Y=x^2-5, solve for x
y+5=x^2
x=√(y+5) so
f^-1(x)=√(y+5)
The constant is -12. I think of constant as the "lonely little number".... Hope this helped =)
Answer:Two such terms are 7x^3*y^9 and -3x*y^5
Their quotient is
7x^3*y^9
--------------
-3x*y^5
This can be simplified as follows:
The numerical coefficients become -7/3.
x^3/x = x^3*x^1 = x^(3 - 1) = x^2 (we subtract the exponent of x in the denominator from the exponent of x in the numerator).
Next, y^9*y^5 = y^4.
The quotient in final reduced form is then (-7/3)x^2*y^4
When we add or subtract polynomials, the addition or subtraction is applied to the coefficients of the terms of equal powers.
For the given polynomials, we are required to do subtraction as follows:
(-4.1 x^2 + 0.9 x - 9.8) - (1.7 x^2 - 2.4 x - 1.6)
= -4.1 x^2 + 0.9 x - 9.8 - 1.7 x^2 + 2.4 x + 1.6
= (-4.1-1.7) x^2 + (0.9+2.4) x + (-9.8+1.6)
= -5.8 x^2 + 3.3 x - 8.2
The final expression (solution) is:
-5.8 x^2 + 3.3 x - 8.2
