Please help with Algebra!!! 10 points!

1 answer:

Tju [1.3M]3 years ago

5 0

4. D
5. A
6. C
7. D
8. D
9. B
10. B

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motikmotik

Look at this and you will be surprised on how it helps many like you that’s a key answer

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2 years ago

Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Algebra II

Calculus

Derivatives

Basic Power Rule:

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

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

Rewrite: $t = ln(\frac{2x-1}{x-1})$
Rewrite [Ln Properties]: $t = ln(2x-1) - ln(x - 1)$
Differentiate [Ln/Chain Rule/Basic Power Rule]: $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
Simplify: $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
Rewrite: $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
Combine: $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
Reciprocate: $\frac{dx}{dt} = -(2x-1)(x-1)$
Distribute: $\frac{dx}{dt} = (1-2x)(x-1)$