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2 years ago

Find the slope of the line.

Mathematics

2 answers:

scoundrel [369]2 years ago

7 0

The points we can clearly see are (-2,-2) and (0,-3). You have to rise up 2 points and got to the left 4, so the rise over run is 2 over -4. Simplify this to -1/2, so the slope is negative 1/2.

Dovator [93]2 years ago

7 0

Answer:

The slope of the line is $-\frac{1}{2}$

Step-by-step explanation:

To know the slope of a line we can have the equation or to know points of the line. In this case from the graph we can get to points (-4,-1) and (0,-3). Having this and using the equation to calculate the slope, we have:

$m=\frac{y2-y1}{x2-x1}$

Defining:

$(x1=-4,y1=-1)$ and $(x2=0,y2=-3)$

Replacing this values in the initial equation, this is:

$m=\frac{-3-(-1)}{0-(-4)}$

$m=\frac{-2}{4} =-\frac{1}{2}$

The slope of the line is $-\frac{1}{2}$

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2 years ago

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In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

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$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

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Which is the required notation.

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Answer:

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2 years ago

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Answer:

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Step-by-step explanation:

As given

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