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Leokris [45]
3 years ago
11

Find the quotient and simplify​

Mathematics
1 answer:
shutvik [7]3 years ago
7 0

Answer:

\large\boxed{\dfrac{9}{s^2-5s}}

Step-by-step explanation:

\dfrac{s+5}{s^2}\div\dfrac{s^2-25}{9s}=\dfrac{s+5}{s\!\!\!\!\diagup_s^2}\cdot\dfrac{9s\!\!\!\!\diagup^1}{s^2-5^2}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{s+5}{s}\cdot\dfrac{9}{(s-5)(s+5)}\\\\\text{cancel}\ (s+5)\\\\=\dfrac{1}{s}\cdot\dfrac{9}{s-5}=\dfrac{(1)(9)}{s(s-5)}\\\\\text{use the distributive property:}\ a(b+c)=ab+ac\\\\=\dfrac{9}{(s)(s)+(s)(5)}=\dfrac{9}{s^2-5s}

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I have corner points of:
WARRIOR [948]
The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10

So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)

Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z

------------------

Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later. 
So let's call it A. Let A = 25

Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later

Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later

Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727

------------------

In summary, we found
A = 25
B = 21
C = 39
D = 22.2727

The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)

Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)

------------------

Final Answer: 39

7 0
4 years ago
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