This quadratic should be solved using the quadratic formula.
The quadratic formula is
(-b +/- sqrt(b2 - 4ac)) / (2a) (Side note: The +/- means that you use the plus sign once, then recalculate again using the minus sign. You should get two answers this way.)
The numbers that we plug into the letters can be found from the general form of a quadratic equation
ax2 + bx + c
In your quadratic a = 3, b = 1, and c = -5. I'll let you try and solve the rest on your own. Whatever your answer is, round it off to two decimal places.
3 (had to feel the space with words)
No. Writing it as a biconditional would mean that "If a number is odd, then it is a prime number" is also true. The original statement is true, but the converse is false.
Answerhold on I got you
Step-by-step explanation:
Answer:
(n^3 + 4 n + 4487)/(n + 1)
Step-by-step explanation:
Simplify the following:
(n^3 + 4 n - 2 + 67^2)/(n + 1)
| | 6 | 7
× | | 6 | 7
| 4 | 6 | 9
4 | 0 | 2 | 0
4 | 4 | 8 | 9:
(n^3 + 4 n - 2 + 4489)/(n + 1)
Grouping like terms, n^3 + 4 n - 2 + 4489 = n^3 + 4 n + (4489 - 2):
(n^3 + 4 n + (4489 - 2))/(n + 1)
4489 - 2 = 4487:
Answer: (n^3 + 4 n + 4487)/(n + 1)
