Answer:
C. <em>c</em> is less than zero
Step-by-step explanation:
The parent radical function y=x^(1/n) has its point of inflection at the origin. The graph shows that point of inflection has been translated left and down.
<h3>Function transformation</h3>
The transformation of the parent function y=x^(1/n) into the function ...
f(x) = a(x +k)^(1/n) +c
represents the following transformations:
- vertical scaling by a factor of 'a'
- left shift by k units
- up shift by c units
<h3>Application</h3>
The location of the inflection point at (-3, -4) indicates it has been shifted left 3 units, and down 4 units. In the transformed function equation, this means ...
The graph says the value of c is less than zero.
__
<em>Additional comment</em>
Apparently, the value of 'a' is 2, and the value of n is 3. The equation of the graph seems to be ...
f(x) = 2(x +3)^(1/3) -4
<span>t=(512/216)^(1/2)
t=((8)(3)^(1/2))/9
t is about 1.539600718 hours
I don't believe the storm will be twice as long because the number multiplied by itself three times will give a much, much higher number than what was your original diameter.</span>
Answer:
Z = a - 2i - 3ib
Step-by-step explanation:
The complex conjugate has the expression with the opposite sign for the imaginary part.
Complex conjugate of Z=a + 2i + 3ib is,
Z = a - 2i - 3ib
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:
The answer is 1.b 2. d 3. c 4. c