log(x+3)(2x+3)+log( x+3)( x+5) =2
logx(2x+3)+3(2x+3 )+log x(x+5)+3(x+5)

Answer:
5( 4w -3)
Step-by-step explanation:
just pulled out a 5
Step-by-step explanation:
hope this image will help u
I have solved this myself
If you mean by a(t)=2t+9 is the acceleration function then we integrate...
v(t)=2t^2/2+9t+c where c is vo and we are told that that is -7 so
v(t)=t^2+9t-7 integrating again we get:
s(t)=t^3/3+9t^2/2-7t+c where c is the initial position which is 8 so
s(t)=t^3/3+9t^2/2-7t+8 which neatened up as
s(t)=(2t^3+27t^2-42t+48)/6