Need some help with these problems please.

Mathematics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

See below.

Step-by-step explanation:

Divide the leading terms:

6x^4 / 2x^2 = 3x^2

3x2 is a parabola so the long run is

as x ---> +/- infinity r(x) ----> + infinity. (answer).

10,000x^3 / 50 x^3

= 200.

So r(x) as a horizontal asymptote at y ( r(x)) = 200. (answer).

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One question simple please help

igor_vitrenko [27]

Answer:

$y = \frac{1}{3}x - 9$

Step-by-step explanation:

Standard equation of a line is y = mx + b, where m is the slope.

Given line y = - 3x + 78, slope, m₁ = -3

<em><u>To find the line perpendicular to the given line.</u></em>

The lines are perpendicular to each other if the product of their slopes = - 1

That is,

$m_1 \times m_2 = -1$

So the slope of new line is

$-3 \times m_2 = -1\\\\m_2 = \frac{-1}{-3} = \frac{1}{3}$

$Therefore , equation \ of \ new \ line \ \\\\(y - y_1) = m_2(x - x_1) , where \ (x_1 , y_1) = (9, -6 )\\\\(y - (-6))= \frac{1}{3}(x -9)\\\\y + 6 = \frac{1}{3} x - 3\\\\y = \frac{1}{3}x -3-6\\\\y = \frac{1}{3}x -9\\\\$

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3 years ago

25 POINTS PLZ ANSWER!!!!!!!!!!!!!!!!!!!!!!!!!!

Tatiana [17]

The answer is c!!!!!!!

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3 years ago

Read 2 more answers

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t

WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

$\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}$

Seperate the differential equation and solve for the constant C.

$\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}$