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3 years ago

I really need help on this one...ASAP!

Mathematics

2 answers:

Agata [3.3K]3 years ago

7 0

$4(a+12)=32\\
a+12=8\\
a=-4$

Leokris [45]3 years ago

6 0

4(a+12)=32
4a+48=32
4a=-16
a=-4

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4x-1divide by 2=x+7<br> what is x

sertanlavr [38]

X= 2.5
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3 years ago

What is the solution to the linear equation? x – StartFraction 2 Over 3 EndFraction x minus StartFraction one-half EndFraction e

Anton [14]

<em>Your question is not well presented</em>

Question:

What is the solution to the linear equation?

$x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x$

Answer:

$x = \frac{-5}{3}$

Step-by-step explanation:

Given

$x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x$

Required

Simplify

$x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x$

Add $\frac{1}{2}$ to both sides

$x - \frac{2}{3}x - \frac{1}{2} + \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x + \frac{1}{2}$

$x - \frac{2}{3}x = \frac{1}{3} + \frac{5}{6}x + \frac{1}{2}$

Subtract $\frac{5}{6}x$ from both sides

$x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{5}{6}x + \frac{1}{2} - \frac{5}{6}x$

$x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{1}{2} + \frac{5}{6}x- \frac{5}{6}x$

$x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{1}{2}$

Take LCMs

$\frac{6x - 4x -5x}{6}= \frac{2+3}{6}$

$\frac{-3x}{6}= \frac{5}{6}$

Multiply both sides by 6

$6 * \frac{-3x}{6}= \frac{5}{6} * 6$

$-3x= 5$

Divide both sides by -3

$\frac{-3x}{-3}= \frac{5}{-3}$

$x = \frac{5}{-3}$

$x = \frac{-5}{3}$

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4 years ago

Read 2 more answers

Anyone knows the answer

vesna_86 [32]

Answer:

it might be 390

Step-by-step explanation:

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3 years ago

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Dmitry_Shevchenko [17]

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3 years ago

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Use a matrix to find the coordinates of the endpoints or vertices of the image of each figure under the given reflection.

hjlf

The vertices of the original quadrilateral can be written in matrix form using the vertex matrix. The vertex matrix is
$\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}$

To find the coordinates of the endpoints or vertices of the image of the given coordinate points reflected about the y-axis, we just need to multiply the transformation matrix by the vertex matrix. The transformation matrix for this particular problem is
$\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}$

Multiplying the two matrices, we have
$\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}=\begin{pmatrix}5&1&3&7\\ 4&-1&-6&-3\end{pmatrix}$

Therefore, the coordinates of the endpoints or vertices of the image are (5,4), (1,-1), (3, -6) and (7, -3).

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3 years ago