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3 years ago

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An epidemic has hit Clinton City. Its population is declining 22% every hour. In just 4 hours, there are only 35,542 people left

in the city. What was the initial population in the city before the epidemic broke out?
14,742
85,688
14,743

1 answer:

guapka [62]3 years ago

5 0

Every hour only 78% survive so we have the equation

35,542 = x(0.78)^4

x = 96,020 answer

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3 years ago

State whether the slope of the line is positive, negative, zero, or undefined.

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The slope of the line is equal to $\tan \alpha,$ where angle $\alpha$ is angle of line incline to the positive x-direction.

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3 years ago

Read 2 more answers

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$ , the slope is again $-\frac{16}{64}=-\frac14$ , so the normal has slope +4, and so on.

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2 years ago