<span>A. cos(x) = cos(-x) [correct, since cos(x) is an even function]
B. Since the cosine function is even, reflection over the x-axis [y-axis] does not change the graph. [false]
C. cos(x) = -cos(x) [ false]
D. The cosine function is odd [even], so it is symmetrical across the origin. [false]</span>
Answer:
Quadrant III
Step-by-step explanation:
The attached picture shows graph of 4 such linear functions with the conditions given in the problem. ALL of them DO NOT pass through Quadrant III.
The graphs shown are of the functions:




<em>So, any linear function of the form
with
and
does not pass through Quadrant III. Answer choice 3 is correct.</em>
The expression for the amount of money, in dollars, Teri has is x-2.
Since Jennie has x dollars while Teri has $6 less than does Jennie, this means that Teri will have: x - 6
Since Teri does not spend any money and earns $4, the the total amount that Teri has will be:
= x - 6 + 4
= x - 2
Therefore, Teri has x-2.
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To find the gradient of the tangent, we must first differentiate the function.

The gradient at x = 0 is given by evaluating f'(0).

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.
The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

So the point (0, -1) lies on the tangent. Substituting into the tangent equation: