Answer:
Exponential
Step-by-step explanation:
By visual inspection the graph generated by the points plotted is an exponential graph as the graph curves upward. The graph is also continous and differs from either a decreasing or increasing Linear graph, which shows a straight best of fit pattern. Hence, the graph most closely represents an exponential graph from visual examination.
Answer:
Points A and C are on the unit circle
Step-by-step explanation:
The unit circle is a circle of radius 1, that is, all the points that are inside the circle have the sum of the squares of its coordinates that are at most 1. Here, we have to test this for each option.
In options B and D, the coordinates 13/7 and 4/3, respectively, means that this sum will be larger than 1, and this points will not be on the unit circle. Now we text for options A and C.
A. ( 5/13, 12/13)
![\sqrt{(\frac{5}{13})^2+(\frac{12}{13})^2} = \sqrt{\frac{25}{169}}+\sqrt{\frac{144}{169}} = \sqrt{\frac{169}{169}} = \sqrt{1} = 1](https://tex.z-dn.net/?f=%5Csqrt%7B%28%5Cfrac%7B5%7D%7B13%7D%29%5E2%2B%28%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B25%7D%7B169%7D%7D%2B%5Csqrt%7B%5Cfrac%7B144%7D%7B169%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B169%7D%7B169%7D%7D%20%3D%20%5Csqrt%7B1%7D%20%3D%201)
Since the sum of the squares is 1, the point is on the unit circle.
C. (1/3, 2/3)
![\sqrt{(\frac{1}{3})^2+(\frac{2}{3})^2} = \sqrt{\frac{1}{9}}+\sqrt{\frac{4}{9}} = \sqrt{\frac{5}{9}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28%5Cfrac%7B1%7D%7B3%7D%29%5E2%2B%28%5Cfrac%7B2%7D%7B3%7D%29%5E2%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7B9%7D%7D%2B%5Csqrt%7B%5Cfrac%7B4%7D%7B9%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B5%7D%7B9%7D%7D)
Since the sum of the squares is less than 1, the point is on the unit circle.
Answer:
{x:x>6}
Step-by-step explanation:
x-3>3
Add 3 to each side
x-3+3>3+3
x>6
{x:x>6}
Answer:
D
Step-by-step explanation: