tex] and  [tex]27a^{2}[/tex] to get [tex]42a^{2}[/tex]

Hope I helped!

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform the augmented matrix to reduced echelon form, and then find all solutions to the corresponding system. (If there are an infinite number of solutions use s1 as your parameter. If there is no solution, enter NO SOLUTION.)

[9 -2 0 -4 8]
[0 7 -1 -1 9]
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(x1, x2, x3, x4)=_______

Answers

Answer:

[tex]x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\[/tex]

Step-by-step explanation:

As the given Augmented matrix is

[tex]\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right][/tex]

Step 1 :

[tex]r_{1}[/tex]↔[tex]r_{1} - r_{2}[/tex]

[tex]\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right][/tex]

Step 2 :

[tex]r_{3}[/tex]↔[tex]r_{3} - 8r_{1}[/tex]

[tex]\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right][/tex]

Step 3 :

[tex]r_{2}[/tex]↔[tex]\frac{r_{2}}{7}[/tex]

[tex]\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right][/tex]

Step 4 :

[tex]r_{1}[/tex]↔[tex]r_{1} + 14r_{2}[/tex] , [tex]r_{3}[/tex]↔[tex]r_{3} - 124r_{2}[/tex]

[tex]\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right][/tex]

Step 5 :

[tex]r_{3}[/tex]↔[tex]\frac{r_{3}. 7}{254}[/tex]

[tex]\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right][/tex]

Step 6 :

[tex]r_{1}[/tex]↔[tex]r_{1} - 4r_{3}[/tex] , [tex]r_{2}[/tex]↔[tex]r_{2} + \frac{1}{7} r_{3}[/tex]

[tex]\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right][/tex]

∴ we get

[tex]x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\[/tex]

NEED HELP!
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x =76° is the answer

Step-by-step explanation:

because being alternate angle.

Answer:

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Step-by-step explanation:

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Therefore,x=104.

Hope it will help you be happy!

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Answers

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Step-by-step explanation:

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Answers

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Answers

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