B1 = 2
b2 = (b1)^2 + 1 = 2^2 + 1 = 5
b3 = (b2)^2 + 1 = 5^2 + 1 = 26
b4 = (b3)^2 + 1 = 26^2 + 1 = 676+1=<span>677</span>
We are given with
dimensions of wheelbarrow = 2ft x 3ft x 1.5 ft
dimensions of truck = 11 ft x 8 ft x 6 ft
The volume of the wheelbarrow is 2(3)(1.5) = 9 ft3
The volume of the truck is 11(8)(6) = 528 ft3
70% of this is
528 ft3 (0.7) = 369.6 ft3
Dividing this by the volume of the wheelbarrow
369.6 ft3 / 9 ft3 = 41.07
He would need to use the wheelbarrow 42 times.
Their sum can be greater than two. There is nothing that states that they couldn't be. All it states it that the two factions sums are less than 1. I hoped this helps
Answer:
C is the answer
Step-by-step explanation:
hope that helped
Answer:
11
Step-by-step explanation:
2x+3
2(4)+3
8+3
11
