In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
We have that discriminant = 3
If
b^2 - 4ac > 0, then the roots are real.
If
b^2 - 4ac < 0 then the roots are imaginary
<span>In
this problem b^2 - 4ac > 0 3 > 0 </span>
then
the two roots must be real
-5║4y-11║-3=12
-5║4y-11║-3 +3=12+3
- 5║4y-11║=15
-5/5║4y-11║=15/-5
║4y-11║=-3
4y-11=-3 -5║4y-11║-3=12
4y-11+11=-3+11 -5║4(2)-11║-3=12
4y=8 -5║-3║-3=12
4y/4=8/4 -5 (3)-3=12
y=2 -18≠12
4y-11=3 -5║4y-11║-3=12
4y-11+11=3+11 -5║4(7/2)-11║-3=12
4y=14 -5(3)-3=12
4y/4=14/4 -18≠ 12
y=7/2
No solutions
Answer:
1 : 64
Step-by-step explanation:
Model:
1/2 in * 1/6 in = 1/12 in^2
Real:
1/3 ft * 1/9 ft = 1/3 * 12 in * 1/9 * 12 in = 16/3 in^2
model : real
1/12 / 16/3 = 1/12 * 3/16 = 3/192 = 1/64
model : real = 1 : 64
F(3k)=3k +7 Equals 9k+7. So 9k+7 would be the Answer.
