Prob 1.

F =120, t =10 put these values in the given equation
F = 120 ( 2 ^ -at ) + 60
Solving this for the above values will give you, a = 1/10

now when t = 30, F = ?
put a = 1/10, t = 30 in F = 120 ( 2 ^ -at ) + 60
F = 75

Ans B

Prob 2.

First machine comepletes the task in 8 hrs so speed is 1/8.
Second machine comepletes the task in 8 hrs so speed is 1/4.

Both machines together will complete the task in 1/4 + 1/8 = 3/8.
So time required will be 8/3 = 2.66 Approx

Percent change = 2.66 - 1/ 1 * 100 = 166.66%
So both machines working together will work 66.66% faster

Ans D.

,
Can you explain the temperature question again please?

F = 120 ( 2 ^ -at ) + 60
Solving this for the above values will give you, a = 1/10
Can you break down how you solved for a?

Also, when I plugged 1/10 into the equation, I got 80, how did you get 75?
put a = 1/10, t = 30 in F = 120 ( 2 ^ -at ) + 60
F = 75

Thanks

Hi again,
for the first question, why did you use percent change to calculate the answer? and what do you get the 1 you're subtracting from 2.66?

F = 120 (2^ -at ) + 60

We know that F is 120 and t =10 ...no forget all exponent rules

Just place the values

120 = 120 (2^ -a10 ) + 60

120 = 120 (1/2^a10 ) + 60

In order to get a 120 on LHS we need 60 + 60

What number will give us a 60 ..120/2 is 60 Hence the powers should cancel each other out

Hence a needs to be 1/10..


NOw if we put 1/30 in the equation

put a = 1/10, t = 30 in F = 120 ( 2 ^ -at ) + 60


F = 120 (1/2 ^1/10 *30) + 60

F = 120/8 + 60
= 15 + 60
= 75

As for the second one ..just pick simple numbers

Say first machine takes 8 hrs
Second machine takes 4 hrs

Lets say k = 32

One machine takes 8 hrs to make 32 items so 4 items in one hour

Second takes 4 hrs to make 32 items so 8 items in one hour

So % = 8/12*100 = 66.66

Hope this helps