Answer:
A
Step-by-step explanation:
In a function, each x value must have a singular y value. This means that one x-value (the values on the left) may not have a line pointing to 2 different y-values (the values on the right). In the first relation, the x-value of 6 points to both 4 and 6. This means that the value repeats, and in a function x-values cannot repeat.
However, y-values can repeat. So, situations like that seen in D, where both the x-values of 2 and 4 equal the y-value of 6 are fine.
To solve for this, we need to find for the value of x
when the 1st derivative of the equation is equal to zero (or at the
extrema point).
So what we have to do first is to derive the given
equation:
f (x) = x^2 + 4 x – 31
Taking the first derivative f’ (x):
f’ (x) = 2 x + 4
Setting f’ (x) = 0 and find for x:
2 x + 4 = 0
x = - 2
Therefore the value of a is:
a = f (-2)
a = (-2)^2 + 4 (-2) – 31
a = 4 – 8 – 31
a = - 35
Answer:
Brainliest pls I never had brain before
Step-by-step explanation:
Just substitute r with 357 and solve.
= 1/7(357) + 3(1/7(357))
= 51 + 3(51)
= 51 + 153
= 204
Therefore, Eliza read 204 in those 3 days.
Best of Luck!
The antiderivate of 5x is 5x^2 / 2 .
