How many real solutions will y=x^2 and y=-2x have? what are the solutions?
2 answers:
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Answer: 1/3
Step-by-step explanation: The probability is given by the number of times the event could happen (in this case get an "e"), over the number of all the possibilities (in this case, the six letters of the word)
Therefore, probability
Factor 16 and see the - factor pairs and add them, see which add to -12
-1-16=-17 nope
-2-8=-10 nope
-4-4=-8
hmm
we will solve with math
xy=16
x+y=-12
minus x both sides
y=-12-x
sub for y in other equation
x(-12-x)=16
-12x-x^2=16
add 12x+x^2 both sides
0=x^2+12x+16
use quadratic formula
if you have
ax^2+bx+c=0
x=
0=x^2+12x+16
a=1
b=12
c=16
x=
x=
x=
x=
x=
x=
or 
aprox
x=-1.52786 or -10.4721
those are the numbers
the numbes are -1.52786 or -10.4721
-1-16=-17 nope
-2-8=-10 nope
-4-4=-8
hmm
we will solve with math
xy=16
x+y=-12
minus x both sides
y=-12-x
sub for y in other equation
x(-12-x)=16
-12x-x^2=16
add 12x+x^2 both sides
0=x^2+12x+16
use quadratic formula
if you have
ax^2+bx+c=0
x=
0=x^2+12x+16
a=1
b=12
c=16
x=
x=
x=
x=
x=
x=
aprox
x=-1.52786 or -10.4721
those are the numbers
the numbes are -1.52786 or -10.4721