11+b-y
Hope this helps
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Answer:
are zeroes of given quadratic equation.
Step-by-step explanation:
We have been a quadratic equation:

We need to find the zeroes of quadratic equation
We have a formula to find zeroes of a quadratic equation:

General form of quadratic equation is 
On comparing general equation with b given equation we get
a=2,b=-10,c=-3
On substituting the values in formula we get


Now substituting D in
we get




Therefore, 
Answer:
50 dimes and 29 nickels
Step-by-step explanation:
Answer:
but its right
Step-by-step explanation:
Step-by-step explanation:
vector AB(3-(-6); 5-7)
vector AB(9;-2)
AB=
=
M is the midpoint of AB
we have B(-5;10) and M(1;7)
let A(x;y)
(x-5)/2 = 1 ⇒ x-5 = 2⇒ x = 7
(10=y)/2 = 7⇒ 10+y = 14 ⇒y= 4
so : A(7;4)
the center of the circle is the midponit of the line joining both ends of the diameter
let A(x;y) be the other end
(-2+x)/2 = 2 ⇒ -2+x = 4⇒ x= 6
(5=y) = -1 ⇒ 5+y = -2 ⇒ y= -7
so the coordinates of the other end are (6; -7)
A,B and C are collinear such as AB=BC so b is the midpoint of AC
(-5+1)/2 = y ⇒ y = -4/2 ⇒ y = -2
((-3=x)/2 = 7 ⇒ -3+x = 14 ⇒ x = 17
so x= 17 and y = -2