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3 years ago

HELP HELP ASAP THANKS

Mathematics

1 answer:

mylen [45]3 years ago

8 0

Answer:

Step-by-step explanation:

the range is written as (min y value, max y value)

the domain is written as (min x value, max x value)

question 6

the min y value on the picture is -3, while the arrows point upward, so the max is infinity, so the domain is [-3,∞), with a bracket on -3 because -3 is included

[-3,∞)

question 7

the min x value is the leftmost point, which is at x = -3, while the max is the rightmost point at x = 3, and both are included in the domain so there should be brackets on both

[-3,3]

question 8

the arrow on the left points to the left and up infinitely, so the min is -∞, the arrow on the right points to the right and up infinitely, so the max x value is ∞

(-∞,∞)

question 9

the min value is the bottommost point at y = -2, and the arrow points upward infinitely so the max y value is ∞

[-2,∞)

question 10

the arrow on the left points to the left infinitely so the min x value is -∞, the arrow on the right points to the right infinitely so the max x value is ∞

(-∞,∞)

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A small source radiates an electromagnetic wave with a single frequency into vacuum, equally in all directions. As the wave move

vlabodo [156]

Answer:

a) Frequency remains constant

b) Wavelength remains constant

c) speed of the wave remains constant

d) Intensity decreases

e) amplitude of its electric field decreases

Explanation:

a) Frequency can be defined as the number of crests that pass a fixed point in the medium in unit time. It is the source of the wave that will determine the frequency. If the small source is changed to a bigger and faster one then the frequency will change. In our case, there is no change of source of wave, so the frequency remains constant.

b) The speed of of the wave is directly proportional to the wavelength. If we double the speed, the wavelength also doubles. Since the speed has not been doubled in our case, the wavelength will remain constant.

c) As indicated in b) since the wavelength is proportional to speed and it has not changed in our case, then the speed remains constant.

d) The intensity of a wave decreases as it moves further away from the source.

e) The intensity is related to the amplitude. Since the intensity decreases, the amplitude also decreases.

5 0

3 years ago

F(x) = x^2 +2x-16/X-2

Lesechka [4]

Answer:

\frac{x^3+2x^2-16-2x}{x}

Step-by-step explanation:

\frac{x^3+2x^2-16-2x}{x}

3 0

3 years ago

NEED HELP !!!!!! READDD THE QUESTION THERE ARE TWO PARTS

mel-nik [20]

We have to find horizontal component of Mayosons velocity.

$\\ \tt\bull\leadsto V_x=vsin\theta$

$\\ \tt\bull\leadsto V_x=2.5sin36$

$\\ \tt\bull\leadsto V_x=2.5(0.58)$

$\\ \tt\bull\leadsto V_x=1.45m/s$

Here 2 vectors A=8 and B=2.5

$\\ \tt\bull\leadsto \overrightarrow{R}$

$\\ \tt\bull\leadsto \sqrt{A^2+B^2+2ABcos\Theta}$

$\\ \tt\bull\leadsto \sqrt{8^2+2.5^2+2(8)(2.5)cos36}$

$\\ \tt\bull\leadsto \sqrt{64+6.25+40(0.8)}$

$\\ \tt\bull\leadsto \sqrt{70.25+32}$

$\\ \tt\bull\leadsto \sqrt{102.25}$

$\\ \tt\bull\leadsto 10.15m/s$

3 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 ∞:

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