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3 years ago

Write an equation for the line parallel to the given line that contains C

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

7 0

Answer:

The equation would be y = -2x + 10

Step-by-step explanation:

To find the line parallel going through the given point, first note that parallel lines have the same slope. This means that we have a slope of -2 in the original line, therefore the parallel line will also have a slope of -2. Now we can use this and the given point to find the line using point-slope form.

y - y1 = m(x - x1)

y - 6 = -2(x - 2)

y - 6 = -2x + 4

y = -2x + 10

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The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$

When t = 0, N = 400

$400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325$

When t = 3, N = 650

$650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49$

The equation for the population becomes:

$N = 725 - 325 e^{-0.49t}$

At t = 5, the population becomes:

$N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697$

N(5) = 700 ( to the nearest 50)

6 0

3 years ago

MAJOR HELP!

Feliz [49]

Answer:

The radius of the cone is 21 units

The volume of the cone is 9702 units³

Step-by-step explanation:

* Lets revise the area of the circle and the volume of the cone

- The base if the cone is a circle

- The area of the circle is πr², where r is the radius of the circle

∵ The area of the base of the cone = 1386 units²

∵ The area of the base = πr²

∵ π = 22/7

∴ 1386 = 22/7 (r²) ⇒ multiply each side by 7

∴ 9702 = 22 r² ⇒ divide both sides by 22

∴ 441 = r² ⇒ take √ for both sides

∴ 21 = r

* The radius of the cone is 21 units

- The volume of the cone = 1/3 πr²h , where h is the height of the cone

∵ The cone's height is equal to the radius of its base

∴ h = r

∵ The volume of the cone = 1/3 πr²h

∴ The volume of the cone = 1/3 πr²(r) = 1/3 πr³

∵ r = 21 units

∴ The volume of the cone = 1/3 (22/7) (21)³

∴ The volume of the cone = 9702 unit³

* The volume of the cone is 9702 units³

3 0

3 years ago

BEST ANSWER GETS BRAINLIEST

Alexandra [31]

Answer:

1. The speed of the truck, S = D/T.

2. The formula that connects D and T is: S = D/T.

3. The coefficient of variation, k, is the ratio of the standard deviation to the mean speed.

Step-by-step explanation:

a) The speed of a truck at a fixed speed is given as the distance covered by the truck divided by the time it takes the truck to cover the said distance. This implies that speed is a function of distance and time. However, this formula represents the mean speed. There are variations in speed.

b) If the truck covers a distance of 60 kilometers, for example, under 3 hours, we can conclude that the speed is 20 kilometers per hour (60/3) or 20 km/hr.

4 0

3 years ago

9. Order the angles of the triangle from smallest to largest

12345 [234]

Answer:

It is b. ACB

Step-by-step explanation:

If draw a triangle, you would give the longest side to BC, because their angles equal 10

And you would give AC the shortest side of the triagle because their angles equal 7

And AB the last side because it is in between the other sides

Since AC have the smallest angle, they would have the smaller numbers

then that leaves B as the biggist number

8 0

3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago