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lys-0071 [83]
2 years ago
14

3. JK Rowling and R.L. Stine are both reading the Hunger Games. J.K. reads 35

Mathematics
1 answer:
Crank2 years ago
5 0

Answer:

Part A:

For J.K. Rowling: 17.5pages/hour

For R.L. Stine: 15pages/hour

Part B:

J.K. Rowling will finish the book first.

Step-by-step explanation:

Part A:

The rate of change in this case represents how much changes a variable in a specific period of time. The variable is the amount of pages.

You have to calculate the amount of pages that J.K. Rowling and R.L. Stine read in one hour.

For J.K. Rowling, 35 pages every 2 hours can be written as:

If you divide 35/2 you can obtain the amount of pages in one hour:

17.5pages/hour (17.5 pages per hour)

For R.L. Stine, 45 pages every 3 hours can be written as:

If you divide 45/3 you can obtain the amount of pages in one hour:

15pages/hour (15 pages per hour)

Part B:

In order to calculate who will finish the hunger games first, you have to calculate the amount of time needed to finish the book.

For J.K. Rowling:

You have to multiply 355 pages by  (because in one hour she reads 17.5 pages) to obtain the amount of hours

For R.L. Stine:

You have to multiply 355 pages by  (because in one hour he reads 15 pages) to obtain the amount of hours

So, J.K. Rowling will finish the book first because she needs less time than R.L. Stine.

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Big Ben (the clocktower) in London has a circumference of 72.22 feet and an area of 415.27 feet^2. What is the diameter AND radi
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Step-by-step explanation:

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Help. <br>Please its urgent show workings.<br>​
laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

\frac{dy}{dx} = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × \frac{d}{dx} (x + 4) ← chain rule

       = 2(x + 4) × 1

        = 2(x + 4)

Then

\frac{dy}{dx} = (2x - 1). 2(x + 4) + (x + 4)². 2

    = 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

    = 2(x + 4) (2x - 1 + x + 4)

    = 2(x + 4)(3x + 3) ← factor out 3

    = 6(x + 4)(x + 1)

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

(b)

y =  x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × \frac{d}{dx} (x² - 1) ← chain rule

        = 3(x² - 1)² × 2x

        = 6x(x² - 1)²

Then

\frac{dy}{dx} = x. 6x(x² - 1)² + (x² - 1)³. 1

    = 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

    = (x² - 1)² (6x² + x² - 1)

     = (x² - 1)²(7x² - 1)

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

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

\frac{dy}{dx} = (x² - 1). 3x² + (x³ + 1), 2x

   = 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

   = x[3x(x² - 1) + 2(x³ + 1) ]

   = x(3x³ - 3x + 2x³ + 2)

   = x(5x³ - 3x + 2) ← distribute

    = 5x^{4} - 3x² + 2x

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

(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × \frac{d}{dx}(x² + 4) ← chain rule

       = 2(x² + 4) × 2x

       = 4x(x² + 4)

Then

\frac{dy}{dx} = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

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    = 3x²(x² + 4)(7x² + 12)

5 0
2 years ago
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