Answer:
I dont know the answer but all I know is that dude broke as heck
Isn’t still 50%? I think I’m wrong someone tell me if I’m wrong
Area de la base es un hexagono: Nota: asumiendo que 39 cm son el lado de un hexagono. Menuda pizza.
A = 3*sqrt(3)/2*a^2 = 3*sqrt(3)/2*39^2 = 3951.67391746839972 cm^2
Hay dos base y tapa, doble:
7903.34783493679944 cm^2
Luego los lados verticales:
6 * 39 * 4.7 = 1099.8 cm^2
TOtal = 9003.14783 = 9003.147 cm^2 = 9003.15 cm^2
If it helps you any



so.. check coefficients C and B for the horizontal shift
Answer: -3/4x -1
Step-by-step explanation: