x²-13x+42=0
Answer:
x1=7,
x2=6
Step-by-step explanation:
x²-13x+42=0
a=1, b=-13, c=42
x1,2=(-b+-sqrt(b^2-4ac)) /2a
x1,2=(-(-13)+-sqrt((-13)^2-4*1*42))/2*1
x1,2=(13+-sqrt (169-168))/2
x1,2=(13+-sqrt(1))/2
x1,2=(13+-1)/2
x1=(13+1)/2, x2=(13-1)/2
x1=14/2, x2=12/2
x1=7, x2 =6
What is the value of y?
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I believe it is but I'm not sure.
Answer:
- 16/5
Step-by-step explanation:
We have point with coordinates (-6, y) which lies ( or belong) on the graph of the straight line 5y=2x-4
When we replace x= -6 in given equation we get 5y = 2·(-6)-4 => 5y= -12-4 =>
5y= - 16 => y= - 16/5
Good luck!!!
x² + y² = 16