All categories

3 years ago

Which graph represents the piecewise-defined function?

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

7 0

Answer:

it can be a pain when nobody helps... ur top right is correct

Step-by-step explanation:

heres a screenshot ..

fiasKO [112]3 years ago

3 0

Answer:

Step-by-step explanation:

A Piecewise function, is a function whose main feature is to be defined by a sequence of intervals. In these graphs we have:

$f(x)=y=\left\{\begin{matrix}-4&&if\: x\leqslant -3\\ 0&&if\: x=2\\3&&if\: x>2\end{matrix}\right.$

So we have three intervals

x=2, for y=0

x >2, for y=3

x ≤ -3, for y=-4

The Piecewise function is putting together these intervals in the coordinate plane. So, it's B.

You might be interested in

2 over 8 divided by 1 over 2 =

Illusion [34]

Answer: 1 over 2 (1/2)

Simplify 2 over 8 to get 1 over 4. Now you have 1 over 4 divided by 1 over 2. Whenever you divide with fractions you flip the fraction and multiply . So it becomes 1 over 4 multiplied by 2 over 1.

2/8 (divided by) 1/2
1/4 (divided by) 1/2
1/4 x 2/1
= 1/2

7 0

3 years ago

What is the measure of the missing angle in the triangle?

mr Goodwill [35]

120+34=154
180-154=26
Angle acb=26

8 0

3 years ago

Read 2 more answers

What is the answer for 5.973/0.11

olya-2409 [2.1K]

That would be 54.3 hope this helps

4 0

3 years ago

Read 2 more answers

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

In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To

Ymorist [56]

Answer:

$t=\frac{78-75}{\frac{7}{\sqrt{20}}}=1.917$

The degrees of freedom are given by:

$df= n-1 = 20-1=19$

The p value would be given by:

$p_v =2*P(t_{19}>1.917)=0.0704$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 75 and the best option would be:

a. No

Step-by-step explanation:

Information given

$\bar X=78$ represent the sample mean