Answer:
between 0&1 and 2&3
Step-by-step explanation:
(Assuming that f(x) = 2x^2 - 5x + 1)
To solve this question we just need to find the zeros of the function f(x), and we do that by making f(x) = 0 and finding the values of x:
2x^2 - 5x + 1 = 0
Now we can use Bhaskara's formula:
Delta = b^2 - 4ac = 25 - 4*2 = 17
sqrt(Delta) = 4.12
x1 = (5 + 4.12) / 4 = 2.28
x2 = (5 - 4.12) / 4 = 0.22
the zero x1 = 2.28 is between 2 and 3, and the zero x2 = 0.22 is between 0 and 1, so the answer is 'between 0&1 and 2&3'
As the options A, B, C and D are not clear in the question text, you can just mark the option that has this result.
Answer:
The desired quotient is 25/12.
Step-by-step explanation:
Rewrite 1 2/3 as an improper fraction: 5/3.
Then divide this 5/3 by 4/5:
5
------ ÷ 4/5
3
It's easier (but completely correct) to invert the divisor (4/5) and then multiply 5/3 by (5/4):
5 5 25
----- * ----- = ------
3 4 12
