HOPE THIS IS THE ANSWER FOR YOUR QUESTION
Answer:
As X → - ∞ , y → ∞ and as x→ ∞ , y → ∞
<h3>
option c is the correct option.</h3>
Step-by-step explanation:
let f(x) = y = 3x² - 5x + 2
y = 3x² - 5x + 2
= x ( 3x - 5 ) + 2
y = ∞ ( 3 ( ∞ - 5 ) ) + 2
= ∞ (∞ ) + 2
y = ∞
y → ∞ as x → ∞
Now,
as x → - ∞
y = x ( 3x - 5 ) + 2
= ∞ ( 3 ( - ∞ ) - 5 ) + 2
= - ∞ ( - ∞ ) + 2
∞² + 2 = ∞
Hence , Option C is the correct answer.
The Taylor Series expansion of f(x) = sin(x) about a = π i given by

where the c's are contants.
That is
f(x) = c₀ + c₁(x-π) +c (x-π)² + c₃ (x-π)³ + ...,
₂
The first few derivatives of f(x) are
f' = c₁
f'' = 2c₂ = 2! c₂
f''' = 3.2c₃ = 3! c₃
f⁽⁴⁾ = 4.3.2c₄ = 4! c₄
and so on.
The pattern indicates that

The derivatives of f(x) are
f' = cos(x)
f'' = -sin(x)
f''' = -cos(x)
f⁽⁴⁾ = sin(x(
and so on
The pattern indicates that
f⁽ⁿ⁾(x) = cos(x), n=1,5,9, ...,
= -sin(x), n=2,6,10, ...,
= -cos(x), n=3,7,11, ...,
= sin(x), n=4,8,12, ...,
The radius of convergence is |x-π|<1 by the ratio test.
So 12 out of 32=12/32=6/16=3/8
percent means parts out of 100 so
x/100=x%
3/8=0.375/1
0.375/1 times 100/100=37.5/100=37.5%
answer is 37.5%
Answer:
Step-by-step explanation:
Notice that, the angle QRS is external to the triangle and adjacent to the angle PRQ. According to the theorem of a external/adjacent angle, we have: m∠QRS = m∠PQR + m∠RPQ, where PQR and RPQ are internal angles.
From the hypothesis, we have:
m∠QRS =(10x−12)∘(10x−12)
m∠PQR = (3x+20)∘(3x+20)
m∠RPQ=(3x−8)∘(3x−8)
Using the first equation and replacing the hypothesis:
m∠QRS = m∠PQR + m∠RPQ
(10x−12)∘(10x−12) = (3x+20)∘(3x+20) + (3x−8)∘(3x−8)
Multiplying and applying the remarkable identity:
Then, we use a calculator to find the roots, which are:
In this case, we will see what root is the right one.
Now, we replace it into m∠QRS =(10x−12)∘(10x−12), because we need to find m∠QRS.
m∠QRS =(10x−12)∘(10x−12) = (10(4.7) - 12) (10(4.7) - 12) = (35) (35) = 1225
Step-by-step explanation: