Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Answer: 2/3
Slope = y2-y1/x2-x1
I’ll take the points (-2;0) and (1;2)
Slope = 0-2/-2-1
=-2/-3=2/3
Answer:
The correct option is D.
Step-by-step explanation:
The given equation is
According to the addition property of equality 
and 
are equivalent equations.
Use addition property of equality, add 3x on both the sides.
Therefore Sam's work is incorrect because he make calculation mistake.
According to the subtraction property of equality and 
are equivalent equations.
Use subtraction property of equality, subtract 5x from both the sides.
Therefore Roy's work is correct because he used subtraction property.
Option D is correct.
Answer:
Step-by-step explanation:
Given the following complex numbers, we are to expressed them in the form of a+bi where a is the real part and b is the imaginary part of the complex number.
1) (2-6i)+(4+2i)
open the parenthesis
= 2-6i+4+2i
collect like terms
= 2+4-6i+2i
= 6-4i
2) (6+5i)(9-2i)
= 6(9)-6(2i)+9(5i)-5i(2i)
= 54-12i+45i-10i²
= 54+33i-10i²
In complex number i² = -1
= 54+33i-10(-1)
= 54+33i+10
= 54+10+33i
= 64+33i
3) For the complex number 2/(3-9i), we will rationalize by multiplying by the conjugate of the denominator i.e 3+9i
= 2/3-9i*3+9i/3+9i
=2(3+9i)/(3-9i)(3+9i)
= 6+18i/9-27i+27i-81i²
= 6+18i/9-81(-1)
= 6+18i/9+81
= 6+18i/90
= 6/90 + 18i/90
= 1/15+1/5 i
4) For (3 − 5i)(7 − 2i)
open the parenthesis
= 3(7)-3(2i)-7(5i)-5i(-2i)
= 21-6i-35i+10i²
= 21-6i-35i+10(-1)
= 21-41i-10
= 11-41i
