-1
No, because it is not an equation (or inequality), but simply two expressions separated by a space.
1.75
10xy3 * 8x5y3 = (10*8)*(x*x5)*(y3*y3) = 80x6y6
4
Yes. You need at least three points with x and y coordinates for this. Let the points be: A(x1, y1), B(x2, y2) and C(x3, y3). For each of the points yi = ax2i + bxi + c, so you have to solve a set of simultaneous equations for i=1 ... 3(the y's and x's are no longer variables! the a, b, c are!): y1 = ax21 + bx1 + c, y2 = ax22 + bx2 + c, y3 = ax23 + bx3 + c
y3 + 9y + 10y - 5/y + 2 can not be solved, as it is not an equation. If you would like to simplify it, you can reduce it a bit: = y3+ 19y - 5/y + 2
Lets start by putting the known number on the other side. Y3 -7500y=125000. There is your start. Take it from here.
72
-y3 + 7y3
One equation with two unknowns usually does not have a solution.
X2+Y3+15 All you can do is simplify it.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
no
y3 x y3 - y (3)3 x 3(3) - 3 9 x 9 - 3 = ? 9 x 9= 81 81 - 3 = 78 I hope that solves your problem
3xyz
It is an algebraic expression.
No.