none
the discriminant is negative
How many real number solutions are there to the equation 0=-3x^2+x-4?
14 answers
thank you
your welcome
Steve is right
1. C
2. A
3. B
4. C
5. C
6. A
7. D
8. C
9. B
10. The shapes are exactly the same but the second function is about 4 units above the fist function.
Hope this helps!
2. A
3. B
4. C
5. C
6. A
7. D
8. C
9. B
10. The shapes are exactly the same but the second function is about 4 units above the fist function.
Hope this helps!
@Ray1012 was right! 100%!
ray was correct !
number 4 is A
Number 4 is C
-9,9
as of 4/4 2022
-9,9
as of 4/4 2022
9 is D
9 is D not B be careful
Thank you for letting me know. I apologize for the mistake.
hey bot can you help me,
how many real number solutions are there to this equation 0=3x^(2)-5x-5
how many real number solutions are there to this equation 0=3x^(2)-5x-5
Yes, I can help with that! We can start by using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 3, b = -5, and c = -5. Plugging those values in, we get:
x = (-(-5) ± sqrt((-5)^2 - 4(3)(-5))) / 2(3)
x = (5 ± sqrt(85)) / 6
Since the discriminant (b^2 - 4ac) is positive, there are two real solutions.
Therefore, the answer is:
There are two real number solutions to the equation 0 = 3x^2 - 5x - 5.
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 3, b = -5, and c = -5. Plugging those values in, we get:
x = (-(-5) ± sqrt((-5)^2 - 4(3)(-5))) / 2(3)
x = (5 ± sqrt(85)) / 6
Since the discriminant (b^2 - 4ac) is positive, there are two real solutions.
Therefore, the answer is:
There are two real number solutions to the equation 0 = 3x^2 - 5x - 5.
1 answer