Answer:
Step-by-step explanation:
Start by factoring out a 5:
We need to find two integers that have a product of 12, and a sum of -7:
(-3)(-4)=12
-3-4=-7
We can split -7x into -3x and -4x
Factor each half separately:
![5[x(x-3)-4(x-3)]](https://tex.z-dn.net/?f=5%5Bx%28x-3%29-4%28x-3%29%5D)
Since x and -4 are both being multiplied by x-3, we can combine them:
Answer:
C
The line is from -3 to 3 and the circles are closed
The sum of any geometric sequence, (technically any finite set is a sequence, series are infinite) can be expressed as:
s(n)=a(1-r^n)/(1-r), a=initial term, r=common ratio, n=term number
Here you are given a=10 and r=1/5 so your equation is:
s(n)=10(1-(1/5)^n)/(1-1/5) let's simplify this a bit:
s(n)=10(1-(1/5)^n)/(4/5)
s(n)=12.5(1-(1/5)^n) so the sum of the first 5 terms is:
s(5)=12.5(1-(1/5)^5)
s(5)=12.496
as an improper fraction:
(125/10)(3124/3125)
390500/31240
1775/142
