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kirill [66]
2 years ago
11

Find the values of x and y which satisfy the following equations

Mathematics
1 answer:
goblinko [34]2 years ago
8 0

Answer:

Step-by-step explanation:

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Answer:

The answer is B

Step-by-step explanation:

I got B because it just looks like its flipped over the y axis. Its also even. Im sorry if you didnt get that but im not the best explainer at these types of problems lol. But its actually b trust me

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What is the equation of the line described below written in slope-intercept form?
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Y=4x+4 


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What is the slope-intercept form of: 4x+6y=2340
valkas [14]

Answer:

Pretty sure its y=-2/3x+390

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%
SVETLANKA909090 [29]

\huge \boxed{\mathbb{QUESTION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)

The remaining elements of the product matrix are found in the same way.

\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)

Simplify each element by multiplying the individual terms.

\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)

Now, sum each element of the matrix.

\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}

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Answer:

2239grams

Step-by-step explanation:

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