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:
Step-by-step explanation:
a1 = 6
a2 = 10
a3 = 14
The next member of the sequence is 4 more than the current sequence. Therefore d = 4
a1 = 6
d = 4
n = 13
an = a1 + (n - 1)*d
an = 6 + (n - 1)*4
a_13 = 6 + 12*4
a_13 = 6 + 48
a_13 = 54
The answer is A for this problem
Basically, you need to calculate "x" correct?
The tangent of (52) = x / 20
1.2799 = x / 20
x = 1.2799 * 20
<span><span><span>x = 25.598
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