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Murrr4er [49]
3 years ago
14

Can someone help me w this question plz.

Mathematics
1 answer:
quester [9]3 years ago
6 0
For any solution it follows that y=y so:

x^2+2x=3x+20  subtract 3x from both sides

x^2-x=20  subtract 20 from both sides

x^2-x-20=0  factor...

x^2-5x+4x-20=0

x(x-5)+4(x-5)=0

(x+4)(x-5)=0

x=-4 and 5  now to find the corresponding y values I'll use y=x^2+2x

y(-4)=16-8=8 and y(5)=25+10=35  so the two solutions are the points:

(-4, 8) and (5, 35)
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ANEK [815]
For the answer to the question above, I'll provide my solutions to my answers for the problem below.

(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)

(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)

(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5

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−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5

(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)

−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= <span><span><span><span><span>−<span><span>6x4</span>y</span></span>−<span><span>2x3</span>y2</span></span>+<span><span>9x2</span>y3</span></span>−<span>3xy4</span></span>+y5</span> 

I hope this helps
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4 years ago
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