Describe the slope of the line.

Mathematics

2 answers:

lesya [120]3 years ago

4 0

Answer:

positive

4/3 is the slope

Step-by-step explanation:

its positive because the line is going up

use a slope formula (y2 - y1) / (x2 - x1) to get 4/3

IgorLugansk [536]3 years ago

3 0

The slope of the line is positive.
the slope is 4/3.

it’s positive because the line is rising/going up
the slope is 4/3 because it rises 4 units and goes to the left 3 units.

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Answer:

$\dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x$

Take partial fractions of the given fraction by writing out the fraction as an identity:

$\begin{aligned}\dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A}{x-4}+\dfrac{B}{x+2}\\\\\implies \dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A(x+2)}{(x-4)(x+2)}+\dfrac{B(x-4)}{(x-4)(x+2)}\\\\\implies x+5 & \equiv A(x+2)+B(x-4)\end{aligned}$

Calculate the values of A and B using substitution:

$\textsf{when }x=4 \implies 9 = A(6)+B(0) \implies A=\dfrac{3}{2}$

$\textsf{when }x=-2 \implies 3 = A(0)+B(-6) \implies B=-\dfrac{1}{2}$

Substitute the found values of A and B:

$\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x = \int \dfrac{3}{2(x-4)}-\dfrac{1}{2(x+2)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies \displaystyle \dfrac{3}{2} \int \dfrac{1}{x-4}- \dfrac{1}{2} \int \dfrac{1}{x+2}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating}\\\\$\displaystyle \int \dfrac{f'(x)}{f(x)}\:\text{d}x=\ln |f(x)| \:\:(+\text{C})$\end{minipage}}$