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trasher [3.6K]
3 years ago
10

An exponential function f(x) = ab^xpasses through the points (0, 2) and (3, 54). What are the values of a and

Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0

Answers:

a = 2

b = 3

=======================================================

Explanation:

Plug in x = 0 and y = 2 to find that

y = a*b^x

2 = a*b^0

2 = a*1

2 = a

a = 2

Then plug in x = 3 and y = 54 to determine the value of b

y = a*b^x

y = 2*b^x

54 = 2*b^3

2b^3 = 54

b^3 = 54/2

b^3 = 27

b = (27)^(1/3)

b = 3

So we have y = a*b^x update to y = 2*3^x

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3 years ago
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Who packed all the pies? Kane’s bakery makes four flavour pies. The weight of each flavour pie is follows:
svet-max [94.6K]
This is an identical problem answered by myself a few days ago.  The answer is reproduced below for your convenience.

It is a very difficult problem if the box sizes are smaller or if there are more pies to pack.
For this particular problem, we first calculate the total weight of pies:
4 apple pies at 700 g = 2800 g
8 banoffee pies at 725 g =5800 g
3 custaard pies at 600 g = 1800 g
5 rhubarb pies at 675 g = 3375 g
for a total of 13775 g

The minimum number of boxes needed will be 13775/4000=3.44, or at least 4 boxes will be required.

We also note that 4 boxes will hold 16000 g, and we only need to pack 13775g, so there is a lot of margin.

We can proceed to do the mechanical way, fill in order of apple, banoffee, custard, and finally rhubarb pies.Following table shows how we fill, naively:      box content               Remaining to pack
Box A B C R     Weight    A B C R   
0                                      4  8  3 5     (original order)
1     4  1  0 0     3525       0  7  3 5
2     0  5  0 0     3625       0  2  3 5
3     0  2  3 0     3250       0  0  0 5
4     0  0  0 5     3375       0  0  0 0
Total weight     13775 g

So that does the job, 4 boxes (minimum) with none exceeding 4000 g to pack a total of 13775 g of pies.
 
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16 4/8 + 23 + 31 2/5 =
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Answer:

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Answer:

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Answer:

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