Answer:
Yes, because the x's do not have more than one y.
Step-by-step explanation:
x y
0 0
5 10
6 12
10 20
(That's meant to be a table). Looking at the x and y table values, you can see that none of the x's have two y's. So, this relation is a function.
Answer:y=2x+1
Step-by-step explanation:
Part A:
3x^2-15x+20=0
ax^2+bx+c=0; a=3, b=-15, c=20
Radicand: R=b^2-4ac
R=(-15)^2-4(3)(20)
R=225-240
R=-15<0, then the equation has two imaginary solutions (no real)
Part B:
3x^2+5x-8=0
a=3, b=5, c=-8
R=(5)^2-4(3)(-8)
R=25+96
R=121>0, then the equation has two different real solutions:
x=[-b+-sqrt(R)] / (2a)
x=[-5+-sqrt(121)] / [2(3)]
x=(-5+-11) / 6
x1=(-5-11)/6=(-16)/6→x1=-8/3
x2=(-5+11)/6=6/6→x2=1
Solutions: x=-8/3 and x=1
I shose this method because I can get the solutions directly, and I don't have to guess the possible solutions
Answer: 
Step-by-step explanation:
