Recent content by Ilikebugs

153-144a<24b<183-144a 259<192a+24b<269 (259)-(183-144a)<192a<(269)-(153-144a) 66+144a<192a<116+144a 66<48a<116 Since a is an integer, a=2 Plugging in a into the 2 equations, -135<24b<-105 -125<24b<-115 If b is an integer, b=-5

a=2 b=-5 c=-2 d=5 f(3)=2(3)^3+(-5)(3)^2+(-2)(3)+5 =54-45-6+5 =8 Is this right?

How do I do that? Is a=2 and b=-5 correct?

51<f(4)=48a+8b<61 259<f(6)=192a+24b<269

-4- 2b- 4a f(0)=5 f(2)=8a+4b+2(-4- 2b- 4a)+5=-3 40<f(4)=64a+16b+4(-4- 2b- 4a)+5<50 240<f(6)=216a+36b+6(-4- 2b- 4a)+5<250 f(0)=5 f(2)=8a+4b+(-8- 4b- 8a)+5=-3 40<f(4)=64a+16b+(-16- 8b- 16a)+5<50 240<f(6)=216a+36b+(-24- 12b- 24a)+5<250 What do I do now?

f(0)=5 f(2)=8a+4b+2c+5=-3 40<f(4)=64a+16b+4c+5<50 240<f(6)=216a+36b+6c+5<250 Is this correct?

8a+4b+2c+5=-3 64a+16b+4c+5=45 216a+36b+6c+5=245 What do I do now?

I don't know, this is hard to do in my head. I got 8a^3+4b^2+2c+5=-3, 64a^3+16b^2+4c+5=45, and 216a^3+36b^2+6c+5=245

I don't know how to solve that

I know that d is 5, but I don't know how to use the inequalities to find f(3)

T1=13.5 T2=16 Trapezoid=height 8, b1= 6 b2=sqr(45) 1/2(6+sqr(45))*8= 24+4(sqr(45)) 43.5+4(sqr(45))? I got 60 through

(12*5)/2=30 ()/2=30 30+30=60?

I think I got 4 points of (1,0) (3,5) (9,8) (13,0) but I don't know how to get the area

The line of the first leg is y=-x The line of the 2nd leg is y=x+4 -x=x+4 x= -2 y=2 Points= 0,0 -2,2 0,4 2 sqr(2)+2 sqr(2)+4= 4+4 sqr(2) km?

Is there a way to determine where Bisector Valley is?