Answer:
Step-by-step explanation:
Since the diameter of Circle Q terminates on the circumference of the circle at (3,0) and (-4,0), the diameter lies on the x-axis.
The midpoint of the ends of the diameter gives the center of the circle.
The radius of this circle is the distance from the center to one of this points on the circumference.
The equation of the circle is given by 
where (a,b)=(-0.5,0) is the center and r=3.5
We substitute to get:
The standard form equation is
Newton’s second law for the system of two boxes in the x-direction is given by:
F - f
k = (m + M )a
F - µk (m + M )g = (m + M )a
⇒ a = (F/m + M) - µkg = (150N / (4kg+10kg)) - 0.6(9.8m/s² ) = 4.83m/s²
For the top box, we get fs = ma = 4kg x 4.83m/s² =19.3N
Find slope first:
m =y2-y1/x2-x1
m = 1--2/2--2
m = 3/4
Select a point, insert your slope, & put it into point slope form:
y-y1 =m(x-x1)
Your final answers are:
y-1=3/4(x-2)
OR
y+2 =3/4(x+2)
The answer is 16,031.
Hope this helps! ;)
Answer:
A'(4,-6) , B'(0,1), C'(-2,-2)
Step-by-step explanation:
From the given graph the coordinates of ΔABC area A (2,-2), B(-2,5) and C(-4,2)
If a translation is applied on ΔABC two units to the right and four units down to create ΔA'B'C'.
Then to find the coordinates of ΔA'B'C' will be we need to apply the translation rule
Now, 
and 
