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jeka94
3 years ago
6

Does anybody know this answer ??6x-12 2x=3x 8x-15

Mathematics
1 answer:
romanna [79]3 years ago
7 0
Hello there.! 

<span>Does anybody know this answer ??6x-12 2x=3x 8x-15

</span><span>0.35775</span>
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Madison can lift 200 kilograms with ease. how much is this in pounds
Firlakuza [10]

Answer:440.925

Step-by-step explanation:

4 0
3 years ago
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Find x<br> 5[x^2]+5[x]-x^2-x=2004
a_sh-v [17]

The answer is either x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\ or x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}

To solve: Let's solve your equation step-by-step.

5x2+5x−x2−x=2004

Step 1: Simplify both sides of the equation.

4x2+4x=2004

Step 2: Subtract 2004 from both sides.

4x2+4x−2004=2004−2004

4x2+4x−2004=0

Step 3: Use quadratic formula with a=4, b=4, c=-2004.

x= −b±√b2−4ac /2a

x= −4±√32080 /8

x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\ or x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}

4 0
3 years ago
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Ax + 4y = 5z, for a​
Stolb23 [73]

Answer:A=5z-4y/x

Step-by-step explanation:

You must bring 4y over first making it negative then to separate the x from the A you have to divide

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3 years ago
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Plz help<br>urgent !!!!<br>will give the brainliest !!​
Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

7 0
3 years ago
Calculate the area of the shaded region. Use 3.14 for pi; round to the nearest tenth, if needed. HELP PLSSS I HAVE TO GET THIS D
icang [17]

Answer:

87.48

Step-by-step explanation:

First find the area of the square

12^2=144

find area of the 2 circles

9*3.14

but theres 2

18*3.14=56.52

144-56.52=87.48

4 0
3 years ago
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