Whitespy001

yes because i understand false the question :p Im sorry :( 

(3k – 100)(k + 35) < 0. 

You want every (...) to have different sign so 

the first () if k>33.333  ()>0 if k<33.333 ()<0                κ              -35           33.333    

the second () if k>-35 ()>0 if k<-35 ()<0                     (1)      -              -                    + 

                                                                                   (2)      -             +                    + 

                                                                              Finally    +              -                    + 

You want  (3k – 100)(k + 35) be <0 so       -35

                          κ  between  -35 and 33.333

Yes thank you i know i find it but becaise the result is 

10e^3 - 5e^2 -49e/2 + 11e from ministry i think i was false but im not .Thank you!

no its e^(x-1) and i write it e^x/e coz i cant write it there like this

Thank you i find the same too but the result say its :

10e^3 - 5e^2 -49e/2 + 11e and i cant find it but is not false because it the result from ministry ... I dont know...Thank you after all!!!

yes (3-3x)= 3(1-x) so if you multiply 3 for 1 and -x you will have the 1st 

tanx=5/10 <=>

tanx=1/2 <=>

tanx=0.5 <=>

x=27

cos35=A/15ft <=>

Α=15*cos35 <=>

A=12.287280664335 so A=12ft

1 year-> 12 months so 3 years->3* 12=36 months 

300*36=10800

so £10800 in 3 years

(12*9)/2

108/2=54

300 is 150%

\(1+1=\sqrt[35]{34359738368}\)

no it is 1 and 1 is not prime