If the weight of the bottom box is b then the weight of a stack of four boxes is b + b/2 + b/4 + b/8
= 8b/8 + 4b/8 + 2b/8 + 1b/8
= 15/8 b
A stack must weigh less than 100 pounds
15/8 b < 100
b < 100 x 8/15 = 53 1/3 pounds
If boxes have a minimum weight of 1 pound then
for stacks that contain 4 boxes the range of b is [8, 53 1/3)
and if stacks contain 1 to 4 boxes the range of b is [1, 53 1/3)
If the only restriction is a maximum of 100 then the range of b is [0, 53 1/3)
Answer:
Other Side (B) = 9.45 inches
Step-by-step explanation:
Given:
Hypotenuse = 10 inches
Other Side (A) = 3.25 inches
Find:
Other Side (B) = ?
Computation:
According to Pythagoras theorem:
Other Side (B) = 9.45 inches
Let student tickets be s and adult tickets be a. The number of tickets sold of both adult and student then is s + a = 396. If each student ticket costs $3, then we represent the money equation by tacking the dollar amount onto the ticket. 3s is the cost of one student ticket. 4a is the cost of an adult ticket. The total money from the sales of both is 4a + 3s = 1385. We now have a system of equations we can solve for a and s. If s+a=396, then s = 396-a. We will sub that into the second equation to get 4a + 3(396-a) = 1385. Distributing we have 4a+1188-3a=1385. a = 197. That means there were 197 adult tickets sold. If s + a = 396, then s + 197 = 396 and s = 199. 197 adult tickets and 199 student tickets. There you go!
Answer:
just wondering if you still need help with this
Step-by-step explanation:
why is this on my timeline?
Answer:
The third answer
Step-by-step explanation:
Remember power over root and the 2 would not be under the root.
