Answer:
a) see the plots below
b) f(x) is exponential; g(x) is linear (see below for explanation)
c) the function values are never equal
Step-by-step explanation:
a) a graph of the two function values is attached
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b) Adjacent values of f(x) have a common ratio of 3, so f(x) is exponential (with a base of 3). Adjacent values of g(x) have a common difference of 2, so g(x) is linear (with a slope of 2).
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c) At x ≥ 1, the slope of f(x) is greater than the slope of g(x), and the value of f(x) is greater than the value of g(x), so the curves can never cross for x > 1. Similarly, for x ≤ 0, the slope of f(x) is less than the slope of g(x). Once again, f(0) is greater than g(0), so the curves can never cross.
In the region between x=0 and x=1, f(x) remains greater than g(x). The smallest difference is about 0.73, near x = 0.545, where the slopes of the two functions are equal.
Answer:
8x -16y = 24
Step-by-step explanation:
8x - 16y= -14
A parallel line (i.e. same slope) will be: 8x - 16y= c
We have to find the value of "c".
If it passes through (x, y) = (3, 0) ⇒ 24 - 0 = c,
so equation is 8x -16y = 24
Answer:
I think it is wrong because it is unlike term
The variables are: x, y
The constant is: 4
The coefficients are: 5, 3
The exponent is: y^2
The general equation for a circle,
, falls out of the Pythagorean Theorem, which states that the square of the hypotenuse of a right triangle is always equal to the sum of the squares of its legs (you might have seen this fact written like
, where <em>a </em>and <em>b</em> are the legs of a right triangle and <em>c </em>is its hypotenuse. When we fix <em /><em>c</em> in place and let <em>a </em>and <em>b </em>vary (in a sense, at least; their values are still dependent on <em>c</em>), the shape swept out by all of those possible triangles is a circle - a shape defined by having all of its points equidistant from some center.
How do we modify this equation to shift the circle and change its radius, then? Well, if we want to change the radius, we simply have to change the hypotenuse of the triangle that's sweeping out the circle in the first place. The default for a circle is 1, but we're looking for a radius of 6, so our equation, in line with Pythagorus's, would look like
, or
.
Shifting the center of the circle is a bit of a longer story, but - at first counterintuitively - you can move a circle's center to the point (a,b) by altering the x and y portions of the equation to read:
