So 2 real solutions
just write 2 factors and expand
(x-3)(x-4)=0
x^2-7x+12=0
discriminant is when you have b^2-4ac
so subsitute
ax^2+bx+c=0
a=1
b=-7
c=12
-7^2-4(1)(12)
49-48=1
discriminant=1
one real solution would be
(x-1)(x-1)=0
x^2-2x+1=0
discriminant=-2^2-4(1)(1)=4-4=0
discriminant=0
no real solutions has the discriminant less than zero
let's say that b=1 and c=20 and a=1
basically, we need a set of numbers where b^2 is less than 4acso
1^2-4(1)(20)=1-80=-79
then we subsitute
x^2+x+20=0
7494 thousand is the answer
Answer:
a is true b is false c is false
Step-by-step explanation:
I used Desmos to graph it but here are the coordinates used to graph it in the pictures below.
Answer: Choice D
In two games, the team lost to the opponent by 1 goal.
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Explanation:
Negative scores indicate the team lost, and the absolute value of those values represent how much of a loss.
So a difference of -1 means the team lost by 1 point.
We have 2 dots over this value, so there are 2 occasions where the team lost to the opponent by 1 goal.
An example would be that say the team scored 3 goals and the opponent scored 4 goals. So we have a differential of 3-4 = -1. The order is important because we would <u>not</u> say 4-3 = 1.
