Slope of the line J
(6-1) / (-3-(-6))
5 / 3
so
m = 5/3
If this line passes through (15,-1)
y = mx + c
-1 = (5/3)*15 + c
-1 = 25 + c
c = -26
equation is
y = (5/3)x - 26
F(x) =x² - 10x, f⁻¹(x) =?
1st find the missing square of x²-10x, ==> (x-5)² - 25
y= (x-5)² - 25; replace x by y and v0ce versa: x= (y-5)² -25
or
(y-5)² = x+25
y-5 = √(x+25) and y = √(x+25) -5
Domain = {x∈R: X>= 25} AND Range ={y∈R: y>= 5}
Answer:
maybe rhombus
Step-by-step explanation:
don't take my word for it
Answer:
I think its 40* because 25% of a 160* circle is 40. idk
Step-by-step explanation:
