$a_1=0.5\\a_2=-0.1\\a_3=0.02\\a_4=-0.004\\a_5=0.0008\\\\a_n=a_1\cdot q^{n-1}\\\\q=a_2:a_1\to q=-0.1:0.5=-0.2\\\\a_n=0.5\cdot(-0.2)^{n-1}=0.5\cdot(-0,2)^n\cdot(-0.2)^{-1}\\\\=0.5\cdot(-0.2)^n\cdot(-5)=-2.5\cdot(-0.2)^n\\\\\boxed{a_n=-2.5\cdot(-0.2)^n}$

6 0

3 years ago

Factor 15x + 35 using the GCF.

stellarik [79]

The answer is A. 5(3x+7)

4 0

2 years ago

if an external point of a circle is at a distance equal to the diameter of the circle from the center of the circle, the length

jeka57 [31]

Let the radius of the circle be r. Then the line from the external point through the center of the circle which extends to the far point on the circle has length 3r .By the tangent - secant theorem

t^2 = 3r * r = 3r^2 ( where t is the length of the tangent).

So t = √(3r^2) = √3r answer.

7 0

2 years ago

In the data set below, what are the lower quartile, the median, and the upper quartile?

ivolga24 [154]

Answer:

its 55

Step-by-step explanation:

6 0

2 years ago