Something funny is that the x value of the vertex lies directl in the middle of the x intercepts
so
we see the x intercepts or 0's at x=8 and 2
the average is x=5
so find f(5) to find the y value of the vertex
f(5)=(5-8)(5-2)
f(5)=(-3)(3)
f(5)=-9
vertex is at (5,-9)
the actual way the teacher wants is to expand then compltete the square to get into the form f(x)=a(x-h)^2+k where the vertex is (h,k)
but whatever
verrtex is at (5,-9)
The solution has to be a value of x. Since you already have x, you just have to plug in the value and see what fits. In this case thats the third from the top:
5^2 - 10*5 + 25 = 0
25 - 50 + 25 = 0
0 = 0
Hope it helps!
F+g(x) = f(x) + g(x) = 3x-1 + x+2 = 4x+1
Suppose m∠1 = x degree
m∠2 = 17 x degree
As angle 1 & 2 are supplementary angles so
m∠1 +m∠2 =180 degree...... eq 1
Substituting the values of angle 1 & 2 in eq 1, we get
x +17x =180
18x=180
x= 180/18 =10 degree
17 x= 170 degree
m∠1 = 10 degree m∠2 = 170 degree.
A prime number is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers
Step-by-step explanation:
