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Setler [38]
3 years ago
14

Y=(2x-3)^5(2-x^4)^3 differentiate the function please

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0
\bf y=(2x-3)^5(2-x^4)^3\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
\\\\\\
\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
\\\\\\
\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)
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The problem based on condition are solved using the unknown variables. The number of tickets sold to the student are 120 and the number of tickets sold to the adults are 200.

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<h3>Variables</h3>

Variables are the unknown and the value of the variables depend on the other variables in the equation. The above problem can be solved defining the variables from the given condition.

Let the total tickets purchased by the students is<em> x</em> and the total tickets porches by the adults is <em>y.</em>

As the total tickets sold for the game is $320. Thus,

\begin{aligned}\\&#10;x+y&=320\\&#10;y&=320-x\\&#10;\end                    .......1

Now as the total money with ticket selling is $1100. Thus,

2.5x+4y=110

Keep the value of u from the equation 1 in the above equation. We get,

\begin{aligned}&#10;2.5x+4(320-x)&=1100\\&#10;2.5x+1280-4x&=1100\\&#10;-1.5x&=1100-1280\\&#10;-1.5x&=-180\\&#10;x&=\dfrac{180}{1.5} \\&#10;x&=120\\&#10;\end

Thus the number of tickets sold to the student are 120.

Keep this value in equation 1,

y=320-x\\&#10;y=320-120\\&#10;y=200

Thus the number of tickets sold to the adults are 200.

Hence the number of tickets sold to the student are 120 and the number of tickets sold to the adults are 200.

brainly.com/question/787279

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