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3 years ago

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

1 answer:

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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Answer:

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Step-by-step explanation:

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Tan 47 = Opp/Adj

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3 years ago

Consider the gender and class make-up of students at a large university by gender and status as given in the table.

azamat

Answer:

0.589

Step-by-step explanation:

THis is a conditional probability question. Let's look at the formula first:

P (A | B) = P(A∩B)/P(B)

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So, it means, the "Probabilty A given that B is equal to Probability A intersection B divided by probability of B."

So we want to know P (Female | Undergraduate ). This in formula is:

P (Female | Undergraduate) = P (Female ∩ Undergraduate)/P(Undergraduate)

Now,

P (Female ∩ Undergraduate) means what is common in both female and undergraduate? There are 43% female that are undergrads. Hence,

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Also,

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this is the answer.

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3 years ago

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borishaifa [10]

Answer with Step-by-step explanation:

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Using identity: $log_x y=\frac{log y}{log x}$

$log y=qlog b$

$q=\frac{log y}{log b}=log_b y$

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We know that

$x^a\cdot x^b=x^{a+b}$

Using identity

$xy=b^{p+q}$

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$log_b(xy)=(p+q)log_b b$

Substitute the values then we get

$log_b(xy)=(log_b x+log_b y)$

By using $log_b b=1$

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3 years ago

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Answer:

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Step-by-step explanation:

13x = -54 -20y give it is A

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Answer:

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Step-by-step explanation:

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