Answer:
x+1 , x<-1
x²-bx-4 , -1≤x≤4
1/2 x +a
substitute x=-1 , continuous everywhere
x+1=x²-bx-4 when x=-1
-1+1=-1²-b-4
<h2>1+b-4=0 ⇒ +b-3 ⇒ b=3</h2>
then for :
x²-bx-4=1/2 x +a for x=4
4²-4b-4=1/2x+a
16-4(3)-4=4/2+a
16-12-4-2=a
<h2>a=-2</h2>
the equations will be :
x+1
x²-3x-4
1/2x-2
I hope this help
Answer:
The graph in the attached figure
Step-by-step explanation:
we have
-----> equation A
-----> equation B
we know that
The solution of the system of equations is the intersection point both graphs
In this problem
The intersection point is 
therefore
The solution is the point
using a graphing tool
see the attached figure
Uhh..not in a mood for a meet, k? Maybe later hun~~
Answer:
y = 6 or 8
Step-by-step explanation:
1. Subtract the constant:
y^2 -14y = -48
2. Add the square of half the y-coefficient:
y^2 -14y +49 = -48 +49
Write as a square, if you like:
(y -7)^2 = 1
3. Take the square root:
y -7 = ±√1 = ±1
4. Add the opposite of the constant on the left:
y = 7 ±1 = 6 or 8
The solution is y = 6 or y = 8.
Answer:
School A
Step-by-step explanation:
School A has a ratio of 18/225 = 2/25 staff to students
School B has a ratio of 23/322 = 2/28 = 1/14 staff to students
Assuming that a better ratio means more staff per students School A has a better ratio
