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Preeti Dabral 1 week, 5 days ago

Given : Three friends Ravi, Raju and Rohit were doing buying and selling of stationery items in market.

To Find : Value of x , y and z

Solution:

The price of per dozen of Pen, Notebook and toys are Rupees x. y and z respectively

Ravi purchases 4 dozens of notebooks and sells 2 dozens of pens and 5 dozens of toys  and earn 1500

=>   2x - 4y + 5z  = 1500

Raju purchases 2 dozens of toy and sells 3 dozens of pens and 1 dozen of notebooks and earn 100

=> 3x + y  - 2z  =  100

Rohit purchases one dozen of pens and sells 3 dozens of notebooks and one dozen of toys and earn 400

-x + 3y + z  = 400

2x - 4y + 5z  = 1500   Eq1

3x + y  - 2z  =  100    Eq2

-x + 3y + z  =  400        Eq3

Eq1 + 2Eq3

=> 2y + 7z = 2300

Eq2 + 3Eq3

=> 10y + z  = 1300

2y + 7z = 2300 => 10y +  35z = 11500

10y + z  = 1300

=> 34z = 10200

=> z  = 300

10y + z  = 1300

=> y = 100

on solving further  x= 200 , y = 100 and z = 300

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Dhairya Sharma 3 weeks, 5 days ago

Plzz check Whether you write the correct question or not because it is very difficult to find 1/1-cot integrating

Himanshu Mishra 3 weeks, 6 days ago

Break the cotx n form of sin and cos and then take lcm and do
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Preeti Dabral 1 week, 3 days ago

  1. (c) Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x2
  2. (b) We have, C(x) = 5000000 + 160x - 0.04x2
    Now, C{x) = 160 - 0.08x
    For maxima/minima, put C'(x) = 0
    {tex}\Rightarrow{/tex} 160 = 0.08x
    {tex}\Rightarrow{/tex} x = 2000
  3. (b) Clearly, from the given condition we can see that we only want critical points that are in the interval [0, 4500]
    Now, we have C(0) = 5000000
    C(2000) = 5160000 and C(4500) = 4910000
    {tex}\therefore{/tex} Maximum value of C(x) would be ₹5160000
  4. (a) The complex must have 4500 apartments to minimise the maintenance cost.
  5. (a) The minimum maintenance cost for each apartment woud be ₹1091.11
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Dhairya Sharma 3 weeks, 5 days ago

Hii rashmi you are from which class

Rashmi Ahuja 1 month ago

Root(1+cos x) =root(2cos² (x/2)) =√2.cos x/2 Ans= Integral of 1/√2 cos x/2 = Integral of sec (x/2)/√2 = (1/√2) log |sec x/2 + tan x/2| ÷ (1/2) = √2 log |sec x/2 + tan x/2|
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Satinder Kaur Sidhu 1 month, 2 weeks ago

1/x
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Karan Chauhan 1 month, 2 weeks ago

x³/3+x²/2+x+c
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The Thakur Shahab 1 month, 3 weeks ago

1
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