I need help no ilnks please and thank you

Can u help me check if this is correct ty

mart [117]

H (4.84) will be the correct answer

4 0

3 years ago

Read 2 more answers

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago

Plz help me quick!!!!!!!!!!!!!!!

Sedbober [7]

Answer:

The answer is 1 over 5

Step-by-step explanation:

10 divided by 5 equals 2. That is your width. The same rule applies to your height. So you would get 10 over 50 which simplif equals 1 over 5.

6 0

3 years ago

Points (0,12) and (1,3) in linear function

puteri [66]

Answer:

y = -9x + 12

Step-by-step explanation:

first find the slope

m= 3 - 12

1 - 0

m = -9

y = mx + b

y = -9x + 12

(the 12 is the y intercept and is given in the question so there is no need to solve for it)

3 0

3 years ago

What is the exact area and arc length of these sectors? Please helppp

Karolina [17]

Answers with step-by-step explanation:

1. Area of sector 1 = $\frac{90}{360} \times \pi \times 12^2 = 36\pi$

2. Area of sector 2 = $\frac{45}{360} \times \pi \times 19^2 = \frac{2527}{8} \pi$

3. Area of sector 3 = $\frac{270}{360} \times \pi \times 15^2 = \frac{675}{4} \pi$

4. Area of sector 4 = $\frac{270}{360} \times \pi \times 6^2 = 27 \pi$

5. Arc length of sector 1 = $\frac{90}{360} \times 2 \times \pi \times 12 = 6\pi$

6. Arc length of sector 2 = $\frac{315}{360} \times 2 \times \pi \times 19 = \frac{133}{4} \pi$

7. Arc length of sector 3 = $\frac{270}{360} \times 2 \times \pi \times 15 = \frac{45}{2}\pi$

8. Arc length of sector 4 = $\frac{270}{360} \times 2 \times \pi \times 6 = 9\pi$

4 0

3 years ago