Deepak Suwalka
Deepak Suwalka answered

Given root of polynomial function x=∛28

g(2)=x3 -28=0

Applying Newton's method

x_(n+1) = x_n - (g(x_n)) / (g'(x_n))

For c₁ ⇒ c₁ = c₀ - (g(c₀)) / (g'(c₀))

Thus, c₀ =3

So, g'(3)=27

g(3)=27-28=-1

c₁ = (3-(-1)) / (27) = (81+1) / (27)

c₁ = (82)/(27)

Deepak Suwalka
Deepak Suwalka answered

To find 'c', solve f'(x)=0

⇒ (x² − 1) × 1 + 2x(x − 2) = 0

⇒ x²− 1 + 2x² − 4x = 0

⇒ 3x2 − 4x−1 = 0

It's a quadratic equation so, solve it using quadratic formula

⇒ [4±√16-(4)(3)(-1)] / (2(3))

⇒x = (2±√7)/(3)

⇒x = (2+√7)/(3) and x = (2-√7)/(3)

x = (2+ √7)/(3) = 1.55 ∈(1,2]

x … Read more

Deepak Suwalka
Deepak Suwalka answered Anonymous' question

geometric return = ⁿ√{(1+r₁) × (1+r₂) × ...} - 1

Where, r = rate of return

n = number of periods

For 1st yr, rate of return = r₁ = 5% = 0.05

For 2nd yr, rate of return = r₂ = -30% = -0.3

Number of years = 2

geometric return = √{(1 + 0.05)(1 - 0.3)}-1= √{(1.05) (0.7)}-1

= -0.143

Geometric … Read more

Deepak Suwalka
Deepak Suwalka answered

Let advance Tickets = A

door tickets = D

Make equations according to the problem

A + D = 120.....(1)

10A + 15D = $1390....(2)

Multiplying eq(1) by 10 and subtracting from eq(2) we get

D = 38(number of tichets at door)

Put value of D in eq(1) we get-

A + 38 = 120

A = 82(number of tichets were purchased in advance)

Deepak Suwalka
Deepak Suwalka answered

16 + (28/2) - (6 /10) - 4 × 2

Use order of operation, i.e., PEMDAS (Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction)

First, simplify parenthesis

= 16 + 14 - 0.6 - 4 × 2

Now, multiplication

= 16 + 14 - 0.6 - 8

Similarly, apply addition and subtraction

= 30 - 0.6 - 8

= 21.4