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3 years ago

|5-x|-|x+8|, if x>5

Mathematics

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

just do it on ur caslc im pretty sure it would work

Step-by-step explanation:

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What is the algebraic description for a transformation that translates the

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Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

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,

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

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

Differentiating with respect to time (t),

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

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

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

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

Here,

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

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

$-6x + y\frac{dy}{dt}=0$

$y\frac{dy}{dt}=6x$

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

Which is the required notation.

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Which equation represents a proportional relationship?

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Answer

y=2/3

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between two variables when their ratios are equivalent

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