Danielle needs to sell candy bars for a club fundraiser. She starts with 72 candy bars and begins selling at a constant rate of
1 answer:
You might be interested in
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
= 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) × (x + 4) ← chain rule
= 2(x + 4) × 1
= 2(x + 4)
Then
= (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 × (x² - 1) ← chain rule
= 3(x² - 1)² × 2x
= 6x(x² - 1)²
Then
= 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
= (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
= 5 - 3x² + 2x
--------------------------------------------------------------------
(d)
y = 3x³(x² + 4)²
f(x) = 3x³ ⇒ f'(x) = 9x²
g(x) = (x² + 4)²
g'(x) = 2(x² + 4) × (x² + 4) ← chain rule
= 2(x² + 4) × 2x
= 4x(x² + 4)
Then
= 3x³. 4x(x² + 4) + (x² + 4)². 9x²
= 12(x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)
= 3x²(x² + 4) [ 4x² + 3(x² + 4) ]
= 3x²(x² + 4)(4x² + 3x² + 12)
= 3x²(x² + 4)(7x² + 12)
Answer:
<em>5 boys play all the two games</em>
<em>25 boys play only one game</em>
Step-by-step explanation:
<u>Sets</u>
There are two sets defined in the question: one for the boys who play hockey (H) and the other for the boys who play volleyball (V).
All the boys play at least one of the two games, so no elements are outside both sets.
There are 30 boys in total. 20 of them play hockey and 15 play volleyball. Since the sum of both numbers is greater than the total of boys, the difference corresponds to the boys who play both games.
Thus 20 + 15 - 30 = 5 boys play both games
Given that 5 boys are shared by both sets, from the 20 playing hockey, 15 play ONLY hockey. From the 15 boys playing volleyball, 10 play ONLY volleyball.
Thus 15 + 10 = 25 boys play only one game.
The Venn diagram is shown in the image.
