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marishachu [46]
1 year ago
12

What is the product?(4y − 3)(2y2 + 3y − 5)8y3 + 3y + 158y3 − 23y + 158y3 − 6y2 − 17y + 158y3 + 6y2 − 29y + 15

Mathematics
1 answer:
motikmotik1 year ago
5 0

We need to find the product of :

\mleft(4y-3\mright)\mleft(2y2+3y-5\mright)

So, the result as following:

\begin{gathered} \mleft(4y-3\mright)\mleft(2y^2+3y-5\mright) \\ =4y\cdot(2y^2+3y-5)-3\cdot(2y^2+3y-5) \\ =8y^3+12y^2-20y-(6y^2+9y-15) \\ =8y^3+12y^2-20y-6y^2-9y+15 \\  \\ =8y^3+6y^2-29y+15 \end{gathered}

So, the answer is the option 4. 8y3 + 6y2 − 29y + 15​

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3 years ago
If A=,[2 5 7],find the value of <br> [1 4 9]<br> [8 3 6]<br>a23+a12-a32 matrix.​
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Answer:

<em><u>T</u></em><em><u>H</u></em><em><u>E</u></em><em><u> </u></em><em><u>C</u></em><em><u>O</u></em><em><u>R</u></em><em><u>R</u></em><em><u>E</u></em><em><u>C</u></em><em><u>T</u></em><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>E</u></em><em><u>R</u></em><em><u> </u></em><em><u>O</u></em><em><u>F</u></em><em><u> </u></em><em><u>T</u></em><em><u>H</u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>Q</u></em><em><u>U</u></em><em><u>E</u></em><em><u>S</u></em><em><u>T</u></em><em><u>I</u></em><em><u>O</u></em><em><u>N</u></em><em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>5</u></em><em><u>.</u></em>

Step-by-step explanation:

Here,

a23=6

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Now,

a23+a12-a32

=6+5-6

=5

6 0
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