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2 years ago

Why is accountability important in your walk as a christian? answer in at least 2-3 complete sentences.

2 answers:

5 0

I do not have a religion, but I study different religions, so here’s my perception!

Accountability is important for Christian’s, because being dishonest is a big sin stated in the Bible, if you don’t take accountability for your actions as a Christian, that is considered lying or being dishonest meaning you would disappoint Jesus. I think taking accountability, for good or bad actions, is very important in life in general, not just for Christian’s.

ad-work [718]2 years ago

4 0

Answer: GOT YA!

Explanation:

Accountability is very important in my walk as a Christian because honestly you can not be a true Christian without it. When you hold yourself or even other accountable it helps you not make those same mistakes or sins again, and when you do not hold yourself or others accountable it can cause you to commit those same sins that you can avoid.

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Differentiate the following functions (i) x(1+x)^3

statuscvo [17]

Answer:

$\displaystyle y' = (1 + x)^2(4x + 1)$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Functions
Function Notation
Factoring

Calculus

Derivatives

Derivative Notation

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

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

Explanation:

Step 1: Define

Identify

y = x(1 + x)³

Step 2: Differentiate

Product Rule [Derivative Rule - Chain Rule]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]$
Derivative Property [Addition/Subtraction]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])$
Basic Power Rule: $\displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})$
Simplify: $\displaystyle y' = (1 + x)^3 + 3x(1 + x)^2$
Factor: $\displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]$
Combine like terms: $\displaystyle y' = (1 + x)^2(4x + 1)$

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

Unit: Derivatives

Book: College Calculus 10e

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3 years ago

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