My feet are asleep,,,.T.T, I'm also stressed and cold and tired

The curve

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

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:

Step 1: Define

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$

y-coordinate

Substitute in x [Function]: $\displaystyle y = \sqrt{4 - 3}$
[√Radical] Subtract: $\displaystyle y = \sqrt{1}$
[√Radical] Evaluate: $\displaystyle y = 1$

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

4 0

3 years ago

Solve the linear equation below for x. * 3x + 4 = 8x – 9 A. 2.6 B. 1 C. 1.09 D. 2

Juli2301 [7.4K]

This answer is A, first you subtract 3x from both sides simplify 8x-9-3x to 5x-9, add 9 to both sides, simplify 4+9 to 13, divided both sides by 5 and you’ll get 2.6

6 0

3 years ago

What is the value of the function at x=−2?

Elza [17]

The value of the function is the Y value where the line crosses the given X value.

When X is at -2, the line is on Y = 3

The answer would be y =3

3 0

3 years ago

Find the gradient of the line segment between the points (3,-2) and (5,-5). Give your answer in its simplest form.

Paraphin [41]

Answer:

2/7

Step-by-step explanation:

gradient y²-y¹/x²-x¹=4-2/4-(-3)

=2/7

{Taking (x¹,y,¹)=(-3,2)

and(x²-y²)=(4,4)

4 0

3 years ago

Write a fraction equivalent of 1/7

Ostrovityanka [42]

Answer:

2/14

Step-by-step explanation:

it is the correct answer for the above mentioned questions

7 0

3 years ago

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My feet are asleep,,,.T.T, I'm also stressed and cold and tired​

My feet are asleep,,,.T.T, I'm also stressed and cold and tired