Hello from MrBillDoesMath!
Answer:
a(n) = (-n)^3 where n = 1,2,3,...
Discussion:
The pattern 1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before each entry, suggesting we try this:
- 1 = (-1)^3
-8 = (-2)^3
-27 = (-3)^3
-64 = (-4)^3
That's what the problem statement asked for
. The answer is equivalently
-1 * (n^3)
Thank you,
MrB
Answer:
a. x = -9 or x = -2
b. -5(x - 4)
c. x = -3 or x = 5
d. x = ±7
Step-by-step explanation:
a. First person;
y = x² + 11x + 18
y = x² + 9x + 2x + 18
y = x(x + 9) + 2(x + 9)
y = (x + 9)(x + 2)
y = x = -9 or x = -2
b. Second person;
y = -5x + 20
The common factor is 5.
y = -5(x - 4)
c. Third person;
y = x² - 2x - 15
y = x² - 5x + 3x - 15
y = x(x - 5) + 3(x - 5)
y = (x + 3)(x - 5)
y = x = -3 or x = 5
d. Fourth person;
y = x² - 49
Applying the difference of squares formula;
(a² - b²) = (a - b)(a + b)
y = x² - 49 = x² - 7² = (x - 7)(x + 7)
y = (x - 7)(x + 7)
y = x = ±7
Answer:
Audrey can afford 4 hours of lesson.
Step-by-step explanation:
Given that:
Amount Audrey has to spend = $111
Cost of racket = $55
Cost per lesson = $14
Let,
y be the total cost
x be the number of hours
According to given statement;
y = 14x + 55
111 = 14x + 55
111 - 55 = 14x
14x = 56
Dividing both sides by 14

Hence,
Audrey can afford 4 hours of lesson.
Step-by-step explanation:
a). A = {x ∈ R I 5x-8 < 7}
5x - 8 < 7 <=> 5x < 8+7 <=> 5x < 15 =>
x < 3 => A = (-∞ ; 3)
A ∩ N = {0 ; 1 ; 2}
A - N* = (-∞ ; 3) - {1 ; 2}
b). A = { x ∈ R I 7x+2 ≤ 9}
7x+2 ≤ 9 <=> 7x ≤ 7 => x ≤ 1 => x ∈ (-∞ ; 1]
A ∩ N = {0 ; 1}
A-N* = (-∞ ; 1)
c). A = { x ∈ R I I 2x-1 I < 5}
I 2x-1 I < 5 <=> -5 ≤ 2x-1 ≤ 5 <=>
-4 ≤ 2x ≤ 6 <=> -2 ≤ x ≤ 3 => x ∈ [-2 ; 3]
A ∩ N = {0 ; 1 ; 2 ; 3}
A - N* = [-2 ; 3) - {1 ; 2}
d). A = {x ∈ R I I 6-3x I ≤ 9}
I 6-3x I ≤ 9 <=> -9 ≤ 6-3x ≤ 9 <=>
-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3 =>
-3 ≤ x ≤ 5 => x ∈ [-3 ; 5]
A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}
A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}

You have a rational expression whose numerator's degree is smaller than the denominator's. This tells you you should consider a partial fraction decomposition. We want to rewrite the integrand in the form


You can use the "cover-up" method here to easily solve for
. It involves fixing a value of
to make 2 of the 3 terms on the right side disappear and leaving a simple algebraic equation to solve for the remaining one.
- If
, then 
- If
, then 
- If
, then 
So the integral we want to compute is the same as

and each integral here is trivial. We end up with

which can be condensed as
