Let's do
Release k for some while
If
So
So vertical asymptote is at origin now
It mentioned that it's at x=-5 so we need to change x
- Vertical asymptote at x=-5
Now
- for k=0 horizontal asymptote at origin
But it's given
Same put y=12 in place of k
Graph attached for verification
Answer:
1/5
Step-by-step explanation:
first write the formula of average speed
- 2) then equate the values of given question
- 3) you will find the solution
Answer:
180
Step-by-step explanation:
all I did was multiply
i don’t know hope this is correct if not let me know so I can fix my mistake
Answer:
A: This polynomial has a degree of 2 , so the equation 12x2+5x−2=0 has two or fewer roots.
B: The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots.
Step-by-step explanation:
f(x) = 12x^2 + 5x - 2.
Since this is a quadratic equation, or a polynomial of second degree, one can easily conclude that this equation will have at most 2 roots. At most 2 roots mean that the function can have either 2 roots at maximum or less than 2 roots. Therefore, in the A category, 2nd option is the correct answer (This polynomial has a degree of 2 , so the equation 12x^2 + 5x − 2 = 0 has two or fewer roots).
To find the roots of f(x), set f(x) = 0. Therefore:
12x^2 + 5x - 2 = 0. Solving the question using the mid term breaking method shows that 12*2=24. The factors of 24 whose difference is 5 are 8 and 3. Therefore:
12x^2 + 8x - 3x - 2 = 0.
4x(3x + 2) -1(3x+2) = 0.
(4x-1)(3x+2) = 0.
4x-1 = 0 or 3x+2 = 0.
x = 1/4 or x = -2/3.
It can be seen that f(x) has two distinct real roots. Therefore, in the B category, 1st Option is the correct answer (The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots)!!!
