This is NOT an arithmetic sequence because there is no common difference. It is a geometric sequence because there is a common ratio. Meaning that each term is a constant ratio or multiple of the previous term. The recursive rule for this geometric sequence is:
a(n)=3*a(n-1), a(1)=3
Answer:
Distance between the points=5
Step-by-step explanation:
The distance between two points in coordinate geometry is measured by using the distance formula:
If we have given two points
![(x_1,y_1)\\\\(x_2,y_2)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29%5C%5C%5C%5C%28x_2%2Cy_2%29)
Distance formula= ![\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
For (0,0) and (4,3)
Distance:
![\sqrt{(4-0)^2+(3-0)^2} \\\\\sqrt{4^2+3^2}\\\\\sqrt{16+9}\\\\\sqrt{25}](https://tex.z-dn.net/?f=%5Csqrt%7B%284-0%29%5E2%2B%283-0%29%5E2%7D%20%5C%5C%5C%5C%5Csqrt%7B4%5E2%2B3%5E2%7D%5C%5C%5C%5C%5Csqrt%7B16%2B9%7D%5C%5C%5C%5C%5Csqrt%7B25%7D)
Or
Distance between the points=5
<u>Given</u>:
The given triangle is a similar triangle.
The length of the hypotenuse is 18 units.
The length of the leg is a.
The length of the part of the hypotenuse is 16 units.
We need to determine the proportion used to find the value of a.
<u>Proportion to find the value of a:</u>
We shall find the proportion to determine the value of a using the geometric mean leg rule.
Applying the leg rule, we have;
![$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Ctext%20%7B%20hypotenuse%20%7D%7D%7B%5Ctext%20%7B%20leg%20%7D%7D%3D%5Cfrac%7B%5Ctext%20%7B%20leg%20%7D%7D%7B%5Ctext%20%7B%20part%20%7D%7D%24)
Substituting the values of hypotenuse, leg and part, we get;
![\frac{18}{a}=\frac{a}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B18%7D%7Ba%7D%3D%5Cfrac%7Ba%7D%7B16%7D)
Thus, the proportion used to find the value of a is ![\frac{18}{a}=\frac{a}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B18%7D%7Ba%7D%3D%5Cfrac%7Ba%7D%7B16%7D)
Hence, Option D is the correct answer.
Here is the answer from the internet