$Method\ 2:\\\\a_n=a_{n-1}+\dfrac{1}{2}\\\\\text{It's an arithmetic sequencion where}\ a_1=-\dfrac{3}{2}\ \text{and common difference}\ d=\dfrac{1}{2}.\\\\\text{The formula of n-th term is}\ a_n=a_1+(n-1)d.\ \text{Substitute}\\\\a_n=-\dfrac{3}{2}+(n-1)\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}+\dfrac{1}{2}n-\dfrac{1}{2}=-\dfrac{4}{2}+\dfrac{1}{2}n=\dfrac{1}{2}n-2\\\\a_9=\dfrac{1}{2}(9)-2=4.5-2=2.5=\dfrac{5}{2}$ $a_1=-\dfrac{3}{2}\\\\a_n=a_{n-1}+\dfrac{1}{2}\\\\Method\ 1:\\\\a_2=a_1+\dfrac{1}{2}\to a_2=-\dfrac{3}{2}+\dfrac{1}{2}=-\dfrac{2}{2}\\\\a_3=a_2+\dfraC{1}{2}\to a_3=-\dfrac{2}{2}+\dfrac{1}{2}=-\dfrac{1}{2}\\\\a_4=a_3+\dfrac{1}{2}\to a_4=-\dfrac{1}{2}+\dfrac{1}{2}=0\\\\a_5=a_4+\dfrac{1}{2}\to a_5=0+\dfrac{1}{2}=\dfrac{1}{2}\\\\a_6=a_5+\dfrac{1}{2}\to a_6=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{2}\\\\a_7=a_6+\dfrac{1}{2}\to a_7=\dfrac{2}{2}+\dfrac{1}{2}=\dfrac{3}{2}$

$a_8=a_7+\dfrac{1}{2}\to a_8=\dfrac{3}{2}+\dfrac{1}{2}=\dfrac{4}{2}\\\\a_9=a_8+\dfraC{1}{2}\to a_9=\dfrac{4}{2}+\dfrac{1}{2}=\dfrac{5}{2}$

5 0

4 years ago

Identify the type of transformation given the rule N(x,y) = (-x,y)

denis23 [38]

Hello!

If the domain of the function is negative, the translation occurring has to be a reflection over the x-axis or y-axis.

Since the translation is affecting the x-values, it has to be a reflection over the y-axis. Similarly, if the translation affects the y-values, it translates over the x-axis.

Therefore, the type of transformation given by the rule is a reflection over the y-axis.

5 0

3 years ago

four congruent circles are packed into a square with the side length of 12 units so that each of the circles in Tang it to two s

nydimaria [60]

Answer:

Do you still need help with this problem?

8 0

3 years ago