All categories

3 years ago

Graphing Do these and you will get brainliest answer!

Mathematics

2 answers:

kompoz [17]3 years ago

6 0

Answer:

y=1

Step-by-step explanation:

il63 [147K]3 years ago

5 0

Here the line is parallel to x axis
Slope (m) will be tan(0)=0
Y intercept is 1.

Y is constant.

The equation of line will be

y=0x+1
y=1

You might be interested in

Which statement best describes f(x) = -2 square root x-7 +1?

nekit [7.7K]

Answer:

-6 is not domain of f(x) but is in range of f(x)

B is correct.

Step-by-step explanation:

Given: $f(x)=-2\sqrt{x-7}+1$

Domain is input value of x where function is well defined.

Range is output value of f(x) for defined value of x.

Here, we have square function. As we know the value inside the square must be positive.

Therefore, x-7≥0

x≥7

Domain: x≥7 or [7,∞)

For range, we will put x=7 into function.

$f(7)$

The maximum value of f(x) is 1.

Range: f(x)<1 or (-∞,1)

-6 is not belongs to domain.

-6 is belongs to range.

Hence, -6 is not domain of f(x) but is in range of f(x)

6 0

3 years ago

Read 2 more answers

Trayvon has a mug shaped like a hemisphere. Its diameter is 18 cm. To the nearest hundredth, what is the capacity of her mug?

xeze [42]

Answer:

1527cm3

Step-by-step explanation:

6 0

3 years ago

Please help me with this! thanks

liraira [26]

Answer:

try looking at a video to see how to solve equations

Step-by-step explanation:

4 0

3 years ago

A brine solution of salt flows at a constant rate of 8L/min into a large tank that initially held 100L of brine solution in whic

GaryK [48]

Answer:

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

The concentration of salt in the tank reach 0.02 kg/L when $t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$

Step-by-step explanation:

Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).

The main equation that we will be using to model this mixing process is:

Rate of change of $\frac{dy}{dt}$ = Rate of salt in - Rate of salt out

We need to determine the rate at which salts enters the tank. From the information given we know:

The brine flows into the tank at a rate of $8\:\frac{L}{min}$
The concentration of salt in the brine entering the tank is $0.05\:\frac{kg}{L}$

The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering)

$(8\:\frac{L}{min}) \cdot (0.05\:\frac{kg}{L})=0.4 \:\frac{kg}{min}$

Next, we need to determine the output rate of salt from the tank.

The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)

The concentration of salt in any part of the tank at time t is just y(t) divided by the volume. From the information given we know:

The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.

$(8\:\frac{L}{min}) \cdot (\frac{y(t)}{100} \:\frac{kg}{L})= \frac{2y(t)}{25} \:\frac{kg}{min}$

At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}, \quad{y(0)=0.5}$

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}\\\\\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}\\\\ \frac{1}{2}\frac{dy}{5-y}=\frac{1}{25}dt \\\\\frac{1}{2}\int \frac{dy}{5-y}=\int \frac{1}{25}dt\\\\\frac{1}{2}\left(-\ln \left|5-y\right|+C\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|+C_1\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|=\frac{1}{25}t+C_2\\\\5-y=C_3e^{-\frac{2t}{25} }\\\\y(t) =5-C_3e^{-\frac{2t}{25} }$

Using the initial condition y(0)=0.5

$y(t) =5-C_3e^{-\frac{2t}{25} }\\y(0)=0.5=5-C_3e^{-\frac{2(0)}{25}} \\C_3=4.5$

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

To determine when the concentration of salt is 0.02 kg/L, we solve for t

$y(t) = 5-4.5e^{-\frac{2t}{25}}\\\\0.02=5-4.5e^{-\frac{2t}{25}}\\\\5\cdot \:100-4.5e^{-\frac{2t}{25}}\cdot \:100=0.02\cdot \:100\\\\500-450e^{-\frac{2t}{25}}=2\\\\500-450e^{-\frac{2t}{25}}-500=2-500\\\\-450e^{-\frac{2t}{25}}=-498\\\\e^{-\frac{2t}{25}}=\frac{83}{75}\\\\\ln \left(e^{-\frac{2t}{25}}\right)=\ln \left(\frac{83}{75}\right)\\\\\frac{2t}{25}=\ln \left(\frac{83}{75}\right)\\\\t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$

5 0

3 years ago

Math help :) pleasee , and thnx

mylen [45]

Answer:

The ball is in the air for approximately 3.27 seconds ⇒ answer A

Step-by-step explanation:

* Lets explain how to solve the problem

- The height of the ball is modeled by the function

h(t) = -4.9 t² + 16 t

- We need to find the time that the ball is in the air

- The ball is in the air from its initial position and then return to the

same position

- That means h(t) = 0 because h(t) represent the height of the ball

from its initial position

∵ h(t) = -4.9 t² + 16 t

∵ h(t) = 0

∴ 0 = -4.9 t² + 16 t

- Add 4.9 t² to both sides

∴ 4.9 t² = 16 t

- Subtract 16 t from both sides

∴ 4.9 t² - 16 t = 0

- Take t as a common factor

∴ t (4.9 t - 16) = 0

- Equate each factor by 0

∴ t = 0 and 4.9 t - 16 = 0

∵ 4.9 t - 16 = 0 ⇒ add 16 for both sides

∴ 4.9 t = 16

- Divide both sides by 4.9

∴ t = 3.2653

∴ t = 0 ⇒ initial position

∴ t = 3.2653 seconds ⇒ final position

* The ball is in the air for approximately 3.27 seconds

6 0

4 years ago