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Rasek [7]
2 years ago
9

There are 3 answers need help to get my grade up to a 70%

Mathematics
1 answer:
svetoff [14.1K]2 years ago
3 0

Answer:

From (2, 1) move up one and over four

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Which are partial products for 68 × 43?
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Partial products are the products obtained during the intermediate stages in order to complete a multiplication process.

Consider 68 \times 43, we have to determine the partial products in this.

Now, 68 \times 43

Expanding this, we get

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= 2400 + 180 + 320 + 24

= 2924

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So, Option 3 and 4 are the correct answers.

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What expression is equivalent to 13-(-21)?
Naddik [55]

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34

Step-by-step explanation:

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5 0
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A company which manufactures computers increases its production
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93312

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5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cleft.%20%5Cbegin%7Bcases%7D%20%7B%208%20x%20%2B%202%20y%20%3D%2046%20%7D%20%5C%5C%20%7B%207
Brut [27]

\huge \boxed{\mathfrak{Question} \downarrow}

\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.

First, write both the equations in its standard form.

8x+2y=46\\ 7x+3y=47

Now, write the equations in form of matrix.

\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)

Then, multiply the equation towards the left by using the inverse of matrix \left(\begin{matrix}8&2\\7&3\end{matrix}\right)

\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

The product of the matrix & its inverse will be the identity matrix.

\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

Now, multiply the matrices that lie on the left-hand side of the equal sign.

\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

For the 2 × 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is ⇨ \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right).

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)

Do the calculations.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)

Multiply the matrices.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)

Do the arithmetics again.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{5}  \\\frac{27}{5}\end{matrix}\right)

Finally, extract the matrix elements x & y & write them separately.

\large \boxed{ \boxed{ \bf \: x=\frac{22}{5},y=\frac{27}{5} }}

3 0
3 years ago
Read 2 more answers
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