Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle
CD is called the middle base and is equal to :(AB + EF)/2
CD = <span>(AB + EF)/2
4x = [(x</span>²+10) + (8-x)]/2
8x = x²+10+8-x
8x= x²-x+18
x²-9x+18 = 0
x' = [-b+√(b²-4.a.c)]/2a & x" = [-b-√(b<span>²-4.a.c)]/2a
x'= 6 and x" = 3, So we have2 values of x</span>
A parabola, a graph of a quadratic function, cannot have a maximum vertex and a minimum vertex at the same time because of the shape of the graph. A parabola is a u-shaped graph. The vertex of the parabola is the point where the u changes direction; if it was increasing, it starts to decrease, and if it was decreasing, it starts to increase. Since a parabola only changes direction once, there will either be a minimum or a maximum, not both.
Answer:
A
Step-by-step explanation:
let's try A. 3(3)+2=9+2=11
11=11
So A is right
Answer: 103
Step-by-step explanation:
8 x 6.5 = 52,
52 + 8 = 60 (pints in total)
there are 2 cups in 1 pint so 60 x 2 = 120
120 - 17 = 103
