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Likurg_2 [28]
3 years ago
13

Maximize Q=3xy^2, where x and y are positive numbers such that x+y^2=2

Mathematics
1 answer:
Ivanshal [37]3 years ago
6 0
Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
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Divide 240g in the ratio 5:3:4​
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Answer:

240g in the ratio of 5:3:4

So we need to divide it evenly.

First add up all the ratios.
5+3+4= 12

Now divide 240 by the sum.

240/ 12 = 20

Now you multiply that into each ratio.

5*20 = 100

3*20 = 60

4*20 = 80
Now put it back in the ratio with the new numbers int he exact order as before.

Final Answer: 100 : 60 : 80

Check:

You can check by adding them all up to see if they all go into 240 OR you can divide them each by the number originaly given and you should get 20.

100 + 60 + 80 = 240

100 / 5 = 20

60/3 = 20

80/4 = 20


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Answer: N=-4

Step-by-step explanation:

STEP

1

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STEP

2

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Pulling out like terms

2.1     Pull out like factors :

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Equation at the end of step

2

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STEP

3

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STEP

4

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Pulling out like terms

4.1     Pull out like factors :

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Equation at the end of step

4

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STEP

5

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Equations which are never true:

5.1      Solve :    -12   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

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One solution was found :

n = -4

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