Answer:
See Explanation
Step-by-step explanation:
<em>The question is incomplete as what is required of the question is not stated.</em>
<em>However, since the question is only limited to distance, a likely question could be to calculate the distance from Bayville to Colleyville.</em>
Represent the distance from Atlanta to Colleyville with AC
Represent the distance from Atlanta to Bayville with AB
Represent the distance from Bayville to Colleyville with BC
So, we have that:
The relationship between AB, AC and BC is:
Make BC the subject of formula:
Convert fraction to decimal
<em>Hence, the distance from Bayville to Colleyville is 14.8 miles</em>
Let's take a look at D:
<span>D) y = (x-1)^2 - 16 Compare this to
y = (x-h)^2 + k This is the std. equation of a parabola in vertex form.
You can see, by comparison, that h=1 and k= -16; these are the coordinates of the vertex, clearly shown in the diagram.
Since the coefficient of (x-h)^2 is +1, the graph opens upward (which the given graph confirms), and is neither compressed nor stretched vertically.</span>
Tex]y= \frac{2}{3}x-5 [/tex]
Y = 4x + 1....slope here is 4. A parallel line will have the same slope
y = mx + b
slope(m) = 4
(3,9)...x = 3 and y = 9
now we sub and find b, the y int
9 = 4(3) + b
9 = 12 + b
9 - 12 = b
-3 = b
so ur parallel equation is : y = 4x - 3
