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
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
−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
The probability that it consists of all boys is 8/14. You're Welcome!
Step-by-step explanation:
Add each term and divide it by 2.
5+7 = 12
12/2 = 6
-2 + 6 = 4
4/2 = 2, then add the i back on to get 2i.
The midpoint is 6 + 2i
2y + 4 + 5y + 8=
7y + 4 + 8=
7y + 12=
