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Alexandra [31]
2 years ago
11

Solve to find X. 1/5(x - 2) = 1/10 (x + 6)

Mathematics
2 answers:
Archy [21]2 years ago
5 0

Answer:

Step-by-step explanation:

1/5(x-2) =1/10(x+6)

1/5x-2 = 1/10x+0.6

1/5x+1/10x= 0.6-2

0.2x+0.1x= -1.4

0.3x= -1.4

x= -4.67

x= -4.7

x= -5

LenaWriter [7]2 years ago
3 0

Answer:

x = 10

Step-by-step explanation:

First, distribute:

\frac{1}{5} x - \frac{2}{5} =\frac{1}{10} x+\frac{6}{10}

Next, subtract \frac{1}{5}x from both sides:

-\frac{2}{5} = \frac{1}{10} x -\frac{1}{5} x+\frac{6}{10}

(multiply \frac{1}{5}x by 2 to get common denominator, \frac{2}{10} x )

-\frac{2}{5} =-\frac{1}{10}x +\frac{6}{10}

Next, subtract \frac{6}{10} from both sides:

-\frac{2}{5} -\frac{6}{10} =-\frac{1}{10} x

(multiply \frac{2}{5} by 2 to get common denominator, \frac{4}{10} )

-\frac{10}{10} =-\frac{1}{10} x ⇒ -1=-\frac{1}{10} x

Next, multiply both sides by 10:

-10=-x

FInally, divide both sides by -1:

x = 10

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Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

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Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

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3 years ago
How do you calculate this?<br> Help
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V = s^3 for a cube

V = (3.5 * 10^-2) ^3

  the numbers are cubed and the exponents are multiplied

V = (3.5)^3 *10^(-2*3)

V= 42.875 * 10^-6  m^3


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