<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u> </u>√15 units
Step-by-step explanation:
Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .
As we know ,
Distance(D) = √(x^1-x^2) +(y^1-y^2)
Now,
D= √(x^1-x^2) +(y^1-y^2)
= √(6-1) +(1+9)
= √5+10
= √15 units
: Therefore the distance between (6,1) and (1,-9) is √15 units.
The answer is
B. -2 and 3
Step-by-step explanation :
Given the expression
x²- 5x + 6 = 0
Firstly we need to find two numbers that when multiplied will give us the constant term 6, and when added will give us the 2nd term 5
These numbers are 3 and 2
Substituting 2x +3x for 5x
x²- (2x+3x)+ 6 = 0
x²-2x-3x+6=0
(x²-2x)-(3x+6)=0
Factoring we have
x(x-2)-3(x-2)=0
x-3=0, x-2=0
x=3, x= 2
