Imaginary numbers are used when a number inside of a square root is negative.
Answer:
32 and 33
Step-by-step explanation:
so , 65-33= 32
32+33 is 65.
A = 6(1/3)²
A = 6(1/9)
A = 6/9 or 2/3
Hope this helps :)
Answer:
The answer is C
Step-by-step explanation:
Trust me I made a graph to organize the numbers
Answer:
y = x + 6
x = 1
y = ¼(x - 5) + 3
Step-by-step explanation:
Vetices are;
A(1,5), B(5,3) and C(-3, -2)
Thus;
Median of AB is; D = (1 + 5)/2, (5 + 3)/2
D = (3, 4)
Median of BC is; E = (5 + (-3))/2, (3 + (-2))/2
E = (1, 0.5)
Median of AC is; F ; (-3 + 1)/2, (-2 + 5)/2
F = (-1, 1.5)
Thus, the median lines will be;
CD, AE & BF.
Thus;
Equation of CD is;
(y - (-3))/(x - (-2)) = (-2 - 4))/(-3 - 3)
(y + 4)/(x + 2) = -6/-6
y - 4 = 1(x + 2)
y = 4 + x + 2
y = x + 6
Equation of AE;
(y - 5)/(x - 1) = (0.5 - 5)/(1 - 1)
(y - 5)/(x - 1) = -4.5/0
Cross multiply to get;
0(y - 5) = -4.5(x - 1)
-4.5x = -4.5
x = 1
Equation of BF;
(y - 3)/(x - 5) = (1.5 - 3)/(-1 - 5)
(y - 3)/(x - 5) = -1.5/-6
(y - 3)/(x - 5) = 1/4
y - 3 = ¼(x - 5)
y = ¼(x - 5) + 3
