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2 years ago

Using the graph of f(x) and g(x), where g(x) = f(k⋅x), determine the value of k.

Mathematics

1 answer:

baherus [9]2 years ago

3 0

Answer:

1/3

Step-by-step explanation:

This is about the interpretation of the graph

From the graph, we can see the 2 lines representing function f(x) and function g(x).

Now for us to find the value of x in g(x) = k⋅f(x), we need to get a mutual x-coordinate where we can easily read their respective y-coordinate values.

We see that the best point for that is where x = -3.

For f(x), when x = -3, y = 1

For g(x), when x = -3, y = -3

we can rewrite them as;

x = -3, f(-3) = 1 and x = -3, g(-3) = -3

Let us plug in the relevant values into g(x) = k⋅f(x) to get;

-3 = k(1)

Thus; k = -1/3

Hope it helps you mark me as brinllent

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(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of

Paraphin [41]

1. dy/dt is the net rate of change of salt in the tank over time. As such, it's equal to the difference in the rates at which salt enters and leaves the tank.

The inflow rate is

(0.4 kg/L) (6 L/min) = 2.4 kg/min

and the outflow rate is

(concentration of salt at time t) (4 L/min)

The concentration of salt is the amount of salt (in kg) per unit volume (in L). At any time t > 0, the volume of solution in the tank is

100 L + (6 L/min - 4 L/min) t = 100 L + (2 L/min) t

That is, the tank starts with 100 L of pure water, and every minute 6 L of solution flows in and 4 L is drained, so there's a net inflow of 2 L of solution per minute. The amount of salt at time t is simply y(t). So, the outflow rate is

(y(t)/(100 + 2t) kg/L) (4 L/min) = 2 y(t) / (50 + t) kg/min

and the differential equation for this situation is

$\dfrac{dy}{dt} = 2.4 \dfrac{\rm kg}{\rm min} - \dfrac{2y}{50+t} \dfrac{\rm kg}{\rm min}$

There's no salt in the tank at the start, so y(0) = 0.

2. Solve the ODE. It's linear, so you can use the integrating factor method.

$\dfrac{dy}{dt} = 2.4 - \dfrac{2y}{50+t}$

$\dfrac{dy}{dt} + \dfrac{2}{50+t} y = 2.4$

The integrating factor is

$\mu = \displaystyle \exp\left(\int \frac{2}{50+t} \, dt\right) = \exp\left(2\ln|50+t|\right) = (50+t)^2$

Multiply both sides of the ODE by µ :

$(50+t)^2 \dfrac{dy}{dt} + 2(50+t) y = 2.4 (50+t)^2$

The left side is the derivative of a product:

$\dfrac{d}{dt}\left[(50+t)^2 y\right] = 2.4 (50+t)^2$

Integrate both sides with respect to t :

$\displaystyle \int \dfrac{d}{dt}\left[(50+t)^2 y\right] \, dt = \int 2.4 (50+t)^2 \, dt$

$\displaystyle (50+t)^2 y = \frac{2.4}3 (50+t)^3 + C$

$\displaystyle y = 0.8 (50+t) + \frac{C}{(50+t)^2}$

Use the initial condition to solve for C :

$y(0) = 0 \implies 0 = 0.8 (50+0) + \dfrac{C}{(50+0)^2} \implies C = -100,000$

Then the amount of salt in the tank at time t is given by the function

$y(t) = 0.8 (50+t) - \dfrac{10^5}{(50+t)^2}$

so that after t = 50 min, the tank contains

$y(50) = 0.8 (50+50) - \dfrac{10^5}{(50+50)^2} = \boxed{70}$

kg of salt.

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2 years ago

The sum of a number and twenty is greater than four times the number decreased by one

Virty [35]

is that the question?

7 0

3 years ago

If logx (8x-3) -logx 4=2, the value of x is

Vaselesa [24]

Answer:

x = 1/2 and 3/2

Step-by-step explanation:

The given equation is logₓ (8x-3) - logₓ 4 = 2

Then we have to determine the value of x.

logₓ $[\frac{(8x-3)}{4}]$ = 2 [ since log a - log b = log $(\frac{a}{b})$ ]

Now $x^{2}=\frac{(8x-3)}{4}$ [if $log_{a}$ b = c then $a^{c}$ = b]

4x² = 8x -3

4x² - 8x + 3 = 0

(2x)² - 2(4x) + (4-4) + 3 = 0

[(2x)² - 2 (4x) + 4 ] - 1 = 0

(2x - 2)² = 1

(2x - 2)² = ± 1 ⇒

2x - 2 = 1 and 2x - 2 = -1

x = 3/2 2x = 1

x = 1/2

Therefore, x = 1/2 and 3/2 are the answers.

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4 years ago

My answers are c,d am I correct

Lady_Fox [76]

Yes you seem to
be correct

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3 years ago

The volume of this cube is 64 cubic units. What's the edge length?

Alona [7]

Volume of a cube = side^3

64 = s^3

So what number when multiplied by itself 3 times gives us 64?

4 !

So the side length is 4

$\boxed{\bf{4 ~units}}$

8 0

3 years ago