Answer:
Third option
Step-by-step explanation:
We can't factor this so we need to use the quadratic formula which states that when ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a. However, we notice that b (which is 6) is even, so we can use the special quadratic formula which states that when ax² + bx + c = 0 and b is even, x = (-b' ± √(b'² - ac)) / a where b' = b / 2. In this case, a = 1, b' = 3 and c = 7 so:
x = (-3 ± √(3² - 1 * 7)) / 1 = -3 ± √2
Yes choice D is the correct answer. Nice job.
If you wanted, you can simplify the linear approximation L(x)
L(x) = f(a) + (x-a)*f'(a)
L(x) = f(-0.5) + (x-(-0.5))*f'(-0.5)
L(x) = 2 + (x+0.5)*4
L(x) = 2+4x+2
L(x) = 4x+4
Then plug in x = 0
L(x) = 4x+4
L(0) = 4(0)+4
L(0) = 4
But your method is a shorter route.
Answer:
It's B. on EtDtGtE
Step-by-step explanation:
A sideways opening parabola is in the form
, so we know from the process of elimination that it will either be b or c. Next we have to realize that if the parabola opens to the left it is a negative parabola, just like if a parabola opens upside down it is a negative parabola. So the one that has the negative out front is b.
0.1000 is the answer sorry if I'm wrong but I know it's timed by a thousand
