Mathematics, 17.01.2021 01:10 guardhic7165

Help me answer this question please

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Answer from: superpuglover5924

9514 1404 393

7953.873

Step-by-step explanation:

The first derivative is ...

f'(x) = 4·3x²·e^x +4x³·e^x = e^x(4x³ +12x²)

Then the second derivative is ...

f''(x) = (12x² +24x)e^x +(4x³ +12x²)e^x

f''(x) = e^x(4x³ +24x² +24x)

So, f''(3) = (e^3)(4·27 +24·9 +24·3) = 396e^3 = 7953.87262158

Rounded to thousandths, this is ...

f''(3) = 7953.873

Answer from: ikepayton

f''(x) ≈ 7953.87

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra I

Factoring

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule: $\displaystyle \frac{d}{dx} [e^x]=e^x$

Step-by-step explanation:

Step 1: Define

f(x) = 4x³eˣ

f''(x) is x = 3 for 2nd Derivative

Step 2: Differentiate

[1st Derivative] Product Rule [Basic Power Rule]: f'(x) = 3 · 4x³⁻¹eˣ + 4x³eˣ[1st Derivative] Simplify: f'(x) = 12x²eˣ + 4x³eˣ[1st Derivative] Factor: f'(x) = eˣ(12x² + 4x³)[2nd Derivative] Product Rule [Basic Power Rule]: f''(x) = eˣ(12x² + 4x³) + eˣ(2 · 12x²⁻¹ + 3 · 4x³⁻¹)[2nd Derivative] Simplify: f''(x) = eˣ(12x² + 4x³) + eˣ(24x + 12x²)[2nd Derivative] Distribute eˣ: f''(x) = 12eˣx² + 4eˣx³ + 24eˣx + 12eˣx²[2nd Derivative] Combine like terms: f''(x) = 24eˣx² + 4eˣx³ + 24eˣx[2nd Derivative] Factor: f''(x) = 4xeˣ(x² + 6x + 6)

Step 3: Evaluate

Substitute: f''(x) = 4(3)e³(3² + 6(3) + 6)Exponents: f''(x) = 12e³(9 + 6(3) + 6)Multiply: f''(x) = 12e³(9 + 18 + 6)Add: f''(x) = 12e³(27 + 6)Add: f''(x) = 12e³(33)Multiply: f''(x) = 396e³Evaluate: f''(x) ≈ 396(20.0855)Multiply: f''(x) ≈ 7953.87

Answer from: Quest

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

Point z is equidistant from the sides of δrst. which must be true? ctzaszaszzsb

Answer from: Quest

Idk i am in 5th grade lol

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