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3 years ago

How can I find slope

1 answer:

5 0

To find the slope of a line, you first chooses 2 points with known coordinates that belongs to the line. Let those points be A(x1,y1) and B(x2,y2) with x2>x1. You then divide the subtraction of their Ys as y2-y1 by the subtractions of their Xs as x2-x1 (the point with the higher x comes first, this is why x2>x1). And let the slope be m. m= (y2-y1) / (x2-x1). If the slope is positive, the line croissant and y2>y1. If the slope is negative, the line is decroissant and y1>y2. Hope this helps! :)

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What is 67 divided by 19

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67 divided by 19 equal 3.52631578947
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What is y+4=9 what is y?

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Answer:

$y = 5$

Step-by-step explanation:

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All of the following were barred from receiving pardons from the Federal government after the Civil War EXCEPT

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2 confederate officers

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3 years ago

Harmonic mean if a number h such that (h-a)/(b-h)=a/b Prove h is H(a,b) iff satisfies either relation a. (1/a)-(1/h) = (1/h) - (

Fudgin [204]

A.

$\displaystyle\frac1a-\frac1h=\frac1h-\frac1b$
$\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac ab$

b.

$\displaystyle h=\frac{2ab}{a+b}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac ab$

The other direction can be proved by following the manipulations in the reverse order.

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3 years ago

Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .

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3 years ago